PHY101 — Midterm Summary (Lectures 1–22)

📘 Lecture 1 — INTRODUCTION TO PHYSICS

📖 Overview: This lecture introduces the fundamental principles of physics as a science, emphasizing the scientific method and the universality of physical laws. It covers the major branches of physics, the concept of dimensions and units, and essential problem-solving strategies including dimensional analysis and estimation techniques.

🗂️ Topics Covered

The lecture covers the scientific method and its application to physics; the universality and consistency of physical laws across space and time; an overview of the four major branches of physics (Classical Mechanics, Electromagnetism, Thermal Physics, and Quantum Mechanics); the three fundamental dimensions of Mass (M), Length (L), and Time (T); dimensional analysis and its importance; the distinction between units and dimensions; unit conversion methodology; order of magnitude estimation; checking equations for dimensional consistency; and techniques for analyzing complex equations through special limits and mental graphing.

📝 Lecture Summary

1. Physics is a science. Science works according to the scientific method.

Physics operates on the scientific method, which accepts only reason, logic, and experimental evidence to determine what is scientifically correct. Scientists do not simply believe authoritative claims—they test and keep testing until satisfied. A discovery becomes a scientific fact only when it is repeatedly established in different laboratories at different times by different people, or when a theoretical result is derived through clear use of established rules. The real strength of science lies in its continuous self-challenge.

2. It is thought that the laws of physics do not change from place to place.

The laws of physics are universal across space and time. Experiments carried out in different countries by scientists of any religion or race have always led to the same results when done honestly and correctly. Evidence from light that left distant stars billions of years ago strongly indicates that physical laws operating at that time were identical to those today. The spectra of different elements then and now are indistinguishable, even after careful examination by physicists.

3. This course will cover the following broad categories:

The course covers four major branches of physics: a) Classical Mechanics (Newtonian mechanics): deals with the motion of bodies under the action of forces. b) Electromagnetism: studies how charges behave under the influence of electric and magnetic fields, and how charges create these fields. c) Thermal Physics: studies the nature of heat and the changes that addition of heat brings about in matter. d) Quantum Mechanics: deals with the physics of small objects such as atoms, nuclei, and quarks (treated only briefly for lack of time).

4. Every physical quantity can be expressed in terms of three fundamental dimensions

All physical quantities can be expressed in terms of three fundamental dimensions: Mass (M), Length (L), and Time (T). Examples:

🔑 Definition — Dimensional Homogeneity: You cannot add quantities that have different dimensions. Force can be added to force, but force can never be added to energy. A formula is definitely wrong if the dimensions on the left and right sides of the equal sign are different.

5. Remember that any function f(x) takes as input a dimensionless number

Any function f(x) takes as input a dimensionless number x and outputs a quantity f (which may or may not have a dimension). For example, consider the function f(θ) = sin θ. Its expansion is: sin θ = θ − θ³/3! + θ⁵/5! − ... If θ had a dimension, then you would be adding quantities of different dimensions, which is not allowed.

💡 Why this matters: This explains why angles are dimensionless—the argument of trigonometric, exponential, and logarithmic functions must always be dimensionless.

6. Do not confuse units and dimensions

Units are different ways to measure the same physical quantity, while dimensions are the fundamental nature of the quantity. For example, mass can be measured in kilograms, pounds, or sair and chatak. This course uses the MKS (Metre-Kilogram-Second) system.

📐 Formula for unit conversion: 1 mi/hr = 1 mi × 5280 ft/mi × 1 m/3.28 ft × 1 hr/3600 s = 0.447 m/s

📌 Example: Converting 1 mile per hour to meters per second:

• Start with: 1 mi/hr
• Multiply by: 5280 ft/mi (cancels miles)
• Multiply by: 1 m/3.28 ft (cancels feet)
• Multiply by: 1 hr/3600 s (cancels hours)
• Result: 0.447 m/s

When written out methodically, quantities cancel cleanly in numerator and denominator, preventing errors.

7. A good scientist first thinks of the larger picture

Order of magnitude estimation is extremely important. Students often make the mistake of focusing on decimal places instead of the first digit, which matters most. If asked to calculate the height of a building and you come up with 0.301219 metres or 4.01219 × 10⁶ metres, the answer is nonsense even if the last six digits are correct. Physics is commonsense first—use your intelligence before submitting any answer.

8. Always check your equations to see if they have the same dimensions

Always check equations for dimensional consistency on both sides. For example, the equation v² = u² + 2at is clearly wrong because the dimensions don't match (the right side has dimensions of L²/T² + LT⁻¹, which is inconsistent). However, v² = u² + 13a²t² could possibly be a correct relation because both sides have dimensions of L²/T².

🔑 Definition — Dimensional Analysis: A method of checking equations by verifying that both sides have the same dimensions. Matching dimensions does not guarantee correctness (the constant 13 might be wrong), but mismatched dimensions guarantee the equation is wrong.

9. Whenever you derive an equation that is a little complicated, see if you can find a special limit

When dealing with complex equations, find special limits where the equation becomes simple and transparent. Imagine that some quantity is very large or very small. Where possible, make a "mental graph" to picture an equation. For example, the formula for molecular speed distribution: f(v) = ve^-(v-v₀)²/a²

Even without knowing the value of a, you can immediately see: a) f(v) goes to zero for large values of v, and at v = 0 b) The maximum value of f(v) occurs at v₀, and the function decreases on both sides of this value

⭐ Key Takeaways

The scientific method requires repeated testing and verification before accepting any discovery as fact. All physical quantities have fundamental dimensions of Mass, Length, and Time, and equations must be dimensionally consistent—you can only add quantities with the same dimensions, and unmatched dimensions on either side of an equation guarantee it is wrong. Units are different from dimensions, and converting between unit systems requires methodical cancellation. Before solving any problem, estimate the order of magnitude for commonsense checking, and always verify dimensional consistency. When analyzing complex equations, look for special limits and create mental graphs to understand the behavior intuitively.

🧠 Quick Revision Questions

1. What are the three fundamental dimensions, and what dimensions do force, energy, and pressure have?
2. Why is the equation v² = u² + 2at dimensionally incorrect? What would make it possibly correct?
3. Why must the argument of a trigonometric function like sin θ be dimensionless?
4. How would you convert 60 miles per hour to meters per second using the methodical cancellation approach?
5. What can you conclude about the molecular speed distribution function f(v) = ve^-(v-v₀)²/a² just by examining its form?

📘 Lecture 2 — KINEMATICS I

📖 Overview: This lecture introduces the fundamental concepts of kinematics, focusing on displacement, velocity, and acceleration. It explains the difference between average and instantaneous quantities, and derives the key equations of motion for bodies moving with constant acceleration. The lecture also introduces vectors and their components, which are essential for describing motion in multiple dimensions.

🗂️ Topics Covered

The lecture covers displacement and position, average and instantaneous velocity, average and instantaneous acceleration (including deceleration), motion with constant acceleration (including derivation of equations v = v₀ + at, x = x₀ + v₀t + ½at², and v² = v₀² + 2a(x - x₀)), the concept of vectors and their components in one, two, and three dimensions, and the geometric addition of vectors using the parallelogram method.

📝 Lecture Summary

Displacement and Position

x(t) is called displacement and it denotes the position of a body at time t. If the displacement is positive, then that body is to the right of the chosen origin, and if negative, then it is to the left.

Average and Instantaneous Velocity

If a body is moving with average speed v, then in time t it will cover a distance d = vt. However, since the speed of a car changes from time to time, one should limit the use of this formula to small time differences only.

🔑 Definition — average speed over a small time interval Δt: the ratio of the change in position Δx to the change in time Δt, formally expressed as (x₂ - x₁) / (t₂ - t₁).

🔑 Definition — instantaneous velocity at any time t: v = lim(Δt→0) Δx / Δt = dx/dt. Here Δx and Δt are both very small quantities that tend to zero, but their ratio v does not.

Acceleration

Just as we have defined velocity as the rate of change of distance, we can define instantaneous acceleration at any time t as: a = lim(Δt→0) Δv / Δt = dv/dt. Here Δv and Δt are both very small quantities that tend to zero but their ratio a is not zero, in general. Negative acceleration is called deceleration. The speed of a decelerating body decreases with time.

💡 Why this matters: Some students are puzzled by the fact that a body can have a very large acceleration but can be standing still at a given time. In fact, it can be moving in the opposite direction to its acceleration. This is not strange because position, velocity, and acceleration are independent quantities.

Motion with Constant Acceleration

If the speed is increasing linearly (i.e., constant acceleration), then the answer is particularly simple: just use the same formula as before but use the average speed.

📐 Formula: For constant acceleration and a body that starts from rest at t = 0, v increases linearly with time, v ∝ t (or v = at). If the body has speed v₀ at t = 0, then at time t: v = v₀ + at.

📌 Example: A car starts from rest (v₀ = 0) with constant acceleration a = 2 m/s². After t = 5 seconds, its velocity is v = 0 + (2)(5) = 10 m/s.

📐 Formula: x = x₀ + v₀t + ½at². This formula tells you how far a body moves in time t if it moves with constant acceleration a, and if started at position x₀ at t = 0 with speed v₀.

Derivation: Using average speed (v₀ + (v₀ + at))/2 = (2v₀ + at)/2 = v₀ + ½at, we get x = x₀ + (v₀ + ½at)t = x₀ + v₀t + ½at².

📐 Formula: v² = v₀² + 2a(x - x₀). This formula tells us what the final speed will be after the body has traveled a distance equal to x - x₀, eliminating the time variable.

Vectors

Vectors: a quantity that has a size as well as direction is called a vector. So, for example, the wind blows with some speed and in some direction. So the wind velocity is a vector.

In one dimension, a vector has only one component (called the x-component). In two dimensions, a vector has both x and y components. In three dimensions, the components are along the x, y, z axes.

🔑 Definition — If we denote a vector r = (x, y), then rₓ = x = r cosθ, and rᵧ = y = r sinθ. Note that x² + y² = r². Also, that tanθ = y/x.

Geometric Addition of Vectors

Two vectors can be added together geometrically. We take any one vector, move it without changing its direction so that both vectors start from the same point, and then make a parallelogram. The diagonal of the parallelogram is the resultant.

For algebraic addition, add the components of the two vectors along each axis separately. So, for example, (1.5, 2.4) + (1, -1) = (2.5, 1.4).

⭐ Key Takeaways

The most critical concepts from this lecture are that displacement, velocity, and acceleration are independent quantities, meaning a body can have high acceleration while stationary or move opposite to its acceleration. For constant acceleration, the three key equations of motion are v = v₀ + at, x = x₀ + v₀t + ½at², and v² = v₀² + 2a(x - x₀), which must be memorized. Vectors have both magnitude and direction, and their components along axes are given by rₓ = r cosθ and rᵧ = r sinθ. Vector addition is performed by adding corresponding components or geometrically using the parallelogram method. Finally, average quantities are defined over finite intervals while instantaneous quantities are limits as time intervals approach zero.

🧠 Quick Revision Questions

1. What is the difference between average velocity and instantaneous velocity, and how is each calculated?
2. A car traveling at 20 m/s brakes with constant deceleration of 4 m/s². How far does it travel before stopping?
3. What are the three equations of motion for constant acceleration, and what variables does each relate?
4. How do you find the components of a vector given its magnitude and direction?
5. Explain how to add two vectors both geometrically and algebraically.

📘 Lecture 3 — KINEMATICS II

📖 Overview: This lecture introduces the concept of the derivative as a fundamental tool for understanding how functions change, with specific application to kinematics. It extends the study of motion from one dimension to two dimensions, covering velocity, acceleration, scalar products, and the important result that the derivative of (t^n) is (nt^{n-1}).

🗂️ Topics Covered

The lecture covers the definition and calculation of derivatives, the derivative of (t^n), application to constant acceleration motion, the concept of second derivative, gravitational acceleration, two-dimensional position and velocity using unit vectors, and the scalar (dot) product of vectors.

📝 Lecture Summary

The Concept of the Derivative

The derivative of a function is exceedingly important. It shows how fast a function changes when its argument is changed. For (f(x)), we say (f) is a function that depends upon the argument (x). Think of (f) as a machine that gives you the value (f) when you input (x). Functions do not always have to be written as (f(x)). For example, (x(t)) is also a function. It tells us where a body is at different times (t).

Definition of the Derivative

The derivative of (x(t)) at time (t) is defined as:

[ \frac{dx}{dt} \equiv \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \lim_{\Delta t \to 0} \frac{x(t + \Delta t) - x(t)}{\Delta t} ]

🔑 Definition — Derivative: The rate of change of a function with respect to its argument (here, time (t)).

Derivative of (x(t) = t^2)

Let's see how to calculate the derivative of a simple function like (x(t) = t^2). We must first calculate the difference in (x) at two slightly different values, (t) and (t + \Delta t), while remembering that we choose (\Delta t) to be extremely small:

[ \Delta x = (t + \Delta t)^2 - t^2 = t^2 + (\Delta t)^2 + 2t\Delta t - t^2 ]

[ \frac{\Delta x}{\Delta t} = \Delta t + 2t ]

[ \frac{dx}{dt} = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = 2t ]

📌 Example: If (x(t) = t^2), then at (t = 3), the instantaneous velocity is (dx/dt = 2(3) = 6).

General Derivative of (t^n)

In exactly the same way you can show that if (x(t) = t^n) then:

[ \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = nt^{n-1} ]

This is an extremely useful result.

🔑 Definition — Power Rule for Derivatives: (\frac{d}{dt}(t^n) = nt^{n-1})

Application to Constant Acceleration Motion

Let us apply the above to the function (x(t)) which represents the distance moved by a body with constant acceleration (see lecture 2):

[ x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 ]

Taking the derivative gives velocity:

[ \frac{dx}{dt} = 0 + v_0 + \frac{1}{2} a (2t) = v_0 + a t ]

Taking the derivative again gives acceleration:

[ \frac{dv}{dt} = 0 + a = a ]

This clearly shows that the acceleration is constant.

💡 Why this matters: The derivative provides a direct mathematical link between position, velocity, and acceleration, allowing us to derive one from another.

The Second Derivative

The second derivative is the rate of rate of change of (x) with respect to (t). It is written as:

[ \frac{d}{dt}\left(\frac{dx}{dt}\right) = \frac{d^2 x}{dt^2} ]

We call (\frac{d^2 x}{dt^2}) the second derivative or the rate of rate of change of (x) with respect to (t).

Gravitational Acceleration

A stone dropped from rest increases its speed in the downward direction according to:

[ \frac{dv}{dt} = g \approx 9.8 \text{ m/sec}^2 ]

This is true provided we are fairly close to the earth, otherwise the value of (g) decreases as we go further away from the earth. Also, note that if we measured distances from the ground up, then the acceleration would be negative.

🔑 Definition — g: The acceleration due to gravity near Earth's surface, approximately (9.8 , \text{m/s}^2) downward.

📐 Formula: (g \approx 9.8 , \text{m/s}^2) → The rate of change of velocity for a freely falling object near Earth's surface.

Position and Velocity in Two Dimensions

It is easy to extend these ideas to a body moving in both the (x) and (y) directions. The position and velocity in 2 dimensions are:

[ \vec{r} = x(t)\hat{i} + y(t)\hat{j} ]

[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} = v_x\hat{i} + v_y\hat{j} ]

Here the unit vectors (\hat{i}) and (\hat{j}) are fixed, meaning that they do not depend upon time.

🔑 Definition — Unit vectors (\hat{i}, \hat{j}): Fixed vectors of magnitude 1 pointing in the (x) and (y) directions respectively, used to express position and velocity components.

Scalar Product of Vectors

The scalar product of two vectors (\vec{A}) and (\vec{B}) is defined as:

[ \vec{A} \cdot \vec{B} = AB \cos\theta ]

You can think of:

• (\vec{A} \cdot \vec{B} = (A)(B \cos\theta)) = (length of (\vec{A})) × (projection of (\vec{B}) on (\vec{A}))
• (\vec{A} \cdot \vec{B} = (B)(A \cos\theta)) = (length of (\vec{B})) × (projection of (\vec{A}) on (\vec{B}))

Remember that for unit vectors:

• (\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = 1)
• (\hat{i} \cdot \hat{j} = 0)

🔑 Definition — Scalar (Dot) Product: (\vec{A} \cdot \vec{B} = AB \cos\theta), where (\theta) is the angle between (\vec{A}) and (\vec{B}). The result is a scalar quantity.

📌 Example: If (\vec{A} = 3\hat{i}) and (\vec{B} = 4\hat{j}), then (\vec{A} \cdot \vec{B} = 3(4)(\cos 90^\circ) = 0). If (\vec{A} = 3\hat{i}) and (\vec{B} = 4\hat{i}), then (\vec{A} \cdot \vec{B} = 3(4)(\cos 0^\circ) = 12).

⭐ Key Takeaways

The derivative is the fundamental tool for analyzing motion — position's derivative gives velocity, and velocity's derivative gives acceleration. For any function (t^n), the derivative is (nt^{n-1}), which is essential for solving kinematics problems. Constant acceleration equations can be confirmed using derivatives. In two dimensions, position and velocity are vectors expressed with fixed unit vectors (\hat{i}) and (\hat{j}), and the scalar product (\vec{A} \cdot \vec{B} = AB \cos\theta) is a key operation for combining vectors. Gravitational acceleration (g) is approximately (9.8 , \text{m/s}^2) downward near Earth's surface.

🧠 Quick Revision Questions

1. Using the definition of the derivative, prove that if (x(t) = t^3), then (\frac{dx}{dt} = 3t^2).
2. Starting from (x(t) = x_0 + v_0 t + \frac{1}{2} a t^2), derive expressions for velocity and acceleration.
3. What is the second derivative of a function, and what does it represent in kinematics?
4. Write the position and velocity vectors for a particle moving in two dimensions, and explain the role of unit vectors (\hat{i}) and (\hat{j}).
5. Calculate the scalar product (\vec{A} \cdot \vec{B}) if (|\vec{A}| = 5), (|\vec{B}| = 10), and the angle between them is (60^\circ). What is the value if they are perpendicular?

📘 Lecture 4 — FORCE AND NEWTON'S LAWS

📖 Overview: This lecture introduces the modern view of force and motion, replacing the ancient belief that force is needed to keep things moving. It explains Newton's three laws of motion, the concepts of inertia, mass, weight, and the vector nature of forces, forming the foundation of classical mechanics.

🗂️ Topics Covered

The lecture covers the ancient versus modern view of motion, inertia, Newton's First Law and inertial frames, the relationship between force and acceleration, Newton's Second Law (F = ma), the meaning and units of force, internal vs external forces, vector addition of forces, the independence of gravitational acceleration from mass, weight versus mass, Newton's Third Law, and the apparent paradox of action-reaction pairs.

📝 Lecture Summary

1. Ancient view: objects tend to stop if they are in motion; force is required to keep something moving.

This was a natural belief because we see objects stop after some time due to friction. Frictionless motion is only observable in special circumstances.

2. Modern view: objects tend to remain in their initial state; force is required to change motion.

Resistance to changes in motion is called inertia. More inertia means it is harder to make a body accelerate or decelerate.

