Rotational Mechanics is considered one of the most difficult topics in JEE Physics. It requires great clarity of concepts and visualization abilities to score marks on questions related to this topic in JEE. Students largely dread this topic, and paper setters love it! It’s a favourite for paper setters because they can test multiple concepts–from the basic F=ma equation to energy conservation (all in a single question from this chapter!)

It is important to study this topic well because questions from it definitely appear on the JEE. Comprehension type questions from this topic are common too. Also, complex questions can be formed when rotational mechanics is combined with other concepts such as electromagnetism, SHM, etc.

To excel in rotational mechanics for IIT JEE Physics, you need to put in a great amount of time and effort. So let’s start by looking at some of the core concepts.

## Rigid Body

When you hear the term “rigid body”, what comes to mind? Well, when I first heard this word I thought it was some object which was very tough to deform–something very hard and “rigid”. I didn’t know that this simple thing was an incredibly importance concept of JEE rotational mechanics! Why is it so important? Because learning the rotational mechanics of a rigid body is much simpler than a non-rigid body. You see, the rotational mechanics of a non-rigid body is very complex–and *not necessary for the JEE*!

### What Exactly Is a Rigid Body?

So what exactly is a “rigid body?” It is basically a system of particles where the distance/separation between each pair of particles remains constant (with respect to time). So it means that the shape and size hardly change during the motion. For every pair of particles in a rigid body, there is zero velocity of separation or approach. Hence, we can say that the deformation of a rigid body during motion is almost zero. A ball bearing made of hardened steel is a good example of a rigid body.

In real life, there are no rigid bodies. They are an assumption made in classical mechanics. This assumption makes the calculations related to the movement of bodies easier. However, in quantum mechanics, point masses (molecules) are often seen as rigid bodies.

Well, now that we know what a rigid body is, let’s study the kinds of motion they can perform. As per the JEE syllabus, we need to learn **Translation** and **Rotation**. Let’s just stick to only these two types of rigid body motion.

## Rigid Body Motion: Pure Translation

Suppose that the car (shown below) is a rigid body and is moving from left to right.

This is a case of pure translation. No rotation is happening here. In a pure translation of a rigid body, the velocity and acceleration of each and every particle of the rigid body is same.

### Visualize the Opposite

This fact can be visualized better if we think of its opposite. So suppose some particles of the car have a greater velocity or acceleration than the remaining particles. The faster particles would move farther and the car would be dismantled. Just imagine the seat of the car having a faster velocity than the remaining parts of the car. Sounds funny, right?

In translation, all the particles of the rigid body move along parallel paths, and if these parallel paths are straight lines, the motion is said to be a **rectilinear translation**.

## Rigid Body Motion: Pure Rotation

Consider the image below in which a rod is rotated around a fixed point O.

If you notice, during the course of one full rotation, the part of the rod at point O remains fixed. However, the tip of the rod farthermost to point O covers the greatest distance (2𝛑l, where l is the length of the rod). Hence, as you can see, the different points of the rod have a different velocity. One very important point to notice here is that all the points on the rod move in concentric circular paths only.

A rigid body is considered to be in pure rotation only if each and every particle of the body moves in a circle, and the centers of all the circles lie on a straight line. This line is known as the **axis of rotation (A.O.R.)**. In pure rotation, all points in the rigid body that are perpendicular to the A.O.R. turn through the same angle in the same interval. In the figure above, the A.O.R. would be the Z axis.

General rigid body motion is a combination of pure translation and pure rotation. So, while solving a rotational mechanics question, try to break the motion of the body into translation and rotation, then solve for each of them.

## Angular Velocity

Have you ever rotated a ball tied to a thread? Think of swinging it in a circular motion above your head. I know, it’s hard to see a connection between concepts in the JEE syllabus and your daily life! But I’m sure we can all imagine swinging a ball tied to a thread. As we spin this ball around, we can observe the fundamentals of angular velocity!

While rotating the ball, if the string is suddenly cut, in which direction will the ball go? Will it still try to complete the circle or will it go tangentially? With what velocity will it go? Is there some relation between the rate of change in the angle subtended at the centre (at the point where you’re holding the string), and the velocity with which the ball goes away when the string is cut suddenly?

We’ll learn the answer to all these questions right here!

### Horizontal and Vertical Planes

Consider the figure below. Notice how the circular motion in the picture is happening in a horizontal plane. A circular motion happening in a vertical plane is an altogether different ball game!

In the figure above, the angle θ is known as **angular displacement**. It is basically the angle covered with respect to a baseline when an object rotates on an axis. The angular displacement happens on the same plane as that of the rotation of the object.

