Rotational Motion JEE Notes, Download PDF & Formulas

Most mechanics problems deal with objects moving in straight lines, but many real-world objects rotate around an axis. A ceiling fan, wheel, and Earth are common examples of rotational motion. When a rigid body rotates about a fixed axis, every particle moves in a circular path. These Rotational Motion JEE Notes cover angular motion, moment of inertia, torque, angular momentum, rotational energy, rolling motion, and important JEE questions for quick JEE revision.

Rotational Motion JEE Notes: Important Concepts

Rotational motion is similar to translational motion. Linear displacement, velocity and acceleration have their corresponding angular quantities. Understanding this relation makes rotational motion formulas easier to apply.

Rigid Body and Axis of Rotation

Rigid Body: An ideal body in which the distance between any two particles remains constant even when forces act on it.

Axis of Rotation: The fixed line about which a body rotates. Every particle moves in a circle around this axis.

Angular Quantities

Angular quantities are measured in radians.

1 revolution = 2π radians = 360°

Angular Kinematics

For constant angular acceleration, rotational equations are similar to linear equations of motion.

JEE Tip: Rotational kinematics is simply the translational formula with:

$$s\rightarrow\theta,\quad u\rightarrow\omega_0,\quad v\rightarrow\omega,\quad a\rightarrow\alpha$$

Worked Example: A wheel starts from rest and rotates with angular acceleration 2 rad/s². Find angular velocity after 5 seconds.

Given:

$$\omega_0=0,\quad \alpha=2,\quad t=5$$

Using:

$$\omega=\omega_0+\alpha t$$

$$\omega=0+2(5)=10\ rad/s$$

Relations Between Linear and Angular Quantities

For a particle at a distance r from the axis of rotation:

Important: Tangential acceleration changes the speed of a particle, while centripetal acceleration changes only the direction of motion.

Moment of Inertia

In translational motion, mass determines the resistance to change in motion. Similarly, in rotational motion, moment of inertia determines the resistance of a body to change its rotational state.

Moment of inertia depends on:

• Mass of the body
• Distribution of mass around the axis
• Position of the axis of rotation

Moment of Inertia (I):

$$I=\sum m_ir_i^2$$

For a continuous body:

$$I=\int r^2dm$$

The SI unit of moment of inertia is:

kg m²

Moment of Inertia of Common Bodies

Important: The moment of inertia of the same body changes when the axis of rotation changes.

Worked Example: Four point masses of 2 kg each are placed at the corners of a square of side 1 m. Find the moment of inertia about one side of the square.

Two masses lie on the axis, so their distance from the axis is zero.

The other two masses are at distance 1 m.

$$I=2(mr^2)$$

$$I=2(2)(1^2)=4\ kg\,m^2$$

Parallel Axis Theorem

The parallel axis theorem helps calculate the moment of inertia about any axis parallel to an axis passing through the centre of mass.

$$I=I_{cm}+Md^2$$

Where:

• $$I$$ = Moment of inertia about the new axis
• $$I_{cm}$$ = Moment of inertia about centre of mass axis
• M = Mass of the body
• d = Distance between the two parallel axes

Worked Example: Find the moment of inertia of a uniform rod about an axis passing through one end.

For a rod about its centre:

$$I_{cm}=\frac1{12}ML^2$$

The distance between the axes:

$$d=\frac L2$$

Using parallel axis theorem:

$$I=I_{cm}+Md^2$$

$$I=\frac1{12}ML^2+M\left(\frac L2\right)^2$$

$$I=\frac1{12}ML^2+\frac14ML^2$$

$$I=\frac13ML^2$$

Perpendicular Axis Theorem

The perpendicular axis theorem is applicable only for plane lamina bodies.

If two perpendicular axes lie in the plane of a body and the third axis is perpendicular to the plane:

$$I_z=I_x+I_y$$

This theorem is applicable for:

• Disc
• Ring
• Rectangular plate

It is not applicable for three-dimensional objects like spheres and cylinders.

Worked Example: The moment of inertia of a disc about an axis perpendicular to its plane is:

$$I_z=\frac12MR^2$$

Find its moment of inertia about a diameter.

By symmetry:

$$I_x=I_y$$

Using perpendicular axis theorem:

$$I_z=I_x+I_y$$

$$\frac12MR^2=2I_x$$

Therefore:

$$I_x=\frac14MR^2$$

JEE Tip: Remember:

• Parallel axis theorem → works for all rigid bodies.
• Perpendicular axis theorem → works only for planar bodies.

Torque and Rotational Motion

In translational motion, force produces acceleration. Similarly, in rotational motion, torque produces angular acceleration.

Torque: The turning effect of a force about an axis is called torque.

$$\vec{\tau}=\vec{r}\times\vec{F}$$

The magnitude of torque is:

$$\tau=rF\sin\theta$$

where:

• r = distance from the axis of rotation
• F = applied force
• θ = angle between force and position vector

Torque can also be written as:

$$\tau=F\times d$$

where d is the perpendicular distance from the axis to the line of action of force.

Unit: N m

Dimension: $$ML^2T^{-2}$$

Important: Torque and work have the same dimensions but they are different physical quantities. Torque is a vector quantity, while work is a scalar quantity.

Newton's Second Law for Rotation

The rotational form of Newton's second law is:

$$\tau_{net}=I\alpha$$

This is the rotational equivalent of:

$$F=ma$$

Here:

• τ = net torque
• I = moment of inertia
• α = angular acceleration

Worked Example: A disc of mass 5 kg and radius 0.2 m rotates about its centre. A tangential force of 10 N acts at its edge. Find angular acceleration.

Torque:

$$\tau=FR$$

$$\tau=10(0.2)=2\ Nm$$

Moment of inertia of disc:

$$I=\frac12MR^2$$

$$I=\frac12(5)(0.2)^2$$

$$I=0.1\ kg\,m^2$$

Using:

$$\tau=I\alpha$$

$$\alpha=\frac{\tau}{I}$$

$$\alpha=\frac2{0.1}=20\ rad/s^2$$

Angular Momentum

Angular momentum is the rotational equivalent of linear momentum.

Angular Momentum of a Particle

$$\vec{L}=\vec{r}\times\vec{p}$$

Magnitude:

$$L=mvr\sin\theta$$

Angular Momentum of a Rigid Body

$$L=I\omega$$

SI Unit:

kg m²/s

Relation between torque and angular momentum:

$$\tau_{net}=\frac{dL}{dt}$$

Conservation of Angular Momentum

If no external torque acts on a system, the angular momentum remains constant.

$$\tau_{ext}=0 \Rightarrow L=I\omega=constant$$

Therefore:

$$I_1\omega_1=I_2\omega_2$$

When moment of inertia decreases, angular velocity increases.

Example: A skater spins faster when pulling her arms closer because reducing the distance of mass from the axis decreases moment of inertia.

Worked Example: A disc with moment of inertia 4 kg m² rotates at 10 rad/s. A ring with moment of inertia 2 kg m² is placed on it. Find the final angular velocity.

Using conservation of angular momentum:

$$I_1\omega_1=(I_1+I_2)\omega_f$$

$$4(10)=(4+2)\omega_f$$

$$40=6\omega_f$$

$$\omega_f=\frac{20}{3}\ rad/s$$

Answer:

$$\omega_f\approx6.67\ rad/s$$

JEE Tip: Before applying conservation of angular momentum, always check that external torque about the chosen axis is zero.