3. Newton's First Law: An object will remain at rest or move with constant velocity unless acted upon by a net external force.

A non-accelerating reference frame is called an inertial frame; Newton's First Law holds only in inertial frames.

4. More force leads to more acceleration: a ∝ F

This means acceleration is directly proportional to the applied force.

5. The greater the mass of a body, the harder it is to change its state of motion.

More mass means more inertia. More mass leads to less acceleration: a ∝ 1/m

6. Newton's Second Law: F = ma

Combine the two observations above: a = F/m, or equivalently F = ma.

7. For it to be useful, we must have separate ways of measuring mass, acceleration, and force.

• Mass is a measure of the amount of matter in a body (e.g. two identical cars have twice the mass of a single one).
• Forces (due to gravity, a stretched spring, repulsion of two like charges, etc.) will be discussed later.

🔑 Definition — Inertia: Resistance to changes in motion; more inertia means harder to accelerate or decelerate.

📐 Formula: F = ma → The net force on a body equals its mass multiplied by its acceleration.

📌 Example: The unit of force is the Newton (N) where: 1 Newton = 1 kilogram·metre/second². Force has dimensions of [mass] × [acceleration] = M LT⁻². In the MKS system, [acceleration] = LT⁻².

8. Forces can be internal or external.

For example, the mutual attraction of atoms within a block of wood are called internal forces. Something pushing the wood is an external force. In the application of F = ma, remember that F stands for the total external force upon the body.

9. Forces are vectors, and so they must be added vectorially: F = F₁ + F₂ + F₃ + ... = ma

This means that the components in the x̂ direction must be added separately, those in the ŷ direction separately, etc.

💡 Why this matters: When considering the acceleration of a body, you must consider only the (net) force acting upon that body.

10. Gravity acts directly on the mass of a body.

This is a very important experimental observation due to Newton. A body of mass m₁ experiences a force F₁ = m₁g, while a body of mass m₂ experiences a force F₂ = m₂g. This means that g is the acceleration with which any body (big or small) falls under the influence of gravity. (Galileo established this when he dropped different masses from the Leaning Tower of Pisa.)

11. The weight of a body W is the force which gravity exerts upon it: W = mg.

Mass and weight are two completely different quantities. For example, if you used a spring balance to weigh a kilo of grapes on Earth, the same grapes would weigh only 1/7 kilo on the Moon.

🔑 Definition — Weight (W): The force of gravity upon a body, given by W = mg.

📐 Formula: W = mg → Weight equals mass times gravitational acceleration.

📌 Example: A 1 kg mass on Earth (g ≈ 9.8 m/s²) weighs 9.8 N. On the Moon (g ≈ 1.6 m/s²), the same mass weighs about 1.4 N, or roughly 1/7 of its Earth weight.

12. Newton's Third Law: For every action there is an equal and opposite reaction.

More precisely, F_AB = -F_BA, where F_AB is the force exerted by body B upon A, whereas F_BA is the force exerted by body A upon B. Ask yourself: what would happen if this was not true? In that case, a system of two bodies, even if completely isolated from the surroundings, would have a net force acting upon it because the net force acting upon both bodies would be F_AB + F_BA ≠ 0.

13. If action and reaction are always equal, then why does a body accelerate at all?

Students are often confused by this. The answer: in considering the acceleration of a body, you must consider only the (net) force acting upon that body. For example, the Earth pulls a stone towards it and causes it to accelerate because there is a net force acting upon the stone. By the Third Law, the stone also pulls the Earth towards it, causing the Earth to accelerate towards the stone. However, because the mass of the Earth is so large, we are only able to see the acceleration of the stone and not that of the Earth.

💡 Why this matters: Action-reaction pairs act on different bodies, so they don't cancel each other when analyzing the motion of a single body.

⭐ Key Takeaways

The most critical concepts from this lecture are: Newton's First Law establishes inertia and inertial frames; Newton's Second Law (F = ma) quantifies how force, mass, and acceleration are related — force is directly proportional to acceleration and inversely proportional to mass; forces are vectors and must be added component-wise; weight (W = mg) is a force due to gravity and is distinct from mass; Newton's Third Law (action-reaction) states that forces always come in equal and opposite pairs acting on different bodies, which explains why a body can accelerate despite the Third Law.

🧠 Quick Revision Questions

1. State Newton's First Law and explain what an inertial frame is.
2. Write Newton's Second Law mathematically. What does each symbol represent?
3. What is the difference between mass and weight? Give an example.
4. State Newton's Third Law and explain why a stone accelerates toward Earth but Earth does not appear to accelerate toward the stone.
5. How do you add forces vectorially when they act in different directions?

📘 Lecture 5 — APPLICATIONS OF NEWTON’S LAWS – I

📖 Overview: This lecture explores how Newton’s Laws apply to both equilibrium and non-equilibrium systems. It introduces the concept of tension in ropes, the nature and mathematics of frictional forces, and presents a general method for solving equilibrium problems by resolving forces into components.

🗂️ Topics Covered

The lecture begins by distinguishing between systems in equilibrium (net force zero) and those out of equilibrium. It then demonstrates how to find force from known acceleration using calculus, and vice versa. The concept of tension in an ideal massless rope is introduced, followed by a detailed discussion of frictional force, including its microscopic origin and the empirical formula linking it to the normal force. Finally, it presents the general principle for solving equilibrium problems by summing forces in chosen directions.

📝 Lecture Summary

Summary of Lecture 5 – APPLICATIONS OF NEWTON’S LAWS – I

An obvious conclusion from F = ma is that if F = 0 then a = 0

An obvious conclusion from F = ma is that if F = 0 then a = 0. This powerful statement means that for any body that is not accelerating, the sum of all the forces acting upon it must vanish.

Examples: equilibrium and non-equilibrium

Examples of systems in equilibrium: a stone resting on the ground; a pencil balanced on your finger; a ladder placed against the wall; an aircraft flying at a constant speed and constant height. Examples of systems out of equilibrium: a stone thrown upwards that is at its highest point; a plane diving downwards; a car at rest whose driver has just stepped on the car's accelerator.

Finding force from acceleration

If you know the acceleration of a body, it is easy to find the force that causes it to accelerate. Example: An aircraft of mass m has position vector, r = (at + bt³)î + (ct² + dt⁴)ĵ. What force is acting upon it? The force F is found from the second derivative of position: F = m (d²x/dt²)î + m (d²y/dt²)ĵ. The solution yields F = 6b mt î + m(2c + 12d t²)ĵ.

The reverse problem: finding position from force

The other way around is not so simple: suppose that you know F and you want to find x. For this you must solve the equation, d²x/dt² = F/m. This may or may not be easy, depending upon F (which may depend upon both x as well as t if the force is not constant).

Tension in a rope

Ropes are useful because you can pull from a distance to change the direction of a force. The tension, often denoted by T, is the force you would feel if you cut the rope and grabbed the ends. For a massless rope (which may be a very good approximation in many situations) the tension is the same at every point along the rope. Why? Because if you take any small slice of the rope it weighs nothing (or very little). So if the force on one side of the slice was any different from the force on the other side, it would be accelerating hugely. All this was for the "ideal rope" which has no mass and never breaks. But this idealization is often good enough.

Frictional force

We are all familiar with frictional force. When two bodies rub against each other, the frictional force acts upon each body separately opposite to its direction of motion (i.e., it acts to slow down the motion). The harder you press two bodies against each other, the greater the friction. Mathematically, F = μN, where N is the force with which you press the two bodies against each other (normal force). The quantity μ is called the coefficient of friction. It is large for rough surfaces, and small for smooth ones. Remember that F = μN is an empirical relation and holds only approximately. 🔑 Definition — Coefficient of friction (μ): a dimensionless scalar value that describes the ratio of the force of friction between two bodies and the force pressing them together. 💡 Why this matters: This simple empirical law is the basis for designing all mechanical systems involving contact and motion, from car brakes to conveyor belts. 📐 Formula: F = μN → The frictional force equals the coefficient of friction multiplied by the normal force.

Origin of friction

Friction is caused by roughness at a microscopic level. If you look at any surface with a powerful microscope you will see unevenness and jaggedness. If these big bumps are levelled somehow, friction will still not disappear because there will still be little bumps due to atoms. More precisely, this is called the electrostatic interaction: the atoms from the two bodies will interact with each other because of the electrostatic interaction between their charges. Even if an atom is neutral, it can still exchange electrons and there will be a force because of surrounding atoms.

Two blocks on a frictionless surface – finding tension and acceleration

Consider the two blocks below on a frictionless surface: We want to find the tension and acceleration. The total force on the first mass is F - T = m₁a. The force on the second mass is simply T and so T = m₂a. Solving the above, we get: a = F/(m₁ + m₂) and T = [m₂/(m₁ + m₂)] F.

General principle for solving equilibrium problems

There is a general principle by which you solve equilibrium problems. For equilibrium, the sum of forces in every direction must vanish. So Fₓ = 0, Fᵧ = 0, and F_z = 0. You may always choose the x, y, z directions according to your convenience. So, for example, as in the lecture problem dealing with a body sliding down an inclined plane, you can choose the directions to be along and perpendicular to the surface of the plane.

⭐ Key Takeaways

The fundamental and most powerful statement from Newton’s First Law is that if the net force on a body is zero, its acceleration is zero. Friction is a complex, microscopic phenomenon approximated by the empirical law F = μN, where μ is the coefficient of friction and N is the normal force. For an ideal, massless rope, tension is constant throughout its length. To solve equilibrium problems, you must resolve all forces into convenient (x, y, z) components and set the sum of forces in each direction to zero. When dealing with connected masses, draw free-body diagrams for each mass and apply Newton’s Second Law to each individually.

🧠 Quick Revision Questions

1. State Newton's First Law in terms of forces and acceleration. What does it imply for a body in equilibrium?
2. A rope is said to be "massless and ideal." What is the key property of the tension in such a rope?
3. Give the mathematical formula for the kinetic frictional force. What do each of the symbols represent?
4. Why is friction not completely eliminated even if a surface is polished to be microscopically smooth?
5. Two masses, m₁ and m₂, are connected by a rope and pulled by a force F on a frictionless surface. Write the expression for the acceleration of the system.

📘 Lecture 6 — APPLICATIONS OF NEWTON’S LAWS – II

📖 Overview: This lecture extends Newton’s laws to real-world scenarios involving fluid resistance, terminal velocity, friction, and accelerated reference frames. It explains how forces like tension, normal force, and friction interact in systems such as elevators, hanging masses, and sliding ropes, and provides systematic problem-solving strategies.

🗂️ Topics Covered

The lecture covers fluid resistance force and terminal speed; a systematic method for solving force problems; weight changes in an accelerating lift; finding acceleration from a hanging mass in a moving wagon; properties and direction of friction; the slipping rope on a table problem; calculating minimum force to prevent a block from slipping; and the Atwood machine problem with two masses over a pulley.

📝 Lecture Summary

1. Fluid Resistance and Terminal Speed

As a body moves through a fluid, it exerts force on the fluid to push it out of the way. By Newton’s third law, the fluid pushes back with an equal and opposite fluid resistance force, always opposite to the body’s velocity relative to the fluid. The magnitude of this force typically increases with speed, following the empirical law: f = kv.

🔑 Definition — Terminal speed: The maximum constant speed a falling object reaches when the downward force of gravity equals the upward fluid resistance force. At terminal speed, mg = kv, so v_final = mg/k.

📐 Formula: v_terminal = mg/k → The terminal velocity equals the weight divided by the fluid resistance constant.

📌 Example: A ball bearing dropped into deep oil. Initially it accelerates downward due to gravity. As speed increases, the upward fluid resistance force increases. Eventually, when mg = kv, the net force is zero and the ball continues at constant terminal speed.

💡 Why this matters: Terminal speed explains why parachutes work and why skydivers reach a constant fall speed.

2. General Problem-Solving Method for Force Problems

When solving Newton’s law problems, follow this systematic approach: a) draw a diagram, b) define an origin for a coordinate system, c) identify all forces (tension, normal, friction, weight, etc.) and their x and y components, d) apply Newton’s law separately along x and y axes, e) find accelerations, then velocities, then displacements.

3. Weight in a Lift (Elevator)

Your apparent weight changes depending on the lift’s acceleration. The normal force N from the floor determines what you feel. If the lift is at rest or moving at constant velocity, a = 0 and N = Mg. If the lift accelerates downwards, then Mg – N = Ma, so N = M(g – a) — you feel lighter. If a = g (cable breaks), N = 0 and you experience weightlessness. If the lift accelerates upwards, N = M(g + a) — you feel heavier.

📐 Formula: N = M(g ± a) where +a for upward acceleration, –a for downward acceleration.

4. Finding Acceleration from a Hanging Mass in a Railway Wagon

Imagine you are in a train with no windows. A mass hangs from the roof at an angle θ. By balancing forces vertically: T cosθ = mg (vertical equilibrium). Horizontally: T sinθ = ma (horizontal acceleration). Dividing the two equations gives tanθ = a/g. The mass m cancels out.

📐 Formula: a = g tanθ → The acceleration of the train equals g times the tangent of the hanging angle.

5. Properties of Friction

Friction is a reaction force that opposes applied motion. If you push a block forward, friction acts backward. If you push left, friction acts right. The direction of the frictional force is always opposite to the direction of the applied force.

6. Slipping Rope on a Table Problem

A rope of total length L and mass per unit length m has length l hanging over the edge. We find when the rope just begins to slip. Vertical balance gives the normal force on the table portion: N = m(L – l)g. The hanging part exerts force mlg to the right, balanced by friction μN to the left. Substituting N: μ m(L – l)g = mlg, simplifying to l = μL/(μ + 1).

📐 Formula: l_critical = μL/(μ + 1) → The critical hanging length for slipping depends on the coefficient of friction and total length.

📌 Example: If μ = 0.5 and L = 3 m, then l = (0.5 × 3)/(0.5 + 1) = 1.5/1.5 = 1 m. So 1 m must hang over the edge for the rope to just begin slipping.

7. Minimum Force to Prevent Block Slipping

A small block of mass m rests on a larger block of mass M. A horizontal force F is applied to the large block. Both accelerate together: F = (m + M)a, so a = F/(m + M). The small block experiences a normal force N = ma = mF/(m+M). To prevent the small block from slipping downward, friction μN must at least equal mg: μ × mF/(m+M) ≥ mg. Solving for F: F_min = (m + M)g/μ.

📐 Formula: F_min = (m + M)g/μ → The minimum horizontal force needed to prevent the small block from slipping.

8. Atwood Machine (Two Masses over a Frictionless Pulley)

Two masses m₁ and m₂ (with m₂ > m₁) are connected by a string over a frictionless pulley. Both masses accelerate with the same magnitude a. For m₂ (heavier, going down): m₂g – T = m₂a. For m₁ (lighter, going up): T – m₁g = m₁a. Adding the equations: m₂g – m₁g = (m₁ + m₂)a, giving a = (m₂ – m₁)g/(m₁ + m₂). Substituting back gives T = 2m₁m₂g/(m₁ + m₂).

📐 Formula: a = (m₂ – m₁)g/(m₁ + m₂) and T = 2m₁m₂g/(m₁ + m₂).

📌 Example: m₁ = 2 kg, m₂ = 3 kg. a = (3 – 2)(9.8)/(3+2) = 9.8/5 = 1.96 m/s². T = 2×2×3×9.8/(5) = 117.6/5 = 23.52 N.

⭐ Key Takeaways

The lecture’s most critical concepts include: fluid resistance force magnitude increases with speed and reaches terminal velocity when gravity balances resistance; apparent weight changes with lift acceleration (N = M(g ± a)); acceleration of a moving vehicle can be found from a hanging mass using tanθ = a/g; friction always opposes applied motion and is central to static equilibrium problems; and for the Atwood machine, both masses have equal acceleration determined by their mass difference over total mass — always memorize the a and T formulas for pulley systems as they appear frequently in exams.

🧠 Quick Revision Questions

1. A lift cable breaks and the lift falls freely. What is the apparent weight of a 70 kg person inside?
2. A 0.5 kg mass hangs from a train roof at 20° to the vertical. Find the train’s acceleration.
3. A rope of length 2 m and μ = 0.3 is on a table. What length must hang over the edge for it to just begin slipping?
4. Two masses 4 kg and 1 kg are connected over a frictionless pulley. Find the acceleration and tension in the string.
5. A 2 kg block sits on a 8 kg block. If μ = 0.4 between them, what minimum horizontal force on the 8 kg block prevents the 2 kg block from slipping?

📘 Lecture 7 — WORK AND ENERGY

📖 Overview: This lecture introduces the fundamental concept of work in physics, distinguishing it from everyday usage, and establishes the critical relationship between work and energy. It explains how work is calculated for constant and variable forces and derives the concept of kinetic energy and power, which are essential for understanding the motion and energy transfer in physical systems.

🗂️ Topics Covered

The lecture begins with the formal definition of work as force times displacement in the direction of motion, explaining it as a scalar quantity. It then covers how to calculate work for non-constant forces using integration. The concept of energy is introduced as the capacity to do work, with a specific derivation of the formula for kinetic energy. Finally, the lecture defines power as the rate of doing work and provides a solved example to tie these concepts together.

📝 Lecture Summary

Definition of work

Work is defined as the force applied in the direction of displacement, multiplied by the displacement. If the force F acts at an angle θ to the direction of motion, only the component of force in the direction of motion does work. The mathematical expression for work is the dot product of the force vector and the displacement vector.

🔑 Definition — Work (W): Work is the product of the component of a force in the direction of motion and the distance moved. 📐 Formula: ( W = \vec{F} \cdot \vec{d} = Fd \cos\theta ) → The work done is equal to the magnitude of the force times the magnitude of the displacement times the cosine of the angle between them. Work is a scalar quantity, possessing magnitude but no direction. Dimensions: Work has dimensions of force times length: ( M \times L^2 \times T^{-2} ). Units: 1 Newton × 1 Metre = 1 Joule (J).

📌 Example: Lifting a 20 kg mass through a distance of 2 metres. The work you do is ( W = 20 \text{ kg} \times 9.8 \text{ m/s}^2 \times 2 \text{ m} = 392 \text{ Joules} ). The force you exert is upward, while the work done by gravity is directed opposite to your motion, resulting in -392 Joules.

Work done by a variable force

If the force varies with distance (e.g., a spring), the simple formula ( W = Fd ) is no longer valid. To calculate the work, we break the total distance into many small intervals, ( \Delta x ), where the force is approximately constant. The total work is the sum of the work done over each small interval. [ W \approx \sum_{n=1}^{N} F_n \Delta x ] As ( \Delta x ) approaches 0 and the number of intervals N approaches infinity, this sum becomes an exact integral. [ W = \lim_{\Delta x \to 0} \sum_{n=1}^{N} F_n \Delta x \equiv \int_{x_i}^{x_f} F(x) dx ] This integral represents the area under the force-versus-position curve, F(x). This method works for both constant and non-constant forces.

