Rotational Motion revision notes for JEE

01Key Concepts & Definitions

Rigid Body

An idealized body with a perfectly definite and unchanging shape. The distances between all pairs of particles within the body do not change under the influence of forces. Real bodies deform, but this deformation is often negligible.

Translational Motion

In pure translational motion, all particles of the body have the exact same velocity at any instant of time.

Rotational Motion (Fixed Axis)

Every particle of the body moves in a circle which lies in a plane perpendicular to the axis of rotation and has its center on the axis. Particles lying directly on the axis remain stationary.

Combined Motion

Most real-world motion (like a rolling cylinder) is a combination of translation of the center of mass and rotation about an axis.

Precession

The movement of the axis of rotation around a vertical line, sweeping out a cone. Example: A spinning top or an oscillating pedestal fan. When a torque acts perpendicular to the angular momentum vector, it causes the axis to sweep horizontally, i.e., to precess.

02Centre of Mass (CM)

The Centre of Mass is the point representing the mean position of the matter in a body or system, calculated as a mass-weighted average.

Two-Particle System: The CM lies on the line joining the particles. Its distance $X$ from the origin is $X = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$ . For equal masses, the CM is exactly midway.
$n$ -Particle System: The position vector of the CM, $\vec{R}$ , is given by $\vec{R} = \frac{\sum m_i \vec{r}_i}{M}$ , where $M = \sum m_i$ is the total mass.
Continuous Mass Distribution: For an extended rigid body, the summation becomes an integral: $\vec{R} = \frac{1}{M} \int \vec{r} \, dm$ , where $dm$ is a small mass element at position $\vec{r}$ .
Symmetry & CM: For homogeneous bodies of regular shapes (rings, discs, spheres, rods), the CM lies exactly at their geometric center due to reflection symmetry. JEE Tip If an object has a point, line, or plane of symmetry, its CM must lie on that point, line, or plane.
The CM does not necessarily lie inside the physical material of the body (e.g., the CM of an L-shaped lamina or a ring lies in the empty space).

03Motion of Centre of Mass & Linear Momentum

Velocity of CM: $\vec{V} = \frac{\sum m_i \vec{v}_i}{M}$ . The total linear momentum of the system is $\vec{P} = M \vec{V}$ .
Acceleration of CM: $\vec{A} = \frac{\sum m_i \vec{a}_i}{M}$ .

Newton's Second Law for a System

$M \vec{A} = \vec{F}_{ext}$ and $\frac{d\vec{P}}{dt} = \vec{F}_{ext}$ .

Internal forces (action-reaction pairs between particles within the system) strictly sum to zero and have absolutely no effect on the motion of the CM.
The CM moves exactly as if the entire mass were concentrated at that point and all external forces were applied directly to it.

Conservation of Linear Momentum

If the net external force $\vec{F}_{ext} = 0$ , the total linear momentum $\vec{P}$ is conserved (constant), and the velocity of the CM remains constant.

Radioactive Decay Example: A moving Radium nucleus decays into Radon and an alpha particle. Since decay forces are internal, the CM of the two products continues along the exact same original path.
Exploding Projectile: If a projectile explodes mid-air, the internal explosive forces do not alter the CM path. The CM of the fragments continues on the original parabolic trajectory. JEE Tip This is a highly tested concept; always trace the original CM path to locate the system's final position after an explosion.
Binary Stars: Moving under mutual internal gravitational forces, their CM moves in a uniform straight line.

04Vector Product (Cross Product)

The moment of force and angular momentum require the vector product: $\vec{c} = \vec{a} \times \vec{b}$ .

Magnitude: $|\vec{a} \times \vec{b}| = ab \sin\theta$ , where $\theta$ is the smaller angle between the vectors.
Direction: Given by the Right-Hand Screw Rule or Right-Hand Rule. Curl fingers from $\vec{a}$ to $\vec{b}$ ; the thumb points along $\vec{c}$ .

Properties

Not commutative: $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ .
Distributive: $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$ .
Self-cross product is zero: $\vec{a} \times \vec{a} = \vec{0}$ .
Reflection Property: Does NOT change sign under reflection (mirror image) because both vectors change sign ( $-\vec{a} \times -\vec{b} = \vec{a} \times \vec{b}$ ).