### Some Definitions

**Angular velocity** (ω) is the rate of change of angular displacement. Depending on the time frame chosen to calculate the rate of change of angular displacement, angular velocity can be of one of the following two types.

**Average angular velocity**: This is simply the rate of change of angular displacement with reference to the time measured over a time interval.**Instantaneous angular velocity**: It is defined as the rate of change of angular position with reference to the time at any instant of time. It is just the limit of the average angular velocity as time “t” approaches zero, that is, at an instant.

### Thumbs UP!

Angular velocity can be thought of as a vector quantity. The direction of ω is obtained using **right-hand thumb rule** — curl up your fingers of the right hand in a manner that the direction of the curl is in the direction of the sense of rotation. Your extended thumb will then give the direction of the angular velocity ω.

The unit of angular velocity is **rad/s** or radians per second.

The *v* mentioned in Figure 3 is known as **tangential velocity**. It is the linear velocity of the particle that is performing a circular motion and is directed along the tangent to the circular path at the given point at that particular instant. It is the initial velocity with which the ball will fall off if the ball is at point P of the circular path (shown in Figure 3) and the string is cut immediately.

The relationship between v, ω, and r is:

*v* = ω × *r*

### Example Problem

Let’s strengthen the concepts we learned above by solving a problem!

**Question**: If a stone is rotating in a circle of diameter 10m with velocity 20m/s, find the angular velocity of the stone.

- a) 14 Hz

b) 4 Hz

c) 2 Hz

d) 16 Hz

**Solution**:

v = ω r

So, ω = v/r

r = 10/2 = 5m and v = 20m/s

Hence, ω = 20/5 = 4 s^{-1} or 4 Hz

**Answer**: b)

Please note that 1 Hz = 2𝜋 rad/s

## Angular Acceleration

In the previous section, we learned about Angular Velocity. Now, let’s talk about the rate of change of Angular Velocity, also known as **Angular Acceleration**, often denoted by 𝛂.

### Merry-Go-Round

Consider the merry-go-round in the figure below. When there’s no one around, the merry-go-round is in a state of rest. Suddenly, the children decide they want to enjoy a ride on the merry-go-round. They select one of their friends (the little girl in the red shirt) and ask her to rotate the merry-go-round fast. She puts in her energy and rotates it hard. The merry-go-round, which was earlier at rest, is now rotating faster and faster (about its centre, of course). The angular velocity (ω) of each point of the merry-go-round is increasing. The merry-go-round not only has Angular Velocity, but also Angular Acceleration. That is the reason for the increasing ω of the merry-go-round.

### Some Definitions

Depending on the time frame chosen to calculate the rate of change of angular velocity, angular acceleration can be of one of the following two types:

**Average angular acceleration**: This is simply the rate of change of angular velocity with reference to the time measured over a time interval.**Instantaneous angular acceleration**: It is defined as the rate of change of angular velocity with reference to the time, at any instant of time. It is just the limit of the average angular acceleration as time “t” approaches zero, that is, at an instant.

The unit of angular acceleration is rad/s^{2} or radians per second squared. The angular acceleration is also a vector quantity. Its direction can be along the direction of angular velocity or opposite to it.

We learned how to find the direction of the angular velocity vector in the previous section using the right-hand thumb rule. If the magnitude of angular velocity ω is increasing, the direction of angular acceleration 𝛼 is along the direction of ω. If the magnitude of ω is decreasing, the direction of angular acceleration 𝛼 is opposite to the direction of ω.

Let’s strengthen our understanding of angular acceleration and angular velocity by solving a problem here!

### Example Problem

**Question**: Suppose when you increase the speed of your room fan from medium to high, the blades accelerate at 1.2 rad/s^{2} for 1.5 seconds. Assume the initial angular speed of the fan blades to be equal to 3.0 rad/s, find the final angular speed of the fan blades in rad/s.

a.) 4.8 rad/s

b.) 9.6 rad/s

c.) 10.0 rad/s

d.) 2.4 rad/s

**Solution:**

𝛼 = (𝛥ω)/(𝛥t)

𝛥ω = ωfinal – ωinitial

Hence, 𝛼 = (ωfinal – ωinitial)/(𝛥t)

Thus, ωfinal = ωinitial + 𝛼(𝛥t)

=> ωfinal = 3.0 + (1.2*1.5) = 4.8 rad/s

**Answer: a)**

In this post, we discussed the fundamentals of rotation mechanics, which are very important in building the foundation to a great understanding of the topic. Hope you found it useful!

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