📌 Example (Constant Force): A constant force, ( F = b ), acting from x=0 to x=a. The work done is the area of a rectangle: ( \int_0^a b , dx = a(b) = ab ).

📌 Example (Linearly Increasing Force): A force that increases linearly with distance, ( F = kx ), acting from x=0 to x=a. The work done is the area under the line, which is a triangle: ( \int_0^a kx , dx = \frac{1}{2} (a)(ka) = \frac{1}{2} k a^2 ).

The concept of kinetic energy

Energy is defined as the capacity of a physical system to do work. It exists in many forms (mechanical, electrical, chemical, nuclear, etc.) and can be stored and converted from one form to another. A fundamental principle is that energy can never be created or destroyed (conservation of energy).

To derive an expression for kinetic energy, consider a constant force accelerating a mass m from rest (speed 0) to a speed v over a distance d.

• The work done by the force is ( W = Fd ).
• Using Newton's second law, ( F = ma ).
• For constant acceleration, the kinematic equation is ( v^2 = 2ad ), which implies ( a = v^2 / 2d ).
• Substituting, the work is ( W = (m \cdot \frac{v^2}{2d}) d = \frac{1}{2} m v^2 ). This work has gone into creating the body's kinetic energy.

🔑 Definition — Kinetic Energy (KE or K): The energy an object possesses due to its motion. 📐 Formula: ( KE = \frac{1}{2} m v^2 ) → Kinetic energy equals one-half times the mass times the square of the velocity.

Power

The work done by a force depends on the force and distance but does not account for the time taken. Power is defined as the rate at which work is done, or the amount of work done per unit time.

🔑 Definition — Power (P): The rate of doing work. 📐 Formula: ( P = \frac{\text{Work}}{\text{Time}} = \frac{W}{\Delta t} ) If the force is constant and does not depend on time, the formula for power can be expressed as: 📐 Formula: ( P = F v ) → Power is equal to the force times the velocity of the object. This is derived from ( P = \frac{F \Delta x}{\Delta t} = F \frac{\Delta x}{\Delta t} = Fv ).

📌 Example (Work-Kinetic Energy Theorem): A constant force accelerates a bus of mass m from speed ( v_1 ) to speed ( v_2 ) over a distance d.

• From kinematics: ( v_2^2 - v_1^2 = 2ad ), so the acceleration is ( a = \frac{v_2^2 - v_1^2}{2d} ).
• The work done by the engine is ( W = Fd = m a d = m \left( \frac{v_2^2 - v_1^2}{2d} \right) d = \frac{1}{2} m v_2^2 - \frac{1}{2} m v_1^2 ). This shows the work done results in a change in the kinetic energy of the bus.

💡 Why this matters: This result, the Work-Energy Theorem, is a powerful principle. It states that the net work done on an object equals its change in kinetic energy (( W_{net} = \Delta KE )), providing a crucial connection between forces, motion, and energy.

⭐ Key Takeaways

1. Work is a scalar quantity defined as the product of the force component in the direction of motion and the displacement, calculated as ( W = Fd \cos \theta ). It is measured in Joules.
2. For a variable force, work is found by calculating the area under the force-versus-displacement graph, which is mathematically equivalent to the integral ( W = \int_{x_i}^{x_f} F(x) dx ).
3. The Work-Energy Theorem is a central concept: the net work done on an object is exactly equal to its change in kinetic energy (( W_{net} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 )).
4. Kinetic energy, the energy of motion, is quantified by ( \frac{1}{2} m v^2 ). Work is the process by which energy is transferred to or from an object, resulting in a change in its kinetic energy.
5. Power measures how quickly work is done, defined as ( P = W / t = Fv ), and it is a critical parameter for describing the performance of engines and machines.

🧠 Quick Revision Questions

1. What is the difference between the scientific definition of work and the everyday usage of the word?
2. How do you calculate the work done by a force that is not constant, such as a spring force?
3. State the Work-Energy Theorem in your own words. What is the mathematical relationship it describes?
4. Derive the formula for kinetic energy ( ( \frac{1}{2} m v^2 ) ) starting from the definitions of work and Newton's second law.
5. A car engine applies a constant force of 1000 N to move a car at a constant velocity of 20 m/s. What is the power output of the engine?

📘 Lecture 8 — Conservation of Energy

📖 Overview: This lecture introduces the concept of potential energy as stored energy of position and explores how it converts to kinetic energy. It explains conservation of mechanical energy through gravitational and spring systems, and distinguishes between conservative and non-conservative forces.

🗂️ Topics Covered

The lecture covers gravitational potential energy and its storage in the gravitational field, elastic potential energy in springs derived from work, kinetic energy's dependence on reference frame, total mechanical energy conservation with examples of thrown balls and rollercoaster motion, the relationship between force and potential energy using calculus, and the definition of conservative forces.

📝 Lecture Summary

1. Potential Energy is Stored Energy of Position

Potential energy is the energy "locked up" somewhere that can do work. An object stores energy as the result of its position. This stored energy of position has the ability or capacity to do work like other forms of energies. Potential energy can be converted into kinetic energy, which follows directly from Newton's Laws.

🔑 Definition — Potential Energy: The stored energy of position possessed by an object.

2. Gravitational Potential Energy

If you lift a stone of mass m from the ground up a distance x, you do work against gravity. The constant force is mg, so W = mgx. By conservation of energy, the work done by you was transformed into gravitational potential energy whose value is exactly equal to mgx.

💡 Why this matters: The energy is stored neither in the mass nor in the earth — it is stored in the gravitational field of the combined system of stone + earth.

📐 Formula: W = mgx → Work done equals gravitational potential energy stored.

3. Elastic Potential Energy in a Spring

Suppose you pull on a spring and stretch it by an amount x away from its equilibrium position. The spring gets harder to pull as it becomes longer. When extended by x and you pull a further distance dx, the small work done is dW = Fdx = kxdx. Adding all small bits of work gives total work:

W = ∫₀ˣ Fdx = ∫₀ˣ kxdx = ½kx²

This work was transformed into energy stored in the spring. The spring contains energy exactly equal to ½kx².

🔑 Definition — Elastic Potential Energy: Energy stored in a deformed spring, equal to the work done to stretch or compress it.

📐 Formula: V = ½kx² → Energy stored in a spring, where k is the spring constant and x is the displacement from equilibrium.

4. Kinetic Energy and Reference Frames

Kinetic energy obviously depends on the frame you choose to measure it in. If you are running with a ball, it has zero kinetic energy with respect to you. But someone standing will see it has kinetic energy.

Example: A box of mass 12 kg is pushed with a constant force so that its speed goes from zero to 1.5 m/s (as measured by a person at rest on the cart) and it covers a distance of 2.4 m. No friction.

📌 Example — Work-Energy Theorem:

• Change in kinetic energy: ΔK = K_f − K_i = ½(12 kg)(1.5 m/s)² − 0 = 13.5 J
• Constant acceleration: a = (v_f² − v_i²) / [2(x_f − x_i)] = (1.5² − 0) / [2(2.4)] = 0.469 m/s²
• Net force: F = ma = (12 kg)(0.469 m/s²) = 5.63 N
• Work done: W = FΔx = (5.63 N)(2.4 m) = 13.5 J (same as ΔK)

This clearly shows that work and energy have different values in different frames.

5. Energy in Different Frames

Suppose somebody is standing on the ground and the trolley moves at 15 m/s relative to the ground. With the box speed relative to ground being 16.5 m/s:

📌 Example — Different Frame Calculation: ΔK' = K'_f − K'_i = ½(12)(16.5)² − ½(12)(15.0)² = 284 J

This demonstrates that the measured kinetic energy change differs between reference frames.

6. Conservation of Mechanical Energy

The total mechanical energy is: E_mech = KE + PE. If there is no friction, then E_mech is conserved. This means the sum does not change with time.

Example: A ball thrown upwards at speed v₀. How high will it go? The loss of kinetic energy equals the gain of potential energy: ½mv₀² = mgh, so h = v₀²/2g.

Example — Frictionless Motion Over Hills: Even though the motion is complicated, total energy is constant to get speeds at points B, C, D:

• At point A: ½mv_A² + 0
• At point B: ½mv_B² + mgh → v_B = √(v_A² − 2gh)
• At point C: ½mv_C² + 0 → v_C = √(v_A² + 2gh)
• At point D: ½mv_D² + mgh/2 → v_D = √(v_A² + gh)

Where h is height, loss of potential energy equals gain in kinetic energy.

7. Conservative Forces and Potential Energy

Potential energy has meaning only for a force that is conservative. Friction is an example of a non-conservative force and a potential energy cannot be defined for it. For a conservative force, F = −dV/dx.

🔑 Definition — Conservative Force: A force for which the work done in going from point A to point B is independent of the path chosen.

For a spring: V = ½kx², so F = −dV/dx = −kx (Hooke's law).

8. Derivation of F = −dV/dx

If a particle moves distance Δx in a potential V, then change in PE is ΔV = −FΔx. From this, F = −ΔV/Δx. Let Δx → 0: F = −lim(ΔV/Δx) as Δx→0 = −dV/dx

⭐ Key Takeaways

Energy conservation is fundamental: gravitational potential energy is mgh and elastic potential energy is ½kx², both derived from work done against forces. Mechanical energy (KE + PE) is conserved only for conservative systems without friction. Kinetic energy values are reference-frame dependent, while changes in potential energy define conservative forces via the relation F = −dV/dx. The work-energy theorem connects net work to changes in kinetic energy across all frames.

🧠 Quick Revision Questions

1. What is potential energy and where is gravitational potential energy actually stored?
2. Derive the expression for the elastic potential energy stored in a spring.
3. Why do measured values of kinetic energy and work differ between reference frames?
4. State the law of conservation of mechanical energy and give an example.
5. How is the force related to potential energy for a conservative force, and how is this derived?

📘 Lecture 9 — MOMENTUM

📖 Overview: This lecture introduces the concept of momentum as the "quantity of motion" of a body and re-expresses Newton's Second Law in terms of momentum. It then extends this idea to systems of many particles, leading to the crucial principle of conservation of momentum, which is independent of the nature of internal forces and is fundamental for analyzing collisions, explosions, and impulse problems.

🗂️ Topics Covered

The lecture defines momentum and its relationship to Newton's Second Law, extends the concept to systems of multiple particles to derive conservation of momentum, and applies it to one-dimensional collisions. It then explores examples including explosions (equal and opposite fragments), projectile motion with an internal explosion, and the concept of impulse—force multiplied by time—including its graphical interpretation and practical implications for reducing impact force.

📝 Lecture Summary

1. Momentum

Momentum is defined as the "quantity of motion" possessed by a body. More precisely, it is: Mass of the body × Velocity of the body. The dimensions of momentum are MLT⁻¹ and the units are kg-m/s. Momentum is a vector quantity, having both magnitude and direction, represented as p = mv.

2. Relation between Momentum and Newton's Second Law

Newton's Second Law can be re-expressed in terms of momentum. Originally written as F = ma, this can be rewritten as: F = dp/dt In words, the rate of change of momentum of a body equals the total force acting upon it.

🔑 Definition — Momentum (p): p = mv (mass × velocity). 📐 Formula: F = dp/dt → The total external force on an object equals the time rate of change of its momentum. 📌 Example: If a constant force of 10 N acts for 2 seconds on a 5 kg object initially at rest, the change in momentum is Δp = FΔt = (10 N)(2 s) = 20 kg·m/s. The final velocity is Δp/m = 20/5 = 4 m/s.

3. Total Momentum of a System of Particles

When there are many particles, the total momentum P of the system is the vector sum of the momenta of all individual particles: P = p₁ + p₂ + ... + pN The rate of change of total momentum is equal to the total external force acting on the system: dP/dt = F_ext This result is independent of what sort of forces act between the bodies (electric, gravitational, etc.) or how complicated these forces are.

🔑 Definition — Total Momentum (P): The vector sum of the momenta of all particles in a system: P = Σ pᵢ. 📐 Formula: dP/dt = F_ext → The rate of change of total momentum equals the net external force. 💡 Why this matters: This means internal forces between particles in a system cancel out in pairs and do not change the total momentum.

4. Conservation of Total Momentum

If no external force acts on a system, its total momentum remains constant (conserved). This is known as the law of conservation of momentum. Even if forces act between the bodies (internal forces), the total momentum is conserved if the net external force is zero.

5. One-Dimensional Collision

Two balls move only along a straight line and collide. Initial momentum Pᵢ = m₁u₁ + m₂u₂ and final momentum P_f = m₁v₁ + m₂v₂. Since no external force acts on the balls during the collision, momentum is conserved; thus: Pᵢ = P_f, or m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

📌 Example: A 2 kg ball moving at 4 m/s collides head-on with a 1 kg ball at rest. After collision, the 2 kg ball moves at 2 m/s. Find the velocity of the 1 kg ball. Initial momentum: Pᵢ = (2)(4) + (1)(0) = 8 kg·m/s Final momentum: P_f = (2)(2) + (1)(v₂) = 4 + v₂ By conservation: 4 + v₂ = 8 → v₂ = 4 m/s.

6. Explosion

Consider an object at rest that explodes into two fragments. Initial momentum Pᵢ = 0. Since no external forces act, final momentum P_f = 0. This gives m₁v₁ + m₂v₂ = 0, or m₁v₁ = -m₂v₂. The fragments fly apart with equal momentum but in opposite directions. The centre-of-mass stays at rest.

📌 Example: A 5 kg bomb at rest explodes into a 2 kg piece and a 3 kg piece. If the 2 kg piece flies to the right at 15 m/s, what is the velocity of the 3 kg piece? Conservation: m₁v₁ + m₂v₂ = 0 → (2)(15) + (3)(v₂) = 0 → 30 + 3v₂ = 0 → v₂ = -10 m/s (to the left).

7. Projectile Explosion Problem

A shell is fired from a cannon with speed 10 m/s at an angle 60° to the horizontal. At the highest point, it explodes into two equal-mass pieces. One piece retraces its path back to the cannon. Find the velocity of the other piece immediately after the explosion.

Solution:

• At highest point, vertical velocity = 0, horizontal velocity = 10 cos 60° = 5 m/s.
• Mass of shell = M, each fragment has mass M/2.
• Initial momentum just before explosion: Pₓ = M × 5 = 5M kg·m/s (horizontal).
• After explosion: One fragment retraces its path, so its horizontal velocity v₁ₓ = -5 m/s (same speed, opposite direction). Its momentum: (M/2)(-5) = -2.5M.
• Let the other fragment's horizontal velocity be v₂ₓ.
• Conservation of horizontal momentum: P_fₓ = Pᵢₓ
• (-2.5M) + (M/2)v₂ₓ = 5M → (M/2)v₂ₓ = 7.5M → v₂ₓ = 15 m/s.
• The second piece moves horizontally at 15 m/s.

8. Impulse

If a large force acts over a short time, a large change in momentum occurs. The impulse I is defined from Newton's Second Law: dp = F dt. Integrating over the time interval t₁ to t₂ gives: I = ∫ F dt = Δp = p_f - p_i In words, the change of momentum equals the impulse, which is the area under the curve of force versus time.

🔑 Definition — Impulse (I): The integral of force over time: I = ∫ F dt. It equals the change in momentum Δp. 📐 Formula: I = ∫ F dt = Δp → Impulse is the area under the F-t curve and equals the momentum change.

📌 Example: A 0.15 kg baseball is pitched at 40 m/s and hit back at 50 m/s. If the bat is in contact for 0.01 s, find the average force. Change in momentum: Δp = (0.15)(50 - (-40)) = (0.15)(90) = 13.5 kg·m/s. From I = F_avg Δt = Δp → F_avg = 13.5 / 0.01 = 1350 N.

9. Force from Impulse: Practical Implications

The average force is F = Δp / Δt. Extending the time (Δt) over which the momentum changes reduces the force. For example, wrapping your thumb with foam increases the collision time, reducing the peak force even though the impulse (Δp) is the same. Force is the rate of change of momentum: F = dp/dt.

💡 Why this matters: The same impulse (same change in momentum) can result in a much smaller peak force if the impact time is increased. This is the principle behind airbags, crumple zones in cars, and padding in sports equipment.

📌 Example: A hammer strikes a table. The F-t curve shows force rising from zero to a peak, then returning to zero. The area under this curve equals the impulse and the momentum change of the hammer.

⭐ Key Takeaways

1. Momentum (p = mv) is a vector quantity, and Newton's Second Law can be written as F = dp/dt, where force equals the rate of change of momentum.
2. Conservation of Momentum is a fundamental law: for any system with no net external force, total momentum P (the vector sum of all individual momenta) remains constant before and after collisions or explosions.
3. Impulse (I = ∫F dt = Δp) equals the change in momentum and is given by the area under the Force vs. Time curve. The same impulse can result from a large force over a short time or a small force over a long time (F = Δp/Δt).
4. In an explosion of an initially stationary object, fragments fly apart with equal and opposite momenta, so their velocities are inversely proportional to their masses (m₁v₁ = -m₂v₂).
5. The center-of-mass of a system under no external forces remains at rest (or moves with constant velocity), even during internal collisions or explosions, because the total momentum of the system is conserved.

🧠 Quick Revision Questions

1. State the law of conservation of momentum in words and write its mathematical form for a system of particles.
2. A 1000 kg car moving at 20 m/s collides with and sticks to a 2000 kg truck at rest. What is their common velocity after the collision?
3. A firecracker at rest explodes into three pieces. If two pieces of equal mass fly off perpendicular to each other at 10 m/s, what can be said about the velocity (magnitude and direction) of the third piece which has twice the mass of each of the first two? (Hint: Use conservation of momentum.)
4. A 0.5 kg ball drops from a height of 10 m and rebounds to 6 m. If the contact time with the ground is 0.05 s, calculate the average force exerted by the ground on the ball. (Hint: First find velocities just before and after impact using kinematics.)
5. Why does catching a fast-moving cricket ball "hurt" less if you move your hands backward with the ball, compared to keeping your hands stationary? Explain using the concept of impulse and force.

📘 Lecture 10 — COLLISIONS

📖 Overview: This lecture explores the physics of collisions, from everyday interactions like a bat hitting a ball to cosmic events like galaxies colliding. The core principle is that momentum is always conserved in any collision, but kinetic energy may or may not be, leading to the distinction between elastic and inelastic collisions. Understanding these concepts is crucial for analyzing a wide range of physical phenomena.

🗂️ Topics Covered

The lecture begins with the fundamental importance of collisions and the law of conservation of momentum. It then derives the equations for elastic collisions in 1D, introducing the concept that relative speed is conserved. Various special cases are examined, including equal masses, a heavy body at rest, a light body at rest, and a light body hitting a heavy target. The practical application of slowing neutrons in a reactor is discussed. Finally, the lecture covers completely inelastic collisions (using a ballistic pendulum example) and the treatment of collisions in 2D or 3D.

📝 Lecture Summary

1. Collisions are extremely important to understand

Collisions are ubiquitous, from electrons with atoms to cars with trucks and galaxies with galaxies. In every case, the sum of the initial momenta equals the sum of the final momenta.

