JEE Important Chapter - Rotational Motion

The rotational motion of the objects plays an important role in understanding dynamics in 3D. In the past chapters, students have learned about the motion in one and two-dimensional planes, where an object will be in a straight line motion. Now the question arises will the same kinematics theory and equations be applicable if the object starts rolling? To answer this question students should study the chapter on rotational motion and understand what is rotational motion.

The rotational motion chapter deals with all the phenomena and theories related to rotation and rotational motion examples. In the rotational motion chapter, we will look at the kinematics and dynamics of a solid body in two different motions. Rotation around a stationary axis, often known as pure rotation, is the first type of motion for a solid body. The second type of solid motion is plane motion, in which the solid body's centre of mass moves in a fixed direction while the body's angular velocity remains perpendicular to that plane at all times.

Furthermore, in the rotational motion chapter, we study the rotational dynamics and the parameters associated with it. The chapter includes many important concepts such as angular velocity, angular acceleration, perpendicular axis theorem, parallel axis theorem, and torque, and we define the moment of inertia of various objects such as a ring, disc, cylinder, etc.. which are all important in exam perspective.

In this article, we will cover the concepts which are useful for the JEE exam and boost their exam preparation.

JEE Main Physics Chapter-wise Solutions 2022-23 

Important Topics of Rotational Motion

• Rigid Bodies

• Rotational Kinematics

• Moment of Inertia

• Parallel Axis Theorem

• Perpendicular Axis Theorem

• Torque

• Angular Momentum

• Angular Displacement

• Conservation of Angular Momentum

• Instantaneous Axis of Rotation

• Rolling motion

• Toppling

Important Concepts of Rotational Motion

List of Important Formulae

Important Comparison of Rotational Kinematics and Linear Kinematics

When we start preparing for our exams its essential to know our understanding with the help of comparisions, differences between the concepts we have previously studied. We know the kinematics of linear motion, now let us have a look at the comparison of linear kinematics with rotational motion.

Solved Examples of Rotational Motion

1. The moment of inertia of a uniform cylinder of length and radius R about its perpendicular bisector is I. What is the ratio l/R such that the moment of inertia is minimum?

Sol:

Given,

We have a uniform cylinder of length l and radius R, such that the pendicular bisector is I. Now we are asked find the ratio $\dfrac {l}{R}$.

We know that, moment of inertia of a cylinder of length l is:

$I=\dfrac {M{R^2}}{4} +\dfrac {M {l^2}}{12}$

$I=\dfrac {M^2}{4{\rho}{\pi}l}+\dfrac {M{l^2}}{12}$

For maximum I, we have $\dfrac {dI}{dl}=0$

$\dfrac {dI}{dl}=\dfrac {-M^2}{4{\pi}{\rho}{l^2}}+\dfrac {M{l}}{6}=0$

$\dfrac {M^2}{4{\pi}{\rho}{l^2}}=\dfrac {M{l}}{6}$

$\dfrac {M}{{4{\pi}{\rho}}}=\dfrac {l^3}{6}$

$\dfrac {4 \pi \rho {R^2}{l}}{{4{\pi}{\rho}}}=\dfrac {l^3}{6}$

$\dfrac {l^2}{R^2}=\dfrac {3}{2}$

$\dfrac {l}{R}=\sqrt {\dfrac {3}{2}}$

Therefore, the ratio of $\dfrac {l}{R}$ is $\sqrt {\dfrac {3}{2}}$.

Key point: Here we should use the formula of volume mass density $\rho=\dfrac {m}{4\pi {R^2} l}$. While solving problems related to cylinder or any other geometrical objects we must remember that the mass density will vary from object to object.

2. Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is?

Sol:

Here, the arrangement of seven identical disks are welded symmetrically, thus all the discs will rotate about center of mass. We know that, according to the parallel axis theorem, The moment of inertia of a body about any axis is equal to the sum of the Moment of Inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

I.e., 

$I={I_C}+M {d^2}$

Where,

d- The perpendicular distance between the two axes.

${I_o}=\dfrac {7M{R^2}}{2}+6M {{\left (2R\right)}^2}=\dfrac {55 M{R^2}}{2}$

Now,  The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is:

${I_P}={I_o}+M{d^2}$

${I_P}=\dfrac {55 M{R^2}}{2}+7M{\left (3R\right )}^2=\dfrac {181}{2} M{R^2}$

Key point: While solving this question, we must first calculate the moment of inertia of the entire system of seven particles, and then we can calculate the moment of inertia at any point on the disk with the help of parallel axis theorem.

Previous Year Questions of Rotational Motion

1. A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in the vertical plane. The upper end is pushed so that the rod falls under gravity, ignoring the friction due to clamping at its lower end, the speed of the free end of the rod when it passes through its lowest position is ____________ ms-1. (Take g = 10 ms-2) (JEE 2021)

Sol:

Given, 

Mass of the steel rod=M= 2 kg

Length of the steel rod=l=0.6 m

The given rod is clamped on a table vertically at its lower end and its free to rotate in a vertical plane. Now, we are asked to determine the speed of the free end of the rod when it passes through its lowest position.

Here, we can use the conservation of energy to obtain the speed. From conservation of energy we get:

P.E=K.E

$mgl=\dfrac {1}{2} I {\omega^2}$

$mgl=\dfrac {1}{2} \dfrac {m {l^2}{\omega^2}}{3}$

$\omega =\sqrt {\dfrac {6g}{l}}$

Speed $v=l\omega=l\times \sqrt {\dfrac {6g}{l}}=\sqrt {6gl}$

$v=6 m/s$

Therefore, the speed of the free end of rod when it passes through its lowest position 6 m/s.

Trick: Here, we can notice that the one end of the rod is fixed, thus we can use the moment of inertia formula when the axis of rotation is perpendicular to the rod at one end i.e., $I=\dfrac {M{l^2}}{3}$.

2. The centre of mass of the solid hemisphere of radius 8 cm is x from the centre of the flat surface. Then value of x is __________. (JEE 2020)

Sol:

We know that centre of mass of a solid hemisphere is,

$x=\dfrac {3R}{8}=\dfrac {3\times 8}{8}=3 cm$

Therefore, the value of x is 3 cm.

Trick: For this kind of problem we need to recall and practice COM of a few geometrical shapes, which will save our time calculation.

Practice Questions

1. A uniform thin bar of mass 6 kg and length of 2.4 meters is bent to make an equilateral hexagon. The moment of inertia about an axis passing through the centre of mass and perpendicular to the plane of a hexagon is _______  10-1 kg m2. 

(Ans: 8)

2. A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string is in the ratio 5: 1. The velocity of the bob at the highest position is ________ m/s. (Take g = 10 m/s2)

(Ans: 5)

3. What is the physical significance of the moment of inertia?

Conclusion

In this article, we discussed one of the important chapters for JEE, rotational motion. We have covered most of the important concepts, formulae, and solved examples along with recently asked questions in JEE exams. In this article, we have also studied how linear kinematics is different from the rotational kinematics and had an overview of the rotational kinematic equations and what rotary motion, translational motion, etc... This article will help students prepare for JEE exam and boost their preparation in one stance.

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