🔑 Definition — Law of Conservation of Momentum: The total momentum of an isolated system (no external forces) remains constant before and after a collision.

2. Momentum conservation

For two bodies of mass (m_1) and (m_2) moving with velocities (u_1) and (u_2) before collision, and (v_1) and (v_2) after collision:

• Formula: (m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2)
• This can be rewritten as: (m_1 (u_1 - v_1) = m_2 (v_2 - u_2))
• This is a single equation with two unknowns, so it is not enough to solve for the final velocities.

3. Elastic collisions

If the collision is elastic, kinetic energy is also conserved.

• Formula: (\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2)
• Combining this with momentum conservation yields:
  • (u_1 - v_1 = v_2 - u_2) or (u_1 + v_1 = u_2 + v_2)

🔑 Definition — Relative speed: The relative speed of the incoming particles equals the relative speed of the outgoing particles in an elastic collision ((u_1 - u_2 = v_2 - v_1)).

Solving for (v_1) and (v_2):

• Formula: (v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u_1 + \left( \frac{2m_2}{m_1 + m_2} \right) u_2)
• Formula: (v_2 = \left( \frac{2m_1}{m_1 + m_2} \right) u_1 + \left( \frac{m_2 - m_1}{m_1 + m_2} \right) u_2)

4. Equal masses

When (m_1 = m_2), the equations simplify drastically.

• Formula: (v_1 = u_2) and (v_2 = u_1)
• Meaning: The two bodies simply exchange their velocities after the collision.

5. Heavy target at rest

What if one body is much heavier and initially at rest? For (m_2 \gg m_1) and (u_2 = 0):

• Formula: (v_1 = -u_1) and (v_2 = 0)
• Meaning: The heavy body remains at rest, and the light body bounces back with the same speed.

6. Light target at rest

What if the lighter body is at rest and is hit by the heavier body? For (m_1 \gg m_2) and (u_2 = 0):

• Formula: (v_1 = u_1) and (v_2 = 2u_1)
• Meaning: The truck's (heavy body) speed is unaffected, but the rickshaw (light body) is thrust forward at twice the truck's speed.

7. Slowing down neutrons

This example shows how to slow down particles (e.g., neutrons in a nuclear reactor) by colliding them with other particles at rest.

• A neutron of mass (m_1) collides head-on with an atomic nucleus of mass (m_2) initially at rest.
• Formula: The fractional decrease in kinetic energy of the neutron is: [ \frac{K_i - K_f}{K_i} = \frac{4 m_1 m_2}{(m_1 + m_2)^2} ]
• Meaning: This formula provides the maximum energy transfer in a head-on elastic collision. This principle is used to moderate (slow down) neutrons in a nuclear reactor.

📌 Example: For a neutron ((m_1 \approx 1) amu) colliding with a carbon nucleus ((m_2 = 12) amu), the fractional decrease in energy is (\frac{4 \times 1 \times 12}{(1+12)^2} = \frac{48}{169} \approx 0.284). So about 28.4% of the neutron's kinetic energy is transferred to the carbon nucleus per head-on collision.

8. Completely inelastic collision (ballistic pendulum)

This is a classic problem where a bullet embeds itself in a block of wood.

• Setup: A bullet of mass (m) is fired into a block of mass (M) suspended as a pendulum. After the collision, the block+bullet swings up to a height (y).
• Solution:
  1. Momentum conservation (inelastic collision): (m v = (m+M) V), where (v) is the bullet's initial speed and (V) is the speed of the block+bullet right after the collision.
  2. Energy conservation (after collision): (\frac{1}{2} (m+M) V^2 = (m+M) g y), giving (V = \sqrt{2gy}).
• Final Formula: The initial speed of the bullet is: [ v = \frac{m+M}{m} \sqrt{2gy} ]

💡 Why this matters: This is a practical method for measuring the speed of a bullet without needing high-speed electronics.

9. Collisions in 2 or 3 dimensions

In higher dimensions, momentum conservation must be applied to each component separately.

• Formula: (\vec{P}_i = \vec{P}_f) is actually three separate equations:
  • (p_{ix} = p_{fx})
  • (p_{iy} = p_{fy})
  • (p_{iz} = p_{fz})
• For an elastic collision in 2D/3D, there is only one extra equation from energy conservation, not three.

10. Energy in inelastic collisions

In an inelastic collision (e.g., a body breaking into 20 pieces), kinetic energy is not conserved.

• Meaning: The initial kinetic energy is used to do work against intermolecular forces, such as breaking bonds. The total energy is conserved, but some is converted into other forms (e.g., heat, sound, deformation energy), so kinetic energy is lost.

⭐ Key Takeaways

The most critical concept is that momentum is always conserved in any collision, regardless of whether it is elastic or inelastic. For elastic collisions, kinetic energy is also conserved, and the relative speed of approach equals the relative speed of separation; this provides a second equation to solve for final velocities. You must be able to apply the derived formulas for special cases (equal masses, heavy vs. light targets) for quick problem-solving. For completely inelastic collisions, objects stick together, and you solve by combining momentum conservation with energy conservation after the collision (e.g., the ballistic pendulum). In 2D or 3D collisions, treat the (x), (y), and (z) components of momentum separately, and add the energy conservation equation only if the collision is elastic.

🧠 Quick Revision Questions

1. State the law of conservation of momentum for a two-body collision.
2. Determine (v_1) and (v_2) for an elastic collision where (m_1 = m_2) and (u_1 = 5 , \text{m/s}), (u_2 = 0 , \text{m/s}).
3. A (10 , \text{kg}) ball moving at (20 , \text{m/s}) hits a (1 , \text{kg}) ball at rest in an elastic collision. What is the speed of the lighter ball after the collision?
4. In the ballistic pendulum setup, a bullet of mass (10 , \text{g}) embeds in a (2 , \text{kg}) block that rises to a height of (0.5 , \text{m}). What was the bullet's initial speed? (Take (g = 10 , \text{m/s}^2))
5. How many conservation equations do you have for a 2D elastic collision, and what are they?

📘 Lecture 11 — ROTATIONAL KINEMATICS

📖 Overview: This lecture introduces rotational kinematics, the description of motion for objects rotating about a fixed axis. It establishes the fundamental angular quantities—displacement, velocity, and acceleration—and their relationships to linear quantities, providing the mathematical framework for analyzing circular motion. This matters because rotational motion is ubiquitous in physics, from planetary orbits to machinery and vehicle engines.

🗂️ Topics Covered

The lecture covers the definition and measurement of angular displacement in radians, the relationship between arc length and angular displacement, average and instantaneous angular speed, angular acceleration, the derivation of tangential velocity and acceleration from angular quantities, a comparison of constant linear and angular acceleration formulas, and detailed problem-solving applications including centripetal acceleration. It concludes with an introduction to vector cross products and their properties.

📝 Lecture Summary

Summary of Lecture 11 – ROTATIONAL KINEMATICS

1. Any rotation is specified by giving two pieces of information: a) The point about which the rotation occurs, i.e. the origin. b) The angle of rotation is denoted by φ and is measured in radians. The maximum value of φ is 2π radians, which corresponds to 360 degrees or one full revolution. 🔑 Definition — 1 radian: 57.3° or 0.159 revolution.

2. The arc length = radius × angular displacement, or s = rφ. From this, it follows that if φ = 2π, then s = 2πr which is the total circumference.

3. Suppose that there is a particle located at the tip of the radius vector. Now we wish to describe its motion of the particle, i.e. as it goes around the circle. So, suppose that the particle moves from angle φ₁ to φ₂ in time t₂ − t₁. Then, the average angular speed ω is defined as: ω = (φ₂ − φ₁) / (t₂ − t₁) = Δφ / Δt Suppose that we look at ω over a very short time. Then: ω = lim(Δt→0) Δφ/Δt = dφ/dt is called the instantaneous angular speed.

4. To familiarize ourselves with the notion of angular speed, let us compute ω for a clock’s second, minute and hour hands:

• ω_second = 2π / 60 = 0.105 rad/s
• ω_minute = 2π / (60 × 60) = 1.75 × 10⁻³ rad/s
• ω_hour = 2π / (60 × 60 × 12) = 1.45 × 10⁻⁴ rad/s

5. Just as we defined acceleration for linear motion, we also define acceleration for circular motion: α ≡ (ω₂ − ω₁) / (t₂ − t₁) = Δω / Δt (average angular speed) Hence, α = lim(Δt→0) Δω/Δt = dω/dt = d²φ/dt² (angular acceleration). Now use s = rφ. Differentiate with respect to time t: ds/dt = r(dφ/dt). The rate of change of arc length s is clearly what we should call the circular speed, v. So v = rω. Since r is held fixed, it follows that dv/dt = r(dω/dt). Now define a_T = dv/dt. Obviously, a_T = rα. Here T stands for tangential, i.e. in the direction of increasing s.

6. Compare the formulae for constant linear and angular accelerations:

💡 Why this matters: The formulae are nearly identical even though they describe two totally different physical situations. Answer: because the mathematics is identical!

7. Problem: The angular speed of a car engine is increased from 1170 rev/min to 2880 rev/min in 12.6 s. a) Find the average angular acceleration in rev/min². b) How many revolutions does the engine make during this time? Solution: α = (ω_f − ω_i) / t = (2880 − 1170) / (12.6/60) = 8140 rev/min² φ = ω_i t + ½αt² = 1170 × (12.6/60) + ½ × 8140 × (12.6/60)² = 425 rev.

8. Wheel A of radius r_A = 10.0 cm is coupled by a chain B to wheel C of radius r_C = 25.0 cm. Wheel A increases its angular speed from rest at a uniform rate of 1.60 rad/s². Determine the time for wheel C to reach a rotational speed of 100 rev/min. Solution: Every part of the chain moves with the same speed, so v_A = v_C. Hence r_A ω_A = r_C ω_C ⇒ ω_A = (r_C / r_A) ω_C = (25.0/10.0) × 100 rev/min = 250 rev/min. Converting: ω_A = 250 × (2π/60) = 26.18 rad/s. Using ω = ω₀ + αt: t = (ω_A − 0) / α = 26.18 / 1.60 = 16.4 s.

9. Imagine a disc going around. All particles on the disc will have same 'ω' and 'α' but different 'v' and 'a'. Clearly 'ω' and 'α' are simpler choices!

10. Now consider a particle going around a circle at constant speed. You might think that constant speed means no acceleration. But this is wrong! It is changing its direction and accelerating. This is called "centripetal acceleration", meaning acceleration directed towards the centre of the circle. Note that the distance between points P₁ and P₂ is Δr = vΔt ≈ rθ. Similarly, Δv ≈ vθ ⇒ a = Δv/Δt ≈ vθ / (rθ/v) = v²/r. More generally, a = v²/r. In vector form: →a = −(v²/r) r̂. The negative sign indicates that the acceleration is towards the centre.

12. Vector Cross Products: The vector cross product of two vectors is defined as: →A × →B = AB sinθ n̂ where n̂ is a unit vector that is perpendicular to both A and B. Apply this definition to unit vectors in 3-dimensions: î × ĵ = k̂, k̂ × î = ĵ, ĵ × k̂ = î.

13. Some key properties of the cross product:

• A × B = −B × A
• A × A = 0
• A × (B + C) = A × B + A × C 🔑 Formula: A × B = (A_y B_z − A_z B_y)î + (A_z B_x − A_x B_z)ĵ + (A_x B_y − A_y B_x)k̂

14. The cross product is only definable in 3 dimensions and has no meaning in 2-d. This is unlike the dot product which has a meaning in any number of dimensions.

⭐ Key Takeaways

Rotational kinematics is mathematically analogous to linear kinematics, with angular displacement φ, angular velocity ω, and angular acceleration α replacing their linear counterparts x, v, and a, and the same constant-acceleration formulas apply. The fundamental relationship between linear and angular quantities is v = rω and a_T = rα, enabling the conversion between rotational and translational motion. A crucial insight is that even uniform circular motion involves acceleration—centripetal acceleration a = v²/r directed toward the center—because the velocity direction changes continuously. The vector cross product, defined as A × B = AB sinθ n̂, is essential for describing rotational quantities in three dimensions and obeys specific algebraic properties including anti-commutativity. Remember that all points on a rotating rigid body share the same ω and α but not the same v and a.

🧠 Quick Revision Questions

1. What is the relationship between arc length s, radius r, and angular displacement φ?
2. Derive the formulas for tangential velocity and tangential acceleration in terms of angular quantities.
3. A wheel accelerates uniformly from rest to 10 rad/s in 5 seconds. How many radians does it turn?
4. Why does a particle moving in a circle at constant speed still have acceleration? What is its magnitude and direction?
5. Compute the cross product A × B if A = (2, 1, 0) and B = (0, 3, 1).

📘 Lecture 12 — PHYSICS OF MANY PARTICLES

📖 Overview: This lecture extends Newtonian mechanics from single particles to systems of many particles. It introduces the concept of the center of mass, showing how it simplifies the description of motion for entire collections of particles, and then develops the rotational motion of rigid bodies, culminating in the rotational analogue of Newton's Second Law.

🗂️ Topics Covered

The lecture defines the center of mass for two and N particles, shows how to calculate its position for symmetrical and discrete objects, and derives Newton's Second Law for a collection of particles. It then introduces the moment of inertia, defines torque, derives the rotational work-energy theorem, and concludes with a comparison of linear and rotational motion, including the separation of kinetic energy into translational and rotational components.

📝 Lecture Summary

Physics of Many Particles

A body is made of a collection of particles. We can think of this body as having a "centre." For two masses, the centre of mass is defined as a vector position that represents the average location of the total mass. In two dimensions, this is expressed by two equations that give the coordinates of the centre of mass of the two-particle system.

🔑 Definition — Centre of Mass (for two masses): The point representing the mean position of the matter in a body, given by (\vec{r}_{cm} = \frac{m_1 \vec{r}1 + m_2 \vec{r}2}{m_1 + m_2}). In components, (x{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}) and (y{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2}).

📌 Example: One mass is placed at (x = 2\text{ cm}) and a second mass, equal to the first, is placed at (x = 6\text{ cm}). The cm position lies halfway between the two: (x_{cm} = \frac{mx_1 + mx_2}{m + m} = \frac{2m + 6m}{2m} = 4\text{ cm}). Note that there is no physical body actually located at (x_{cm} = 4\text{ cm})! The centre of mass can be a point where there is no matter. If the first mass is three times bigger than the second: (x_{cm} = \frac{(3m)x_1 + mx_2}{3m + m} = \frac{2(3m) + 6m}{4m} = 3\text{ cm}). This shows that the cm lies closer to the heavier body. 💡 Why this matters: The centre of mass is a weighted average, meaning heavier masses have a greater "pull" on its location.

Centre of Mass for N Masses and Newton's Second Law

For N masses, the centre of mass position is the obvious generalization. In words, it says: choose any origin and draw vectors (\vec{r}_1, \vec{r}_2, ... \vec{r}_N) that connect to the masses (m_1, m_2, ... m_N). Heavier masses get more importance in the sum. So if (m_2) is much larger than any of the others, the cm is very close to the position vector of (m_2). For symmetrical objects like a sphere or circle, the cm lies at the centre; for a cylinder it is on the axis halfway between the two faces.

The definition of the cm allows Newton's Second Law to be written for an entire collection of particles. The total external force (\vec{F}{ext}) on the system equals the total mass (M) times the acceleration of the centre of mass (\vec{a}{cm}). This is because the sum of internal forces (\vec{F}{int}) cancel out due to Newton's Third Law ((\vec{F}{12} + \vec{F}_{21} = 0)).

🔑 Definition — Centre of Mass (for N masses): (\vec{r}_{cm} \equiv \frac{1}{M} \sum_i m_i \vec{r}_i), where (M = \sum_i m_i).

📐 Formula: (\vec{v}{cm} = \frac{d\vec{r}{cm}}{dt} = \frac{1}{M} \sum_i m_i \vec{v}i) and (\vec{a}{cm} = \frac{d\vec{v}{cm}}{dt} = \frac{1}{M} \sum_i m_i \vec{a}i). Newton's Second Law for a system of particles is: (\sum \vec{F}{ext} = M \vec{a}{cm}).

Rotational Kinetic Energy and Moment of Inertia

Now consider rotational motion for a rigid system of N particles. Rigid means that all particles have a fixed distance from the origin. The kinetic energy is (K = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 + ...). Since (v_i = r_i \omega), this becomes (K = \frac{1}{2} (\sum_i m_i r_i^2) \omega^2). If we define the moment of inertia (I \equiv \sum_i m_i r_i^2), then the kinetic energy is (K = \frac{1}{2} I \omega^2). This is analogous to (K = \frac{1}{2} M v^2) for linear motion.

📐 Formula: Rotational Kinetic Energy: (K = \frac{1}{2} I \omega^2), where Moment of Inertia (I = \sum_i m_i r_i^2).

📌 Example: Two particles (m_1) and (m_2) are connected by a light rigid rod of length (L). Neglecting the mass of the rod, the rotational inertia (I) about an axis perpendicular to the rod and at a distance (x) from (m_1) is (I = m_1 x^2 + m_2 (L - x)^2). To find the value of (x) for which (I) is largest, calculate (\frac{dI}{dx} = 0): (\frac{dI}{dx} = 2m_1 x - 2m_2 (L - x) = 0 \implies x_{max} = \frac{m_2 L}{m_1 + m_2}).

Continuous Mass Distributions

Although matter is made up of discrete atoms, it is useful to think of matter as being continuously distributed. Since a sum (\sum) becomes an integral (\int), the definitions for moment of inertia and centre of mass become:

• (I = \int r^2 dm)
• (\vec{R}_{cm} = \frac{1}{M} \int \vec{r} dm)

📌 Example (Hoop): A hoop with mass distributed uniformly over it. The moment of inertia is (I = \int r^2 dm = R^2 \int dm = MR^2).

📌 Example (Solid Plate): For a solid plate, (I = \int r^2 dm), where (dm = 2\pi r dr \rho_0). The integral becomes (I = \frac{1}{2} (\pi R^2 \rho_0) R^2 = \frac{1}{2} M R^2).

Torque and Rotational Work

It is easier to turn things when the applied force acts at a greater distance. This is because torque (\vec{\tau} = \vec{r} \times \vec{F}). The magnitude of torque is (\tau = r F \sin \theta), where (\theta) is the angle between the radius vector and the force.

When a force (\vec{F}) acts through a distance (d\vec{s}), it does work (dW = \vec{F} \cdot d\vec{s}). For rotational motion, the small amount of work done by a torque is (dW = \tau d\phi). The net work done by all particles is (dW_{net} = (\sum \tau_{ext}) d\phi = (\sum \tau_{ext}) \omega dt). The change in kinetic energy is (dK = I \omega d\omega = (I \alpha) \omega dt). By conservation of energy, (dW_{net} = dK), which leads to the rotational analogue of Newton's Second Law: (\sum \tau_{ext} = I \alpha).

🔑 Definition — Torque: A measure of the force that can cause an object to rotate about an axis. (\vec{\tau} = \vec{r} \times \vec{F}).

📐 Formula: Rotational Work-Energy Theorem leads to (\sum \tau_{ext} = I \alpha).

Comparison of Linear and Rotational Motion

Total Kinetic Energy of a Rolling Object

Rotational and translational motion can occur simultaneously, like a car's wheel rotating and translating. The total kinetic energy is the sum of the energies of the two motions: (K = \frac{1}{2} M v_{cm}^2 + \frac{1}{2} I \omega^2). For a system of N particles, the total kinetic energy divides up neatly into the kinetic energy of rotation and translation. This can be derived by writing the velocity of each particle as (\vec{v}i = \vec{v}{cm} + \vec{v}_i'), where (\vec{v}i') is the velocity of a particle with respect to the cm frame. The cross term (\vec{v}{cm} \cdot \sum m_i \vec{v}_i') is zero because the total momentum is zero in the cm frame.

📐 Formula: (K = \frac{1}{2} M v_{cm}^2 + \frac{1}{2} I \omega^2).

⭐ Key Takeaways

The centre of mass is a powerful concept that allows us to treat the motion of an entire system as if all its mass were concentrated at a single point, with the net external force applied there. The moment of inertia is the rotational analogue of mass, quantifying an object's resistance to changes in rotational motion. Torque is the rotational analogue of force, and the equation (\tau_{net} = I \alpha) is the rotational analogue of Newton's Second Law. The total kinetic energy of a rolling or moving rigid body can be cleanly separated into translational energy of the centre of mass and rotational energy about the centre of mass. For any collection of particles, the total kinetic energy is the sum of the kinetic energy of the centre of mass and the kinetic energy of motion relative to the centre of mass.

🧠 Quick Revision Questions

1. Define the center of mass for a two-particle system. How does the position of the centre of mass shift if one mass is much larger than the other?
2. Write Newton's Second Law for a collection of particles. Why do internal forces not appear in this equation?
3. What is the moment of inertia? How does it relate to rotational kinetic energy? Calculate the moment of inertia of a thin hoop of mass M and radius R about its centre.
4. Define torque. What is the rotational analogue of Newton's Second Law, and how is it derived from the work-energy theorem?
5. A solid sphere is rolling without slipping on a flat surface. How is its total kinetic energy expressed? Explain the origin of both terms.

📘 Lecture 13 — ANGULAR MOMENTUM

📖 Overview: This lecture introduces the concept of angular momentum, beginning with its definition and calculation using a projectile as an example. It then establishes the fundamental relationship between torque and the rate of change of angular momentum, applying these principles to explain the precession of a spinning top and the conservation of angular momentum in a system of particles.

🗂️ Topics Covered

The lecture covers the definition and various forms of angular momentum; calculating angular momentum for a projectile; deriving the relationship between torque and the rate of change of angular momentum; applying this to explain the precession of a spinning top; extending the concept to a system of particles and showing that internal torques cancel; applying conservation of angular momentum; and contrasting linear and rotational motion equations.

📝 Lecture Summary

1. Recall the definition of angular momentum:

The angular momentum L of a particle is defined as L = r × p. The magnitude can be expressed in several equivalent ways: (a) L = rp sinθ (b) L = r(p sinθ) = r p⟂ (c) L = p(r sinθ) = p r⟂

🔑 Definition — Angular Momentum: A vector quantity representing the rotational analog of linear momentum, defined as the cross product of the position vector (r) and the linear momentum vector (p).

2. Angular Momentum of a Projectile

We can calculate the angular momentum of a projectile thrown from the ground at an angle θ. The initial angular momentum is zero because the position vector r is zero.

The projectile's coordinates at time t after launch are: x = (v₀ cosθ) t, y = (v₀ sinθ) t - ½ g t²

The velocity components are: vₓ = v₀ cosθ, vᵧ = v₀ sinθ - g t

Hence, L⃗ = r⃗ × p⃗ = (x î + y ĵ) × (vₓ î + vᵧ ĵ) m = m (x vᵧ - y vₓ) k̂ = m [ (v₀ cosθ t)(v₀ sinθ - g t) - (v₀ sinθ t - ½ g t²)(v₀ cosθ) ] k̂ = - ½ m g t² v₀ cosθ k̂

📌 Example: For a projectile, the angular momentum increases as t².

3. Relation between Torque and Rate of Change of L

The lecture establishes that τ = dL/dt. Just as a particle's momentum changes with time because of a force, a particle's angular momentum changes with time because of a torque.

The derivation starts with: ΔL = L(t + Δt) - L(t) = (r + Δr) × (p + Δp) - r × p = r × Δp + Δr × p (ignoring second-order term Δr × Δp)

Dividing by Δt and taking the limit as Δt → 0: dL/dt = r × (dp/dt) + (dr/dt) × p = r × F + v × p

Since v × p = v × (m v) = 0, the final fundamental equation is: dL/dt = τ

🔑 Definition — Torque (τ): The rotational analog of force, defined as τ = r × F, which causes a change in angular momentum.

📐 Formula: dL/dt = τ → The time rate of change of angular momentum of a particle equals the net torque acting on it.

4. The Spinning Top

The spinning top is an excellent application of torque and angular momentum. The torque due to gravity (τ = r × F) is perpendicular to L. Since τ = dL/dt, torque can only change the direction of L, not its magnitude.

The precession speed ω (the rate at which the axis of the top rotates) is given by: ω = Mgr / L

As the top slows down due to friction, L decreases, and the top precesses faster and faster.

🔑 Definition — Precession: The slow, conical rotation of the axis of a spinning object, such as a top, due to an external torque.

📐 Formula: ω = Mgr / L → The precession speed equals the torque due to gravity divided by the angular momentum of the spinning top.

5. Torque on a System of Particles

The torque on a system of particles can come from both external and internal forces. However, if internal forces between particles are equal, opposite, and directed along the line joining them, the total internal torque is zero.

For two particles: ∑ τ_int = τ₁ + τ₂ = r₁ × F₁₂ + r₂ × F₂₁ Since F₂₁ = -F₁₂ = F₁₂ r̂, then ∑ τ_int = (r₁ - r₂) × (F₁₂ r̂) = r₁₂ × (F₁₂ r̂) = 0

Thus, the net external torque acting on a system of particles equals the time rate of change of the total angular momentum of the system: ∑ τ_ext = dL_total/dt.

📐 Formula: ∑ τ_int = 0 → Internal torques cancel out, so only external torques change the total angular momentum of a system.

6. Conservation of Angular Momentum

It follows from ∑ τ_ext = dL/dt that if no net external torque acts on a system, then the total angular momentum is constant.

📐 Formula: ∑ τ_ext = 0 → L = constant (Conservation of Angular Momentum)

📌 Example: When two stationary discs, each with ½ MR², fall on top of a rotating disc, the total angular momentum is unchanged. I_i ω_i = I_f ω_f If the initial moment of inertia is I_i = ½ MR² and the final moment of inertia is I_f = ½ MR² + 2(½ MR²) = (3/2) MR², then: ω_f = ω_i (I_i / I_f) = ω_i (½ MR² / (3/2) MR²) = ω_i / 3

7. Similarities and Differences

Linear momentum: F = dp/dt, p = m v Rotational momentum: τ = dL/dt, L = I ω

Key difference: For linear momentum p, the origin does not matter. But L depends on the choice of origin. If you shift the origin by vector c, the new angular momentum L' is related to the old L by: L' = r' × p = (c + r) × p = c × p + L

8. Linear and Angular Acceleration

The derivative of velocity v = ω × r gives acceleration: a = dv/dt = d(ω × r)/dt = (dω/dt) × r + ω × (dr/dt) = α × r + ω × v

So the acceleration has a tangential part (a_T = α × r) and a radial part (a_R = ω × v).

⭐ Key Takeaways

The most critical takeaway is that torque is the rotational equivalent of force, causing a change in angular momentum via the fundamental equation τ = dL/dt. Key applications include the conservation of angular momentum (when net external torque is zero) which is used to solve problems like objects falling on a rotating disc, and the precession of a spinning top where a torque perpendicular to L changes its direction, not magnitude. Internal torques cancel in a system of particles, meaning only external torques can alter the total angular momentum. Finally, while linear and rotational motion equations are analogous, angular momentum is origin-dependent, unlike linear momentum.

🧠 Quick Revision Questions

1. What is the definition of angular momentum L of a particle, and what are three equivalent ways to write its magnitude?
2. Derive the relationship between torque and angular momentum. What does this equation imply about the effect of torque on angular momentum?
3. Using conservation of angular momentum, if a rotating disc with moment of inertia I and angular speed ω has two additional stationary discs dropped onto it, what is the final angular speed?
4. Explain why the axis of a spinning top precesses rather than falls over. How does the precession speed change as the top's spin slows down?
5. Why does the total internal torque on a system of particles equal zero? What does this imply about the net torque that can change the system's total angular momentum?

📘 Lecture 14 — EQUILIBRIUM OF RIGID BODIES

📖 Overview: This lecture introduces the conditions for mechanical equilibrium of rigid bodies, covering both translational and rotational equilibrium. It explains how to analyze static systems using force and torque balance, with practical applications including beam supports, balance problems, and finding the center of gravity.

🗂️ Topics Covered

Definition of rigid bodies and mechanical equilibrium conditions; translational and rotational equilibrium equations; static equilibrium analysis of beams on supports; torque independence of origin choice for equilibrium bodies; balance problems and center of gravity; finding center of gravity by suspension; non-uniform bar equilibrium with cords; rod leaning without slipping; stable, unstable, and neutral equilibrium types.

📝 Lecture Summary

1. Rigid Body Definition

A rigid body is one where all parts of the body are fixed relative to each other (for example, a pencil). Fluids and gases are non-rigid.

2. Translational and Rotational Motion

The translational motion of the centre of mass of a rigid body is governed by:

• Net external force: $\frac{d\vec{P}}{dt} = \vec{F}$ where $\vec{F} = \sum \vec{F}_{ext}$
• Net external torque: $\frac{d\vec{L}}{dt} = \vec{\tau}$ where $\vec{\tau} = \sum \vec{\tau}_{ext}$

🔑 Definition — Mechanical Equilibrium: A rigid body is in mechanical equilibrium if both the linear momentum $\vec{P}$ and angular momentum $\vec{L}$ have a constant value, i.e., $\frac{d\vec{P}}{dt} = 0$ and $\frac{d\vec{L}}{dt} = 0$. Static equilibrium refers to $\vec{P} = 0$ and $\vec{L} = 0$.

3. Example: Beam on Supports

Consider a beam resting on supports with force $F_1$ and $F_2$ pushing upward:

Step 1 — Force balance in vertical y-direction: $$\sum F_y = F_1 + F_2 - Mg - mg = 0$$

Step 2 — Torque balance: $$\sum \tau_y = (F_1)(0) + (F_2)(L) - (Mg)(L/4) - (mg)(L/2) = 0$$

From these two conditions: $$F_1 = \frac{3M + 2m}{4}g, \quad F_2 = \frac{M + 2m}{4}g$$

4. Choice of Origin and Torque

Torque depends on where you choose the origin of coordinates. For a body in equilibrium, the choice of origin does not matter.

Proof: Torque about point O: $\vec{\tau}_O = \vec{r}_1 \times \vec{F}_1 + \vec{r}_2 \times \vec{F}_2 + \cdots + \vec{r}_N \times \vec{F}_N$

Torque about point P (shifted by $\vec{r}_P$): $$\vec{\tau}_P = [\vec{r}_1 \times \vec{F}_1 + \vec{r}_2 \times \vec{F}_2 + \cdots + \vec{r}_N \times \vec{F}_N] - [\vec{r}_P \times (\vec{F}_1 + \vec{F}_2 + \cdots + \vec{F}_N)]$$ $$\vec{\tau}_P = \vec{\tau}O - [\vec{r}P \times (\sum F{ext})]$$ Since $\sum F{ext} = 0$ for a body in translational equilibrium, $\vec{\tau}_P = \vec{\tau}_O$.

💡 Why this matters: This shows the torque about any point is the same for a body in equilibrium, making calculations more flexible.

5. Practical Balance Problem

Consider a balance in equilibrium with two known weights. To find mass $m$ in terms of $m_1$ and $m_2$:

Taking torques about the knife edge in two cases:

• $mgx = m_1 g (L - x)$
• $m_2 g x = mg (L - x)$

This gives: $\frac{m}{m_1} = \frac{m_2}{m}$ or $m = \sqrt{m_1 m_2}$

🔑 Remarkably: We do not need the values of $x$ or $L$ to find $m$.

6. Centre of Gravity

The centre of gravity (CG) is the average location of the weight of an object. The net force on the whole body equals the sum of forces over all individual particles: $\vec{F} = \sum m_i \vec{g} = M\vec{g}$ where $M$ is the total mass.

If $\vec{g}$ has the same value at all points of the body, then:

1. The weight is equal to $M\vec{g}$
2. The centre of gravity coincides with the centre of mass

The net torque about the centre of mass due to gravity about the centre of mass of a body is zero.

7. Finding CG by Suspension

The CG of an irregular object can be found by suspending it on a pivot. The object will hang such that the CG is directly below the pivot point.

8. Non-Uniform Bar Problem

A non-uniform bar of weight $W$ is suspended at rest horizontally by two light cords. Find distance $x$ from left-hand end to centre of gravity.

Solution:

• Horizontal: $T_2 \sin\phi - T_1 \sin\theta = 0$
• Vertical: $T_2 \cos\phi + T_1 \cos\theta - W = 0 \Rightarrow T_2 = \frac{W}{\sin(\theta + \phi)}$

Taking torque about one end: $$-Wx + (T_2 \cos\phi)L = 0 \Rightarrow x = \frac{(T_2 \cos\phi)L}{W} = \frac{L \cos\phi}{\sin(\theta + \phi)}$$

9. Rod Leaning Without Slipping

Find the least angle $\theta$ at which the rod can lean to the horizontal without slipping.

Translational equilibrium:

• $R_1 = \mu_2 R_2$
• $R_2 + \mu_1 R_1 = W$

This gives: $R_2 = \frac{W}{(1 + \mu_1 \mu_2)}$

Rotational equilibrium about point A: $R_2 \times OB = W \times OD + \mu_2 R_2 \times OA$ $R_2 \times AB \cos\theta = W \times \frac{AB \cos\theta}{2} + \mu_2 R_2 \times AB \sin\theta$

This gives: $\cos\theta(R_2 - \frac{W}{2}) = \mu_2 R_2 \sin\theta \Rightarrow \tan\theta = \frac{1 - \mu_1 \mu_2}{2\mu_2}$

10. Types of Equilibrium

Stable equilibrium: Object returns to its original position if displaced slightly. Unstable equilibrium: Object moves farther away from its original position if displaced slightly. Neutral equilibrium: Object stays in its new position if displaced slightly.

⭐ Key Takeaways

For equilibrium of rigid bodies, both the sum of all external forces and the sum of all external torques must equal zero. Torque calculations give the same result regardless of the chosen reference point when translational equilibrium holds. The centre of gravity coincides with the centre of mass when gravitational acceleration is uniform. Practical problems involving beams, balances, and leaning rods are solved by applying force and torque balance simultaneously. Understanding the three types of equilibrium (stable, unstable, neutral) is essential for predicting the behavior of objects when displaced.

🧠 Quick Revision Questions

1. What are the two conditions for a rigid body to be in mechanical equilibrium?

2. In the beam-on-supports example, what are the expressions for $F_1$ and $F_2$ in terms of $M$, $m$, and $g$?

3. Why does the choice of origin not affect torque calculations for a body in equilibrium?

4. In the balance problem, what is the mass $m$ in terms of $m_1$ and $m_2$?

5. What is the difference between stable, unstable, and neutral equilibrium?

📘 Lecture 15 — OSCILLATIONS: I

📖 Overview: This lecture introduces the fundamental concepts of oscillatory motion, including period, frequency, and amplitude. It derives the equation of motion for a simple harmonic oscillator (SHO) using a mass-spring system, solves the differential equation, and explores energy conservation in oscillatory systems. Understanding oscillations is crucial because they appear throughout physics—from pendulums to molecular vibrations and electromagnetic waves.

🗂️ Topics Covered

The lecture covers the definition and characteristics of oscillations (period, frequency, amplitude), restoring forces and Hooke's Law, deriving the simple harmonic oscillator equation from Newton's second law, differentiating trigonometric functions to solve the SHO equation, interpreting the solution in terms of amplitude and phase, calculating energy (kinetic and potential) in oscillatory motion, speed as a function of displacement, and the effective spring constant for springs in parallel and series.

📝 Lecture Summary

1. Basic Characteristics of Oscillations

An oscillation is any self-repeating motion. This motion is characterized by three key quantities: the period ( T ), which is the time for completing one full cycle; the frequency ( f = 1/T ), which is the number of cycles per second (also denoted by ( \nu )); and the amplitude ( A ), which is the maximum displacement from equilibrium (or the size of the oscillation).

2. Why Systems Oscillate

A system oscillates because a force is always directed towards a central equilibrium position. In other words, the restoring force always acts to return the object to its equilibrium position. So the object will oscillate around the equilibrium position. The restoring force depends on the displacement: ( F_{\text{restore}} = -k\Delta x ), where ( \Delta x ) is the distance away from the equilibrium point, the negative sign shows that the force acts towards the equilibrium point, and ( k ) is a constant that gives the strength of the restoring force.

🔑 Definition — Restoring Force: A force that acts to return a system to its equilibrium position, proportional to displacement and opposite in direction.

3. The Simple Harmonic Oscillator (SHO) Equation

For a spring tied to a mass that can move freely over a frictionless surface: ( F(x) = -kx ) (where ( x ) is the extension). The energy stored in the spring is ( U(x) = \frac{1}{2}kx^2 ). Using Newton's second law: ( m\frac{d^2x}{dt^2} = -kx ), which rearranges to: [ \frac{d^2x}{dt^2} + \omega^2 x = 0 \quad \text{where} \quad \omega^2 \equiv \frac{k}{m} ] This is the equation of motion of a simple harmonic oscillator (SHO).

🔑 Definition — Simple Harmonic Oscillator (SHO): A system where the restoring force is directly proportional to displacement and acts in the opposite direction, described by ( \frac{d^2x}{dt^2} + \omega^2 x = 0 ).

💡 Why this matters: The SHO equation ( \frac{d^2x}{dt^2} + \omega^2 x = 0 ) appears throughout physics. Only the definition of ( \omega ) changes depending on the physical situation.

📐 Formula: ( \frac{d^2x}{dt^2} + \omega^2 x = 0 ) → The acceleration is proportional to negative displacement, producing oscillatory motion.

4. Differentiating Trigonometric Functions

To solve the SHO equation, we first learn to differentiate trigonometric functions. Starting with the derivative of ( \cos\omega t ): [ \frac{d}{dt}\cos\omega t = -\omega\sin\omega t ] (Using ( \sin\theta \approx \theta ) for small ( \theta )). And the derivative of ( \sin\omega t ): [ \frac{d}{dt}\sin\omega t = \omega\cos\omega t ] (Using ( \cos\theta \approx 1 ) for small ( \theta )).

📐 Formula: ( \frac{d}{dt}\cos\omega t = -\omega\sin\omega t ) → The derivative of cosine is negative sine times omega. 📐 Formula: ( \frac{d}{dt}\sin\omega t = \omega\cos\omega t ) → The derivative of sine is cosine times omega.

5. Second Derivatives

Differentiating twice gives back the original function with a negative factor: [ \frac{d^2}{dt^2}(\sin\omega t) = \frac{d}{dt}(\omega\cos\omega t) = -\omega^2\sin\omega t ] [ \frac{d^2}{dt^2}(\cos\omega t) = \frac{d}{dt}(-\omega\sin\omega t) = -\omega^2\cos\omega t ] So twice differentiating either ( \sin\omega t ) or ( \cos\omega t ) gives the same function back!

📐 Formula: ( \frac{d^2}{dt^2}(\sin\omega t) = -\omega^2\sin\omega t ) and ( \frac{d^2}{dt^2}(\cos\omega t) = -\omega^2\cos\omega t ).

6. Solution of the SHO Equation

Any function of the form ( x(t) = a\cos\omega t + b\sin\omega t ) satisfies ( \frac{d^2x}{dt^2} = -\omega^2 x ).

Interpretation of ( \omega ), ( a ), and ( b ):

a) Significance of ( \omega ): If we replace ( t ) by ( t + \frac{2\pi}{\omega} ) in either ( \sin\omega t ) or ( \cos\omega t ), the function repeats itself. So ( \frac{2\pi}{\omega} ) is the period ( T ): [ T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}} ] The frequency ( \nu = 1/T = \frac{1}{2\pi}\sqrt{\frac{k}{m}} ), so ( \omega = 2\pi\nu = \sqrt{\frac{k}{m}} ). ( \omega ) is called the angular frequency.

🔑 Definition — Angular Frequency (( \omega )): The rate of change of phase per unit time, measured in radians/second, related to frequency by ( \omega = 2\pi\nu ).

📐 Formula: ( T = 2\pi\sqrt{\frac{m}{k}} ) → Period depends on mass and spring constant but NOT on amplitude.

b) Meaning of ( a ) and ( b ): From ( x(t) = a\cos\omega t + b\sin\omega t ):

• At ( t=0 ): ( x(0) = a ) — so ( a ) is the initial position
• ( \frac{dx}{dt} = -\omega a\sin\omega t + \omega b\cos\omega t ), so at ( t=0 ): ( \frac{dx}{dt} = \omega b ) — so ( b ) is the initial velocity divided by ( \omega )

c) Amplitude form: The solution can also be written as: [ x(t) = x_m \cos(\omega t + \phi) ] where ( x_m ) is the amplitude (maximum displacement), and ( \phi ) is the phase constant. Since ( -x_m \leq x \leq +x_m ), ( x_m ) is called the amplitude of the motion. The frequency of simple harmonic motion is independent of the amplitude.

d) The quantity ( \theta = \omega t + \phi ) is called the phase of the motion. A different value of ( \phi ) just means the origin of time has been chosen differently.

🔑 Definition — Amplitude (( x_m )): The maximum displacement from equilibrium in oscillatory motion. 🔑 Definition — Phase (( \omega t + \phi )): The argument of the sine or cosine function in oscillatory motion, determining the state of the oscillator at any time ( t ). 🔑 Definition — Phase Constant (( \phi )): The initial phase at ( t=0 ), determining the starting point in the oscillation cycle.

7. Energy of Simple Harmonic Motion

With ( \phi = 0 ) (so ( x = x_m \cos\omega t )):

Potential Energy: [ U = \frac{1}{2}kx^2 = \frac{1}{2}kx_m^2 \cos^2\omega t ]

Kinetic Energy: [ K = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{dx}{dt}\right)^2 = \frac{1}{2}m\omega^2 x_m^2 \sin^2\omega t = \frac{1}{2}kx_m^2 \sin^2\omega t ]

Total Energy (sum of potential + kinetic): [ E = K + U = \frac{1}{2}kx_m^2 (\cos^2\omega t + \sin^2\omega t) = \frac{1}{2}kx_m^2 ]

The total energy is independent of time and energy goes from kinetic to potential, back to kinetic, etc.

📐 Formula: ( E = \frac{1}{2}kx_m^2 = \frac{1}{2}m\omega^2 x_m^2 ) → Total energy is constant and proportional to the square of the amplitude.

💡 Why this matters: Energy conservation in SHM means the oscillator continuously converts between kinetic and potential energy without loss.

8. Speed as a Function of Displacement

From energy conservation: velocity ( v = \frac{dx}{dt} = \pm\sqrt{\frac{k}{m}(x_m^2 - x^2)} )

• Speed is maximum at ( x = 0 ) (equilibrium position)
• Speed is zero at ( x = \pm x_m ) (extreme positions)

📐 Formula: ( v = \pm\sqrt{\frac{k}{m}(x_m^2 - x^2)} ) → Speed depends on position, being maximum at equilibrium and zero at maximum displacement.

9. Springs in Parallel and Series

Parallel combination: Makes it harder to stretch them. The effective spring constant is: [ k_{\text{eff}} = k_1 + k_2 ]

Series combination: Makes them easier to stretch. The effective spring constant is: [ k_{\text{eff}} = \left(\frac{1}{k_1} + \frac{1}{k_2}\right)^{-1} ]

So a mass will oscillate faster in the parallel case compared to the series case.

📐 Formula (Parallel): ( k_{\text{eff}} = k_1 + k_2 ) → Stiffer spring, higher frequency. 📐 Formula (Series): ( k_{\text{eff}} = \left(\frac{1}{k_1} + \frac{1}{k_2}\right)^{-1} ) → Softer spring, lower frequency.

⭐ Key Takeaways

The simple harmonic oscillator is the fundamental model for all periodic motion, characterized by a restoring force proportional to displacement (( F = -kx )). Its equation of motion is ( \frac{d^2x}{dt^2} + \omega^2 x = 0 ) with solution ( x(t) = x_m \cos(\omega t + \phi) ), where ( \omega = \sqrt{k/m} ). Critically, the period ( T = 2\pi\sqrt{m/k} ) depends only on mass and spring constant, not on amplitude. The total mechanical energy ( E = \frac{1}{2}kx_m^2 ) remains constant, oscillating between kinetic and potential forms. Finally, springs in parallel produce a stiffer system (higher ( k ), faster oscillation) while springs in series produce a softer system (lower ( k ), slower oscillation).

🧠 Quick Revision Questions

1. What are the three basic characteristics that describe any oscillation?
2. Write down the differential equation for a simple harmonic oscillator and define the angular frequency ( \omega ).
3. Show that ( x(t) = x_m \cos(\omega t + \phi) ) satisfies the SHO equation.
4. Derive expressions for kinetic energy, potential energy, and total energy in SHM. Prove that total energy is constant.
5. Two identical springs of spring constant ( k ) are connected (a) in parallel and (b) in series. Find the effective spring constant in each case and state which combination makes a mass oscillate faster.

📘 Lecture 16 — Oscillations: II

📖 Overview: This lecture extends the concept of simple harmonic motion (SHO) to physical pendulums, damped oscillations, and forced oscillations. It explores how real systems behave under damping and driving forces, culminating in the critical concept of resonance, which has profound applications across physics and engineering.

🗂️ Topics Covered

The lecture begins by revisiting the simple pendulum and showing it approximates an SHO for small angles. It then generalizes to a physical pendulum, introducing the moment of inertia and center of gyration. Next, it covers the superposition of two SHMs along the same line, leading to constructive and destructive interference. This is followed by the composition of perpendicular SHMs, yielding elliptical and Lissajous figures. Finally, it introduces damped harmonic motion and forced oscillations with resonance.

📝 Lecture Summary

1. The Simple Pendulum Revisited

For a mass suspended from a string, the restoring force is ( F = -mg\sin\theta ). For small (\theta), (\sin\theta \approx \theta), and using ( x = L\theta ) we get ( F = -\left(\frac{mg}{L}\right)x ). This is proportional to displacement, so it is an SHO with angular frequency ( \omega = \sqrt{g/L} ). Even without the small-angle approximation, the motion is periodic but mathematically complex.

🔑 Definition — Simple Harmonic Oscillator (SHO): A system where the restoring force is directly proportional to displacement from equilibrium. 📐 Formula: ( \omega = \sqrt{g/L} ) → The angular frequency of a simple pendulum depends only on length and gravity. 📌 Example: For a pendulum of length ( L = 1.0\ \text{m} ) and ( g = 9.8\ \text{m/s}^2 ), ( \omega = \sqrt{9.8/1.0} = 3.13\ \text{rad/s} ).

2. The Physical Pendulum

For an extended object pivoted at a point, torque is ( \tau = -Mgd\sin\theta ). For small (\theta), (\tau = -Mgd\theta). Using ( \tau = I\alpha ) and ( \alpha = d^2\theta/dt^2 ), we get ( I\frac{d^2\theta}{dt^2} = -Mgd\theta ) or ( \frac{d^2\theta}{dt^2} = -\left(\frac{Mgd}{I}\right)\theta ). Hence the oscillation frequency is ( \omega = \sqrt{Mgd/I} ). If the pivot is at the center of mass, torque vanishes and the object does not oscillate.

🔑 Definition — Physical Pendulum: Any rigid body pivoted so it can swing freely under gravity. 📐 Formula: ( \omega = \sqrt{Mgd/I} ) → Frequency depends on mass, gravity, distance from pivot to cm, and moment of inertia. 📌 Example: For a uniform rod of length ( L ) pivoted at one end, ( d = L/2 ) and ( I = \frac{1}{3}ML^2 ), so ( \omega = \sqrt{\frac{Mg(L/2)}{(1/3)ML^2}} = \sqrt{\frac{3g}{2L}} ).

3. Centre of Gyration

If a physical pendulum is pivoted at a point P at distance ( L ) from the cm, and we want it to behave like a simple pendulum, we set ( T = 2\pi\sqrt{L/g} = 2\pi\sqrt{I/(Mgd)} ). Solving gives ( L = I/(Md) ). This point P is called the centre of gyration; from this pivot, the mass appears concentrated at the cm.

🔑 Definition — Centre of Gyration: The point on a physical pendulum from which, if suspended, it oscillates as if all mass were at the cm. 📐 Formula: ( L = I/(Md) ) → Equivalent simple pendulum length. 📌 Example: For the rod above, ( I = \frac{1}{3}ML^2 ), ( d = L/2 ), so ( L = \frac{(1/3)ML^2}{M(L/2)} = \frac{2}{3}L ).

4. Sum of Two SHMs Along the Same Line

Given ( x_1 = A_1\sin\omega t ) and ( x_2 = A_2\sin(\omega t + \phi) ), their sum is ( x = R\sin(\omega t + \theta) ) where ( R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi} ) and ( \tan\theta = \frac{A_2\sin\phi}{A_1 + A_2\cos\phi} ). For ( \phi = 0 ), ( R = A_1 + A_2 ) (constructive interference). For ( \phi = \pi ), ( R = |A_1 - A_2| ) (destructive interference).

🔑 Definition — Superposition: Adding two waves of same frequency yields a wave with amplitude dependent on phase difference. 📐 Formula: ( R = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos\phi} ) 📌 Example: If ( A_1 = 3\ \text{cm} ), ( A_2 = 4\ \text{cm} ), ( \phi = 0 ), then ( R = 3+4 = 7\ \text{cm} ); if ( \phi = \pi ), ( R = |3-4| = 1\ \text{cm} ).

5. Composition of Two Perpendicular SHMs

For ( x = A\sin\omega t ) and ( y = B\sin(\omega t + \phi) ), eliminating ( t ) gives the ellipse equation: ( \frac{x^2}{A^2} + \frac{y^2}{B^2} - \frac{2xy}{AB}\cos\phi = \sin^2\phi ). This is generally an ellipse; for special (\phi), it becomes a line or circle.

🔑 Definition — Lissajous Figures: Curves formed by combining perpendicular oscillations; periodic only if frequency ratio is rational. 📐 Formula: ( \frac{x^2}{A^2} + \frac{y^2}{B^2} - \frac{2xy}{AB}\cos\phi = \sin^2\phi )

6. Damped Harmonic Motion

Damping force is ( -b(dx/dt) ). Newton's law gives ( m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = 0 ). For weak damping (( k/m \geq (b/(2m))^2 )), the solution is ( x = x_0 e^{-bt/(2m)} \cos(\omega't + \phi) ) where ( \omega' = \sqrt{k/m - (b/(2m))^2} ). Amplitude decays to ( 1/e ) of initial value when ( bt/(2m) = 1 ).

🔑 Definition — Damped Harmonic Motion: Oscillation with amplitude decreasing over time due to a dissipative force. 📐 Formula: ( \omega' = \sqrt{k/m - (b/(2m))^2} ) 📌 Example: For a mass ( m = 0.2\ \text{kg} ), spring constant ( k = 5\ \text{N/m} ), damping constant ( b = 0.1\ \text{kg/s} ), undamped ( \omega_0 = \sqrt{5/0.2} = 5\ \text{rad/s} ), damped ( \omega' = \sqrt{25 - (0.1/(0.4))^2} = \sqrt{25 - 0.0625} \approx 4.99\ \text{rad/s} ).

💡 Why this matters: Damping explains why real oscillations eventually stop, crucial for shock absorbers, musical instruments, and building stability.

7. Forced Oscillation and Resonance

For a driving force ( F_0\cos\omega t ), the equation is ( m\frac{d^2x}{dt^2} + kx = F_0\cos\omega t ). The solution is ( x = \frac{F_0}{m(\omega_0^2 - \omega^2)}\cos\omega t ), where ( \omega_0 = \sqrt{k/m} ) is the natural frequency. Amplitude "blows up" (goes to infinity) as ( \omega \to \omega_0 ) in the absence of damping. With damping, amplitude remains finite but is maximized when driving frequency equals natural frequency—this is resonance.

🔑 Definition — Resonance: Condition where a driving frequency matches the natural frequency, maximizing oscillation amplitude. 📐 Formula: Amplitude = ( \frac{F_0}{m(\omega_0^2 - \omega^2)} ) (undamped) 📌 Example: A child's swing (( \omega_0 \approx 1\ \text{rad/s} )) pushed at the same frequency results in large amplitude swings.

💡 Why this matters: Resonance can be destructive (e.g., Tacoma Narrows Bridge collapse) or useful (e.g., tuning a radio, MRI imaging).

⭐ Key Takeaways

The simple pendulum approximates an SHO only for small angles, with frequency ( \sqrt{g/L} ). A physical pendulum's frequency ( \sqrt{Mgd/I} ) generalizes this, introducing the center of gyration. Superposition of two SHMs produces interference—constructive when in phase, destructive when out of phase. Perpendicular SHMs combine into elliptical or Lissajous patterns. Damping reduces amplitude over time, described by an exponential decay factor ( e^{-bt/(2m)} ). Forced oscillations exhibit resonance, where amplitude peaks dramatically when the driving frequency matches the natural frequency—a critical phenomenon with vast practical implications.

🧠 Quick Revision Questions

1. What is the angular frequency of a simple pendulum of length 0.5 m on Earth?
2. For a physical pendulum, what happens to the oscillation if the pivot is at the center of mass?
3. Two SHMs of amplitudes 2 cm and 5 cm, same frequency, have a phase difference of 60°. What is the resultant amplitude?
4. In damped harmonic motion, what does the damping constant ( b ) represent physically?
5. What condition must the driving frequency satisfy for resonance to occur in an undamped forced oscillator?

📘 Lecture 17 — PHYSICS OF MATERIALS

📖 Overview: This lecture introduces the fundamental concepts of elasticity and plasticity in materials, covering how materials deform under various types of forces. It explains the key parameters that describe material behavior, including stress, strain, and elastic moduli, and extends these principles to fluids, pressure, and Pascal's Principle, which are essential for understanding material science and fluid mechanics.

🗂️ Topics Covered

The lecture covers elasticity and plasticity, definitions of stress (longitudinal, volume, and shearing) and strain (longitudinal, volume, and shearing), Hooke's Law and the three moduli of elasticity (Young's, Bulk, and Shear), Poisson's ratio, the work done in stretching a wire, the fundamental properties of fluids, the concept of pressure and density, the variation of pressure with depth in a fluid, and Pascal's Principle.

📝 Lecture Summary

1. Elasticity

If a body completely recovers its original shape and size when external forces are removed, it is called perfectly elastic. Quartz, steel and glass are very nearly elastic.

🔑 Definition — Elasticity: the property by virtue of which a body tends to regain its original shape and size when deforming forces are removed.

2. Plasticity

If a body has no tendency to regain its original shape and size, it is called perfectly plastic. Common plastics, kneaded dough, solid honey, etc are plastics.

3. Stress

Stress characterizes the strength of the forces causing the stretch, squeeze, or twist. It is defined usually as force/unit area but may have different definitions to suit different situations. We distinguish between three types of stresses: a) If the deforming force is applied along some linear dimension of a body, the stress is called longitudinal stress or tensile stress or compressive stress. b) If the force acts normally and uniformly from all sides of a body, the stress is called volume stress. c) If the force is applied tangentially to one face of a rectangular body, keeping the other face fixed, the stress is called tangential or shearing stress.

4. Strain

Strain: When deforming forces are applied on a body, it undergoes a change in shape or size. The fractional (or relative) change in shape or size is called the strain. Strain = change / original dimension. Strain is a ratio of similar quantities so it has no units. There are 3 different kinds of strain: a) Longitudinal (linear) strain is the ratio of the change in length (ΔL) to original length (l), i.e., the linear strain = Δl / l. b) Volume strain is the ratio of the change in volume (ΔV) to original volume (V). Volume strain = ΔV / V. c) Shearing strain: For small θ, the angular deformation (θ) in radians is called shearing strain. As shown in the diagram, θ ≈ tan θ = Δx / l, so shearing strain = θ = Δx / l.

5. Hooke's Law

Hooke's Law: for small deformations, stress is proportional to strain. Stress = E × Strain. The constant E is called the modulus of elasticity. E has the same units as stress because strain is dimensionless. There are three moduli of elasticity. (a) Young's modulus (Y) for linear strain: Y = longitudinal stress / longitudinal strain = (F/A) / (Δl/l). (b) Bulk Modulus (B) for volume strain: Let a body of volume V be subjected to a uniform pressure ΔP on its entire surface and let ΔV be the corresponding decrease in its volume. Then, B = Volume Stress / Volume Strain = - ΔP / (ΔV/V). 1/B is called the compressibility. A material having a small value of B can be compressed easily. Its value lies between 0 and 0.5. (c) Shear Modulus (η) for shearing strain: Let a force F produce a strain θ. Then, η = shearing stress / shearing strain = (F/A) / θ = F / (A * tan θ) = F / (A * Δx/l).

6. Poisson's Ratio

When a wire is stretched, its length increases and radius decreases. The ratio of the lateral strain to the longitudinal strain is called Poisson's ratio, σ = (Δr / r) / (Δl / l).

7. Work Done in Stretching a Wire

We can calculate the work done in stretching a wire. Obviously, we must do work against a force. If x is the extension produced by the force F in a wire of length l, then F = (YA/l) x. The work done in extending the wire through Δl is given by, W = ∫₀^(Δl) F dx = ∫₀^(Δl) (YA/l) x dx = (YA/l) (Δl²/2) = (1/2) * Volume * Stress * Strain. We can also write this as, W = 1/2 * Load * Extension.

📌 Example: The work done in stretching a wire is calculated by integrating the force from 0 to the final extension Δl. Since force F = (YA/l)x, the integral yields W = (YA/l)(Δl²/2).

8. Fluids

A fluid is a substance that can flow and does not have a shape of its own. Thus all liquids and gases are fluids. Solids possess all the three moduli of elasticity whereas a fluid possess only the bulk modulus (B). A fluid at rest cannot sustain a tangential force. If such force is applied to a fluid, the different layers simply slide over one another. Therefore the forces acting on a fluid at rest have to be normal to the surface. This implies that the free surface of a liquid at rest, under gravity, in a container, is horizontal.

9. Pressure and Density

The normal force per unit area is called pressure, P = ΔF / ΔA. Its unit is Newtons/metre², or Pascal (Pa). Another scalar is density, ρ = Δm / ΔV, where Δm is the mass of a small piece of the material and ΔV is the volume it occupies. Pressure is a scalar quantity.

10. Variation of Pressure with Depth

Let us calculate how the pressure in a fluid changes with depth. Take a small element of fluid volume submerged within the body of the fluid: dm = ρ dV = ρ A dy, so (dm)g = ρgA dy. Now let us require that the sum of the forces on the fluid element is zero: pA - (p+dp)A - ρgAdy = 0 ⇒ dp/dy = -ρg. Note that we are taking the origin (y=0) at the bottom of the liquid. Therefore as the elevation increases (dy positive), the pressure decreases (dp negative). The quantity ρg is the weight per unit volume of the fluid. For liquids, which are nearly incompressible, ρ is practically constant. ⇒ dp/dy = Δp/Δy = (p₂ - p₁) / (y₂ - y₁) = -ρg ⇒ p₂ - p₁ = -ρg(y₂ - y₁).

📐 Formula: dp/dy = -ρg → The rate of change of pressure with height is equal to the negative of the fluid's weight per unit volume.

11. Pascal's Principle

Pascal's Principle: Pressure applied to an enclosed fluid is transmitted to every portion of the fluid and to the walls of the containing vessel. This follows from the above: if h is the height below the liquid's surface, then p = p_ext + ρgh. Here p_ext is the pressure at the surface of the liquid, and so the difference in pressure is Δp = Δp_ext + Δ(ρgh). Therefore, Δ(ρgh) = 0 ⇒ Δp = Δp_ext (because liquids are incompressible). So the pressure is everywhere the same. The pressure in liquids is everywhere the same.

💡 Why this matters: Pascal's Principle is fundamental to the operation of hydraulic systems, such as car brakes and lifts, where a small force applied at one point creates a large force at another.

⭐ Key Takeaways

You must understand the distinction between elasticity and plasticity and memorize the three types of stress and strain. Hooke's Law with its three moduli (Young's, Bulk, and Shear) is critical, including their formulas and physical meanings, and you should know that Poisson's ratio relates lateral and longitudinal strain. The work done in stretching a wire can be expressed as half the product of load and extension, or half the product of volume, stress, and strain. Finally, for fluids, pressure increases linearly with depth (dp/dy = -ρg), and Pascal's Principle states that pressure is transmitted undiminished throughout an enclosed fluid.

🧠 Quick Revision Questions

1. What is the definition of stress, and what are the three types of stress described in the lecture?
2. State Hooke's Law and write the formula for Young's modulus, Bulk modulus, and Shear modulus.
3. A fluid at rest cannot sustain which type of force, and what is the consequence of this for its free surface?
4. Derive the equation dp/dy = -ρg and explain what happens to pressure as you increase in elevation within a fluid.
5. State Pascal's Principle and explain why the pressure is the same everywhere in an enclosed liquid.

📘 Lecture 18 — Physics of Fluids

📖 Overview: This lecture introduces the physics of fluids, covering surface tension, pressure differences in bubbles and drops, and the principles of fluid flow including the continuity equation and Bernoulli's equation. It explains fundamental concepts that are essential for understanding everyday phenomena like liquid surfaces, soap bubbles, and aircraft lift.

🗂️ Topics Covered

The lecture begins with the definition of fluids and molecular interactions, then explores surface tension as a property of liquids, its measurement using a sliding wire apparatus, and its representation as potential energy per unit area. It then examines pressure differences in soap bubbles and liquid drops due to surface tension. The discussion transitions to fluid dynamics with the continuity equation for incompressible flow and concludes with Bernoulli's equation, including its application to flow in pipes and the principle of aircraft lift.

📝 Lecture Summary

1. A fluid is matter that has no definite shape

A fluid, including both gases and liquids, adjusts to the shape of its container. All fluids are composed of molecules that attract each other.

2. Liquids exhibit surface tension

A liquid's free surface tends to contract to the minimum possible area, indicating a state of tension. This arises from cohesive forces, the attractive forces between molecules of the liquid. Deep inside the liquid, a molecule experiences equal forces from all directions, resulting in no net force. At the surface, a molecule is surrounded by only half as many molecules (since there are none above), leading to a net inward force.

3. The force of contraction

The surface tension, denoted by γ (gamma), is the force of contraction per unit length acting at right angles to an imaginary line on the liquid's surface. It is defined as γ = F / L, where F is the force exerted by the liquid's "skin" along a line of length L. The SI unit of surface tension is N/m.

4. Quantitative measurement of surface tension

In a setup with a soap film on a wire frame and a sliding wire of weight w, the force T pulls the wire downward. The net downward force is F = w + T. As the film has both a front and back surface, this force acts along a total length of 2L. Therefore, surface tension is: γ = F / (2L) ⇒ F = 2γL Hence, the surface tension is γ = (w + T) / (2L).

5. Work done to stretch the skin of a liquid

When the sliding wire is moved by a displacement Δx, the work done is FΔx. This is a conservative force, so the change in potential energy is ΔU = FΔx = 2γLΔx. The product LΔx equals the change in surface area, ΔA. Thus, γ = ΔU / ΔA. This demonstrates that surface tension is the surface potential energy per unit area.

7. Surface tension and pressure differences

Surface tension causes a pressure difference between the inside and outside of a soap bubble or a liquid drop. A soap bubble has two spherical surfaces with a thin liquid layer between. For a bubble of radius r, the net force due to surface tension is 2(2πrγ). In equilibrium, the forces are equal, leading to an excess pressure of p – p₀ = 4γ / r. For a liquid drop, which has only one surface, the excess pressure is p – p₀ = 2γ / r.

💡 Why this matters: This explains why smaller soap bubbles have higher internal pressure than larger ones.

8. Continuity Equation

Since liquids are incompressible, equal volumes must flow through different sections of a pipe in the same time. This gives V₁ = V₂. As volume V = A * L = A * v * t, we have A₁v₁t = A₂v₂t, simplifying to A₁v₁ = A₂v₂. This means the fluid flows faster where the tube is narrower and slower where it is wider.

9. Bernoulli's Equation

For a steadily flowing fluid, Bernoulli's equation states: p + (1/2)ρv² + ρgy = constant This equation relates pressure (p), kinetic energy per unit volume ((1/2)ρv²), and gravitational potential energy per unit volume (ρgy).

🔑 Definition — Bernoulli's Equation: The sum of the pressure, kinetic energy per unit volume, and gravitational potential energy per unit volume is constant along a streamline for an ideal fluid.

📐 Formula: p₁ + (1/2)ρv₁² + ρgy₁ = p₂ + (1/2)ρv₂² + ρgy₂ → The pressure energy plus kinetic energy plus potential energy per unit volume is constant.

10. Application of Bernoulli's Equation

For water flowing in a horizontal pipe where the cross-section decreases (y₁ = y₂), Bernoulli's equation simplifies to: p₁ + (1/2)ρv₁² = p₂ + (1/2)ρv₂² Using the continuity equation A₁v₁ = A₂v₂, we find: v₂ = (A₁/A₂)v₁ and therefore: p₂ = p₁ – (1/2)ρ(v₂² – v₁²)

This shows pressure is smaller where the fluid flows faster.

💡 Why this matters: This principle explains how aircraft achieve lift. The wing shape is curved, causing air to move faster over the top than the bottom. The lower pressure on top and higher pressure below creates a net upward force called lift.

📌 Example – Aircraft Lift: Air moves faster over the curved top of an aircraft wing than the flatter bottom. According to Bernoulli's principle, the faster-moving air on top exerts less pressure than the slower air below. This pressure difference results in a net upward force, lifting the aircraft.

⭐ Key Takeaways

Surface tension is a property of liquids that causes the surface to contract, defined as force per unit length and also as surface potential energy per unit area. The pressure inside a soap bubble (4γ/r) is double that inside a liquid drop (2γ/r) due to the bubble having two surfaces. For incompressible fluid flow, the continuity equation A₁v₁ = A₂v₂ states that flow speed increases as the tube's cross-sectional area decreases. Bernoulli's equation, p + (1/2)ρv² + ρgy = constant, is a statement of energy conservation for flowing fluids, explaining that where flow speed is high, pressure is low, which is the fundamental principle behind aircraft lift.

🧠 Quick Revision Questions

1. What is the definition of surface tension, and what are its two equivalent forms?
2. Why does the pressure inside a soap bubble differ from that inside a liquid drop of the same radius?
3. A liquid flows from a wide pipe of diameter 4 cm into a narrow pipe of diameter 2 cm. If the speed in the wide pipe is 1 m/s, what is the speed in the narrow pipe? (Assume incompressible flow)
4. State Bernoulli's equation for an ideal fluid and explain the physical meaning of each term.
5. How does Bernoulli's principle explain the generation of lift on an aircraft wing?

📘 Lecture 19 — PHYSICS OF SOUND

📖 Overview: This lecture introduces the physics of sound waves, which are longitudinal oscillations of density that carry energy. It covers how sound intensity is measured in decibels, the mathematical description of sound waves, and the Doppler Effect, which explains why the frequency of sound changes when the source or observer is in motion.

🗂️ Topics Covered

Sound waves as longitudinal density oscillations with measurable intensity using decibels; the mathematical wave equation for density including wavelength, time period, frequency, wavenumber, and angular frequency; and the Doppler Effect with three cases: moving observer with source at rest, moving source with observer at rest, and both moving simultaneously.

📝 Lecture Summary

1. Sound Waves and Intensity

Sound waves correspond to longitudinal oscillations of density. As sound waves move, you find the density of air greater in some places (compressions) and less in others (rarefactions). Sound waves carry energy. The minimum energy that humans can hear is about 10⁻¹² watts per cm³, which is called the threshold of hearing (I₀).

💡 Why this matters: Understanding that sound is a density wave helps explain why you can hear around corners (diffraction) and why sound travels at different speeds through different materials.

2. Decibels (Intensity Measurement)

To measure the intensity of sound, we use decibels (db) as the unit. Decibels are a relative measure to compare the intensity of different sounds with one another.

🔑 Definition — Intensity Level (R) in decibels: R = 10 log₁₀ (I / I₀) where I is the intensity of the sound and I₀ is the threshold of hearing (10⁻¹² W/cm³).

📌 Example: On a street without traffic, the sound level is about 30 db. A pressure horn creates about 90 db. Serious ear damage happens around 120 db.

3. Mathematical Description of a Sound Wave

A sound wave moving in the x-direction with speed v is described by: ρ(x, t) = ρₘ sin [2π/λ (x − vt)] where ρ(x, t) is the density of air at a point x at time t.

a) Constant Phase Motion: Suppose as time t increases, we move in such a way as to keep (x − vt) constant. To keep the density ρ(x, t) constant, we would have to move with the speed of sound, i.e., v.

b) Wavelength (λ): In the expression for ρ(x, t), replace x by x + λ. Nothing changes because sin[2π/λ (x + λ − vt)] = sin[2π/λ (x − vt)]. This is why we call λ the wavelength, meaning the length after which a wave repeats itself.

c) Time Period (T) and Frequency (ν): Replace t by t + T where T = λ/v. Nothing changes. T is called the time period of the sound wave, meaning the time after which it repeats itself. The frequency is the number of cycles per second and is related to T through ν = 1/T.

d) Wavenumber (k) and Angular Frequency (ω):

🔑 Definition — Wavenumber: k = 2π/λ

🔑 Definition — Angular Frequency: ω = 2π/T = 2πν

4. Doppler Effect

The Doppler Effect describes how relative motion between source and observer causes the observer to receive a frequency different from that emitted by the source. One must distinguish between two cases.

Case 1: Moving Observer, Source at Rest: If the observer was at rest, the number of waves received in time t would be t/T (or vt/λ). If she is moving towards the source with speed v₀, the additional number of waves received is v₀t/λ.

🔑 Definition — Doppler Formula (Moving Observer): ν' = (v + v₀)/λ = (v + v₀)/(v/ν) = ν (v + v₀)/v

Conclusion: As the observer runs towards the source, she hears a higher frequency (higher pitch).

Case 2: Moving Source, Observer at Rest: As the source runs towards the observer, more waves must be packed together. Each wavelength is reduced by vₛ/ν.

🔑 Definition — Doppler Formula (Moving Source): The wavelength seen by the observer is λ' = (v − vₛ)/ν. The frequency heard is ν' = v/λ' = v / [(v − vₛ)/ν] = ν [v / (v − vₛ)]

Case 3: Moving Source and Moving Observer: The general formula combines both effects.

🔑 Definition — General Doppler Formula: ν' = ν [(v + vₒ) / (v − vₛ)]

The above two results are special cases of this formula.

⭐ Key Takeaways

Sound waves are longitudinal density oscillations measurable in decibels using a logarithmic scale relative to the threshold of hearing. The mathematical wave equation uses wavelength, time period, frequency, wavenumber, and angular frequency. The Doppler Effect formula changes depending on whether the observer, source, or both are moving; the general formula is ν' = ν[(v + vₒ)/(v − vₛ)], where the observer moving toward the source and the source moving toward the observer both increase the observed frequency. Understanding decibels is crucial for hearing safety (120 db causes ear damage).

🧠 Quick Revision Questions

1. What is the threshold of hearing in watts per cm³, and what is it called?
2. If a sound has an intensity of 10⁻⁶ W/cm³, what is its intensity level in decibels?
3. What happens to the density wave if you replace x by x + λ in the wave equation ρ(x, t) = ρₘ sin[2π/λ (x − vt)]?
4. A police siren emitting 800 Hz is moving toward you at 30 m/s. If the speed of sound is 340 m/s, what frequency do you hear?
5. If both source and observer are moving, which formula correctly gives the observed frequency?

📘 Lecture 20 — WAVE MOTION

📖 Overview: This lecture explores the fundamental principles of wave motion, which is any self-repeating periodic motion that transports energy from one point to another. It covers the classification of waves, their mathematical representation, the concept of phase, and how waves superpose and interfere. Understanding wave motion is crucial because it explains phenomena from sound and light to earthquakes.

🗂️ Topics Covered

The lecture begins by defining and classifying wave motion into longitudinal and transverse waves, emphasizing that waves transport energy without transporting matter. It then explains amplitude and its relation to power, including the inverse square law for spherical waves. The mathematical representation of a wave with a phase constant is introduced, followed by a detailed analysis of interference from two sources. The formula for the speed of sound in a medium is stated, and finally, the concept of a pulse and its propagation speed (phase velocity) is derived.

📝 Lecture Summary

1. Wave Motion

Wave motion is any kind of self-repeating (periodic, or oscillatory) motion that transports energy from one point to another. Waves are of two basic kinds:

• Longitudinal Waves: the oscillation is parallel to the direction of wave travel. Examples: sound, spring, "P-type" earthquake waves.
• Transverse Waves: the oscillation is perpendicular to the direction of wave travel. Examples: radio or light waves, string, "S-type" earthquake waves.

2. Waves Transport Energy, Not Matter

Taking the vibration of a string as an example, each segment of the string stays in the same place, but the work done on the string at one end is transmitted to the other end. Work is done in lifting the mass at the other end. 💡 Why this matters: This fundamental principle distinguishes wave motion from the bulk motion of matter.

3. Amplitude and Power

The height of a wave is called the amplitude. The average power (or intensity) in a wave is proportional to the square of the amplitude. So if ( a(t) = a_0 \sin(\omega t - kx) ) is a wave of some kind, then ( a_0 ) is the amplitude and ( I \propto a_0^2 ).

4. Spherical Waves and the Inverse Square Law

A sound source placed at the origin will radiate sound waves in all directions equally. These are called spherical waves. For spherical waves, the amplitude ( \propto \frac{1}{r} ) and so the power ( \propto \frac{1}{r^2} ).

Consider a source of sound and draw two spheres. Let ( P_1 ) be the total radiated power and ( I_1 ) the intensity at ( r_1 ), etc. All the power (and energy) that crosses ( r_1 ) also crosses ( r_2 ) since none is lost in between. We have that, ( 4 \pi r_1^2 I_1 = P_1 ) and ( 4 \pi r_2^2 I_2 = P_2 ). But ( P_1 = P_2 = P ), and so ( \frac{I_1}{I_2} = \frac{r_2^2}{r_1^2} ) or ( I \propto \frac{1}{r^2} ).

🔑 Definition — Spherical Waves: Waves that radiate outwards equally in all directions from a point source.

📐 Formula: ( \frac{I_1}{I_2} = \frac{r_2^2}{r_1^2} ) → The intensity of a spherical wave is inversely proportional to the square of the distance from the source.

5. The Phase Constant

We have encountered waves of the kind ( y(x,t) = y_0 \sin(kx - \omega t) ) in the previous lecture. Obviously ( y(0,0) = 0 ). But what if the wave is not zero at ( x = 0, t = 0 )? Then it could be represented by ( y(x,t) = y_m \sin(kx - \omega t - \phi) ), where ( kx - \omega t - \phi ) is called the phase and ( \phi ) is called the phase constant.

Note that you can rewrite ( y(x,t) ) either as: a) ( y(x,t) = y_m \sin \left[ k \left( x - \frac{\phi}{k} \right) - \omega t \right] ), or as b) ( y(x,t) = y_m \sin \left[ kx - \omega \left( t + \frac{\phi}{\omega} \right) \right] ).

The two different ways of writing the same expression can be interpreted differently. In (a) ( x ) has effectively been shifted to ( x - \frac{\phi}{k} ) whereas in (b) ( t ) has been shifted to ( t + \frac{\phi}{\omega} ). So the phase constant only moves the wave forward or backward in space or time.

6. Interference of Two Waves

When two sources are present, the total amplitude at any point is the sum of the two separate amplitudes, ( y(x,t) = y_1(x,t) + y_2(x,t) ). The power is proportional to the square of the amplitude, so ( P \propto (y_1 + y_2)^2 ). This is why interference happens. Suppose the two waves have equal amplitude. Let the two waves be: ( y_1(x,t) = y_m \sin(kx - \omega t - \phi_1) ) and ( y_2(x,t) = y_m \sin(kx - \omega t - \phi_2) )

The total amplitude is: ( y(x,t) = y_1(x,t) + y_2(x,t) ) ( = y_m \left[ \sin(kx - \omega t - \phi_1) + \sin(kx - \omega t - \phi_2) \right] )

Using the trigonometric formula, ( \sin B + \sin C = 2 \sin \frac{1}{2} (B+C) \times \cos \frac{1}{2} (B-C) ), we get: ( y(x,t) = \left[ 2 y_m \cos \left( \frac{\Delta \phi}{2} \right) \right] \times \sin(kx - \omega t - \phi') )

Here ( \Delta \phi = \phi_2 - \phi_1 ) is the difference of phases, and ( \phi' = \frac{\phi_1 + \phi_2}{2} ) is the sum.

So what we learn from this? That if ( \phi_2 = \phi_1 ), then the two waves are in phase and the resultant amplitude is maximum (because ( \cos 0 = 1 )). And that if ( \phi_2 = \phi_1 + \pi ), then the two waves are out of phase and the resultant amplitude is minimum (because ( \cos \pi/2 = 0 )). The two waves have interfered with each other and have increased/decreased their amplitude in these two extreme cases. In general, ( \cos \left( \frac{\Delta \phi}{2} \right) ) will be some number that lies between -1 and +1.

🔑 Definition — Interference: The phenomenon that occurs when two or more waves overlap, resulting in a new wave pattern with an amplitude that is the sum of the individual amplitudes.

7. Speed of Sound in a Medium

The speed of sound in a medium is given by the formula: ( v = \sqrt{\frac{B}{\rho}} ), where ( B ) is the bulk modulus and ( \rho ) is the density of the medium.

📐 Formula: ( v = \sqrt{\frac{B}{\rho}} ) → The speed of sound increases with the stiffness (bulk modulus) of the medium and decreases with its density.

8. Speed of a Pulse

A pulse is a burst of energy (sound, electromagnetic, heat, ...) and could have any shape. Mathematically any pulse has the form ( y(x,t) = f(x - vt) ). Here ( f ) is any function (e.g., sin, cos, exp, ...). Note that at time ( t=0 ), ( y(x,0) = f(x) ) and the shape would look as on the left in the diagram below. At a late time ( t ), it will look just the same, but shifted to the right. In other words, at time ( t ), ( y(x,t) = f(x') ) where ( x' = x - vt ).

Fix your attention on any one point of the curve and follow it as the pulse moves to the right. From ( x - vt = \text{constant} ) it follows that ( \frac{dx}{dt} - v = 0 ), or ( v = \frac{dx}{dt} ). This is called the phase velocity because we derived it using the constancy of phase.

🔑 Definition — Pulse: A single, non-repeating disturbance that travels through a medium.

📐 Formula: ( v = \frac{dx}{dt} ) → The speed of a wave pulse is the rate at which a fixed point of constant phase moves.

⭐ Key Takeaways

Wave motion is the transfer of energy through a medium without the net transfer of matter, classified into longitudinal and transverse waves based on the direction of oscillation relative to wave travel. The intensity or power of a wave is proportional to the square of its amplitude, and for spherical waves, this intensity follows an inverse square law with distance from the source. A wave can be represented by a sinusoidal function containing a phase constant, which simply shifts the wave in space or time. When two waves overlap, they undergo interference, with the resultant amplitude determined by the phase difference between them: maximum for waves in phase and minimum for waves out of phase. The speed of any waveform, whether a continuous wave or a single pulse, is given by the phase velocity, ( v = \frac{dx}{dt} ), derived from the condition of constant phase.

🧠 Quick Revision Questions

1. What are the two basic categories of waves, and how do the oscillations differ relative to the direction of wave travel?
2. State the relationship between the amplitude of a wave and its average power or intensity.
3. For spherical waves radiating from a point source, how does the intensity change as the distance from the source is doubled?
4. What is the condition on the phase difference ((\Delta \phi)) for two waves of equal amplitude to exhibit fully constructive interference?
5. Write the mathematical form of a wave pulse traveling in the positive x-direction and derive an expression for its speed.

📘 Lecture 21 — GRAVITY

📖 Overview: This lecture covers the fundamental law of universal gravitation, including Newton's formulation, experimental determination of the gravitational constant G, and its application to Earth's mass and density. It also explores gravitational potential energy, escape velocity, satellite orbits, and Kepler's laws of planetary motion.

🗂️ Topics Covered

Newton's law of universal gravitation and the direction of gravitational forces; experimental determination of G using the Cavendish torsion balance; calculation of Earth's mass, gravity, and density; gravitational potential and potential energy; escape velocity from Earth and other bodies; satellite circular orbit conditions, orbital velocity, and period; total energy of orbiting satellites; Kepler's equal area law derived from angular momentum conservation.

📝 Lecture Summary

1. Newton's Law of Universal Gravitation

Newton's law of universal gravitation states that the force of attraction between two masses m₁ and m₂ is proportional to the product of their masses and inversely proportional to the square of the distance between them. Putting in a constant of proportionality, F = G (m₁m₂)/r². The force is directed along the line joining the two bodies. Looking at the diagram, F₂₁ = Force on m₂ by m₁, and F₁₂ = Force on m₁ by m₂. By Newton's Third Law, F₁₂ = -F₂₁.

2. The Gravitational Constant G

The gravitational constant G is a very small quantity and needs a very sensitive experiment. An early experiment to find G involved suspending two masses and measuring the deflection. A thread provides the restoring torque κθ. From the figure you can see that the gravitational torque is 2(GmM/r²)L. The deflection θ can be measured by observing the beam of the light reflected from the small mirror. In equilibrium the torques balance, (GmM/r²)L = κθ. Hence G = (κθ r²)/(GmML). How is κ found? It can be found from observing the period of free oscillations, T = 2π√(I/κ) ⇒ κ = (4π²I)/T² with I = (mL²)/2.

🔑 Definition — Gravitational Constant G: The proportionality constant in Newton's law of universal gravitation. The modern value is G = 6.67259 × 10⁻¹¹ N·m²/kg².

3. Earth's Gravity and Mass

The magnitude of the force with which the Earth attracts a body of mass m towards its centre is F = GmMₑ/Rₑ², where Rₑ = 6400 km is the radius of the Earth and Mₑ is the mass. The material does not matter — iron, wood, leather, etc. all feel the force in proportion to their masses. If the body can fall freely, then it will accelerate. So, F = mg = GmMₑ/Rₑ². We measure g, the acceleration due to gravity, as 9.8 m/s². From this we can immediately deduce Earth's mass: Mₑ = gRₑ²/G = 5.97 × 10²⁴ kg. What a remarkable achievement! The volume of the Earth is Vₑ = (4/3)πRₑ³ = 1.08 × 10²¹ m³. Hence the density of the Earth is ρₑ = Mₑ/Vₑ = 5462 kg/m³. So this is 5.462 times greater than the density of water and tells us that the Earth must be quite dense inside.

📐 Formula: g = GMₑ/Rₑ² → The acceleration due to gravity at Earth's surface equals the gravitational constant times Earth's mass divided by the square of Earth's radius.

4. Gravitational Potential

The gravitational potential is an important quantity. It is the work done in moving a unit mass from infinity to a given point R, and equals V(r) = -GM/R. Proof: Conservation of energy says dV = -Fdr ⇒ ∫dV = -∫F(r)dr. Integrate both sides: 0 - V(R) = GM∫(1/r²)dr from R to ∞ = -GM[1/r] from R to ∞ = -GM(0 - 1/R) = GM/R. Therefore, V(R) = -GM/R.

5. Change in Potential Energy Near Earth's Surface

Using the above formula, let us calculate the change in potential energy ΔU when we raise a body of mass m to a height h above the Earth's surface. ΔU = GMm(1/Rₑ - 1/(Rₑ + h)) = GMm(1 - (1 + h/Rₑ)⁻¹). Now suppose that the distance h is much smaller than the Earth's radius. So, for h ≪ Rₑ, (1 + h/Rₑ)⁻¹ ≈ 1 - h/Rₑ. So we find ΔU = GMm(1 - (1 - h/Rₑ)) = GMm(h/Rₑ²) = mgh.

🔑 Definition — Gravitational Potential Energy near Earth: For small heights h ≪ Rₑ, the change in potential energy ΔU = mgh.

6. Escape Velocity

We can use the expression for potential energy and the law of conservation of energy to find the minimum velocity needed for a body to escape the Earth's gravity. Far away from the Earth, the potential energy is zero, and the smallest value for the kinetic energy is also zero. Requiring that (KE + PE) at r = R equals (KE + PE) at r = ∞ gives (1/2)mvₑ² - GMm/R = 0 + 0. From this, vₑ = √(2GM/R) = √(2gRₑ). Putting in some numbers we find that for the Earth vₑ = 11.2 km/s. For the Sun, vₑ = 618 km/s. For a Black Hole, the escape velocity is so high that nothing can escape, even if it could move with the speed of light! (Nevertheless, Black Holes can be observed because when matter falls into them, a certain kind of radiation is emitted.)

📐 Formula: vₑ = √(2GM/R) → Escape velocity equals the square root of twice the gravitational constant times the mass divided by the radius.

7. Satellite Orbits

A satellite is in circular orbit over the Earth's surface. The condition for equilibrium: mv²/r = GMm/r² ⇒ v₀ = √(GM/r). If Rₑ is the Earth's radius, and h is the height of the satellite above the ground, then r = Rₑ + h. For h ≪ Rₑ, we can approximate v₀ = √(GM/Rₑ) = √(gRₑ). We can easily calculate the time for one complete revolution, T = 2π/ω = 2πr/v₀ = 2πr/√(GM/r) = 2πr³/²/√(GM). This gives the important result, observed by Kepler nearly 3 centuries ago that T² ∝ r³, or T² = (4π²/GM)r³.

📐 Formula: v₀ = √(GM/r) → The orbital velocity of a satellite equals the square root of the gravitational constant times Earth's mass divided by the orbital radius.

📐 Formula: T = 2πr³/²/√(GM) → The orbital period equals 2π times the 3/2 power of orbital radius divided by the square root of GM.

8. Total Energy of a Satellite

What is the total energy of a satellite moving in a circular orbit around the Earth? Clearly, it has two parts, kinetic and potential. Remember that the potential energy is negative. So, E = KE + PE = (1/2)mv₀² - GMₑm/r. But v₀² = GM/r as we saw earlier, and therefore E = (1/2)(GMm/r) - GMm/r = -(1/2)GMm/r. Note that the magnitude of the potential energy is larger than the kinetic energy. If it wasn't, the satellite would not be bound to the Earth!

📐 Formula: E = -(1/2)GMm/r → The total energy of a satellite in circular orbit is negative one-half times the gravitational constant times the masses divided by the orbital radius.

💡 Why this matters: The negative total energy indicates a bound system — the satellite cannot escape Earth's gravity without additional energy input.

9. Kepler's Equal Area Law

A famous discovery of the astronomer Johann Kepler some 300 years ago says that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time. We can easily see this from the conservation of angular momentum. Call ΔA the area swept out in time Δt. Then from the diagram you can see that ΔA = (1/2)r(rΔθ). Divide this by Δt and then take the limit where it becomes very small: dA/dt = lim(Δt→0) ΔA/Δt = (1/2)r²(lim(Δt→0) Δθ/Δt) = (1/2)r²ω = L/(2m). Since L is a constant, we have proved one of Kepler's laws (with so little effort)!

⭐ Key Takeaways

The gravitational force is always attractive and follows the inverse-square law. The gravitational constant G is extremely small, making Cavendish's measurement a remarkable achievement. The acceleration due to gravity g = 9.8 m/s² allows direct calculation of Earth's mass and density. Escape velocity depends only on the mass and radius of the celestial body, not on the mass of the escaping object. For satellites in circular orbits, total energy is negative and half the magnitude of potential energy, indicating a bound system. Kepler's equal area law follows directly from conservation of angular momentum.

🧠 Quick Revision Questions

1. Write the mathematical expression for Newton's law of universal gravitation, identifying all variables.
2. How is the gravitational constant G determined experimentally using the Cavendish torsion balance?
3. Derive the escape velocity formula from conservation of energy.
4. Show that for a satellite in circular orbit, T² ∝ r³.
5. Prove Kepler's equal area law using angular momentum conservation.

📘 Lecture 22 — ELECTROSTATICS I

📖 Overview: This lecture introduces electrostatics, the study of stationary electric charges. It covers Coulomb's Law for the force between charges, the concepts of electric field and field lines, and the behavior of electric dipoles. Understanding these fundamentals is crucial for analyzing electric forces, fields, and the structure of matter at a microscopic level.

🗂️ Topics Covered

This lecture begins with Coulomb's Law describing the force between point charges, and the constants involved. It then explains the quantization and conservation of charge, introduces the concept of scalar and vector fields, and defines the electric field. The lecture continues with typical electric field magnitudes, a method for measuring charge, and the superposition principle for calculating fields from multiple charges. Finally, it derives the electric field of a dipole and the torque it experiences in a uniform field.

📝 Lecture Summary

1. Coulomb's Law

Like charges repel, unlike charges attract. Coulomb's Law quantifies this force. It states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. In mathematical terms: ( F \propto \frac{q_1 q_2}{r^2} ), which becomes an equality with a constant of proportionality ( k ). In MKS units, charge is measured in Coulombs (C) and ( k = \frac{1}{4 \pi \varepsilon_0} ), where the permittivity of free space is ( \varepsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2 ) and ( k = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 ). The vector form is ( \vec{F}{12} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r{12}^2} \hat{r}{12} ), where ( \hat{r}{12} ) is the unit vector from charge 2 to charge 1, representing the force on charge 1 due to charge 2.

🔑 Definition — Coulomb's Law: The magnitude of the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. 📐 Formula: ( F = k \frac{q_1 q_2}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} ) → The electric force between two charges decreases with the square of the distance between them. 📌 Example: For many charges, the total force on a charge like ( q_1 ) is the vector sum of the forces from all other individual charges: ( \vec{F}1 = \vec{F}{12} + \vec{F}{13} + \vec{F}{14} + \cdots ).

2. Charge is Quantized

Charge is quantized, meaning it comes only in discrete units. The size of any charge can only be an integer multiple of the fundamental charge, ( e = 1.602 \times 10^{-19} \text{ C} ). This is expressed as ( q = ne ), where ( n ) is an integer.

🔑 Definition — Charge Quantization: The principle that electric charge exists in discrete amounts, which are integer multiples of the elementary charge, ( e ).

3. Charge is Conserved

Charge is conserved, meaning it is never created or destroyed in any process. The total electric charge in an isolated system remains constant over time. In any reaction, the initial total charge equals the final total charge.

📌 Example: In electron-positron annihilation, ( e^- + e^+ \rightarrow \gamma + \gamma ), the initial total charge is 0 (from +1 and -1) and the final total charge is 0 (neutral photons).

4. Field

A field is a quantity that has a definite value at any point in space and at any time. A scalar field is described by a single number, such as temperature ( T(x, y, z, t) ). A vector field is described by three numbers at each point in space and time, such as velocity, pressure, or the electric field.

🔑 Definition — Field: A physical quantity that has a value at every point in space and time.

5. The Electric Field

The electric field ( \vec{E} ) is a vector field. It is defined as the electric force on a unit test charge placed at that point. To avoid disturbing the field, the test charge ( q_0 ) must be very small. The electric field due to a point charge ( q ) is ( \vec{E} = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} \hat{r} ). Electric fields can be visualized using field lines that start on positive charges and end on negative charges, with the density of lines indicating the field's magnitude.

🔑 Definition — Electric Field: The force per unit charge experienced by a stationary test charge: ( \vec{E} = \frac{\vec{F}}{q_0} ). 📌 Example: Typical magnitudes include ( 10^{11} \text{ N/C} ) inside an atom and ( 10^{-2} \text{ N/C} ) inside a wire.

6. Measuring Charge

One method to measure charge is by balancing the gravitational force on a charged particle with the electric force from a known field. For a particle of mass ( m ) in equilibrium, ( mg = qE ), so the unknown charge can be found as ( q = \frac{mg}{E} ).

📐 Formula: ( q = \frac{mg}{E} ) → The charge on a particle can be determined by finding the electric field required to balance its weight.

7. The Superposition Principle

The total electric field at a point due to a collection of charges is the vector sum of the fields produced by each charge individually. This is the superposition principle: ( \vec{E} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 + \cdots = k \sum_i \frac{q_i}{r_i^2} \hat{r}_i ), where ( \hat{r}_i ) is the unit vector pointing from charge ( q_i ) to the point of observation.

8. The Electric Field of a Dipole

An electric dipole consists of two equal and opposite charges, ( +q ) and ( -q ), separated by a distance ( d ). For a point on the perpendicular bisector of the dipole at a large distance ( x \gg d ), the vertical components of the fields from the two charges cancel. The net field is directed along the dipole axis. The magnitude is calculated by summing the horizontal components: ( E = 2E_+ \cos \theta ), where ( \cos \theta = \frac{d/2}{\sqrt{x^2 + (d/2)^2}} ). This yields ( E = \frac{1}{4 \pi \varepsilon_0} \frac{qd}{[x^2 + (d/2)^2]^{3/2}} ).

📐 Formula: For ( x \gg d ), the dipole field is ( E = \frac{1}{4 \pi \varepsilon_0} \frac{p}{x^3} ), where the dipole moment is ( p = qd ).

9. Torque on a Dipole

When an electric dipole is placed in a uniform electric field ( \vec{E} ), the forces on the two charges are equal and opposite, creating a torque but no net force. The magnitude of the torque is ( \tau = F \frac{d}{2} \sin \theta + F \frac{d}{2} \sin \theta = Fd \sin \theta ). Since ( F = qE ), the torque is ( \tau = qE d \sin \theta = pE \sin \theta ). The direction of the torque is perpendicular to the plane containing the dipole and the field.

📐 Formula: ( \tau = pE \sin \theta ) → The torque on a dipole in a uniform electric field is the product of its dipole moment, the field strength, and the sine of the angle between them.

⭐ Key Takeaways

Coulomb's Law is the fundamental equation for the electrostatic force, being inversely proportional to the square of the distance. Charge is both quantized, existing only in integer multiples of the elementary charge, and conserved, meaning the total charge in an isolated system remains constant. The electric field is the force per unit charge and is a vector field; the field from multiple charges is found using the superposition principle. An electric dipole produces an electric field that falls off as ( 1/r^3 ) along its axis, and it experiences a torque in a uniform field, given by ( \tau = pE \sin \theta ).

🧠 Quick Revision Questions

1. State Coulomb's Law in words and give its mathematical form.
2. What is the value of the elementary charge, ( e ), in Coulombs?
3. What does the principle of charge conservation imply for an isolated system?
4. How is the electric field defined, and what is the formula for the field due to a point charge?
5. For an electric dipole far away, how does the electric field magnitude depend on distance?