Orthogonal Unit Vectors: $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0$ . In cyclic order: $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ .

05Angular Velocity & Kinematics

Angular Velocity ( $\vec{\omega}$ ): Defined as $\vec{\omega} = \frac{d\theta}{dt}$ . For a fixed axis, $\vec{\omega}$ points along the axis (determined by the right-hand rule). Every particle in a rigid body shares the identical angular velocity $\vec{\omega}$ at any instant.

Relation to Linear Velocity

The velocity of a particle at position $\vec{r}$ from the origin on the axis is given by the cross product: $\vec{v} = \vec{\omega} \times \vec{r}$ .

Magnitude: $v = \omega r_{\perp}$ , where $r_{\perp}$ is the perpendicular distance from the axis.

Angular Acceleration ( $\vec{\alpha}$ ): Time rate of change of angular velocity, $\vec{\alpha} = \frac{d\vec{\omega}}{dt}$ .

Kinematics of Rotational Motion (Fixed Axis)

Analogous to linear 1D kinematics:

$\omega = \omega_0 + \alpha t$

$\theta = \theta_0 + \omega_0 t + \frac{1}{2}\alpha t^2$

$\omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)$

06Torque & Angular Momentum

Moment of Force (Torque,

\vec{\tau}

)

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

Magnitude: $\tau = r F \sin\theta = r_{\perp}F = rF_{\perp}$ .

Torque is zero if $F=0$ , $r=0$ , or the force acts directly along the position vector (line of action passes through the origin).

Fixed Axis vs. Fixed Point Kinematics: The relation $\vec{v} = \vec{\omega} \times \vec{r}$ applies to both fixed-axis rotation and general rotation with one fixed point (like a spinning top). For a fixed point, $\vec{r}$ is simply the position vector of the particle with respect to that fixed point.

Angular Momentum of a Particle (

\vec{l}

)

$\vec{l} = \vec{r} \times \vec{p}$ .

Time derivative: $\frac{d\vec{l}}{dt} = \vec{\tau}$ .
JEE Tip A particle simply moving in a straight line with constant velocity has a non-zero, CONSTANT angular momentum about any point not on its line of motion ( $l = mv r_{\perp}$ ).

Angular Momentum of a System (

\vec{L}

)

$\vec{L} = \sum (\vec{r}_i \times \vec{p}_i)$ .

Newton's Second Law for Rotation: $\frac{d\vec{L}}{dt} = \vec{\tau}_{ext}$ . Internal torques sum to zero (assuming central forces between particles).

Conservation of Angular Momentum

If total external torque $\vec{\tau}_{ext} = 0$ , then $\vec{L} = \text{constant}$ .

If moment of inertia changes (e.g., acrobat folding arms), $I_1 \omega_1 = I_2 \omega_2$ . JEE Tip Shrinking $I$ leads to a higher $\omega$ , resulting in an INCREASE in rotational kinetic energy. The extra energy comes from the internal work done by the acrobat's muscles to pull their arms inwards.

General Angular Momentum Vectors: For a single particle rotating about the z-axis, $\vec{l}$ has a z-component ( $\vec{l}_z$ ) parallel to the axis and a perpendicular component ( $\vec{l}_\perp$ ). $\vec{l}$ is NOT necessarily parallel to $\vec{\omega}$ .

Symmetric vs. Asymmetric Rigid Bodies

For an entire rigid body, $\vec{L} = \vec{L}_z + \vec{L}_\perp$ . If the body is strictly symmetric about the axis of rotation, diametrically opposite mass elements cancel out each other's perpendicular components, resulting in $\vec{L}_\perp = 0$ . ONLY then is $\vec{L} = \vec{L}_z = I\vec{\omega}$ valid as a parallel vector relationship.

JEE Tip For an asymmetric body rotating about a fixed axis, $\vec{L}$ and $\vec{\omega}$ are NOT parallel ( $\vec{L}_{\perp} \neq 0$ ).

07Equilibrium of a Rigid Body

A rigid body is in full mechanical equilibrium if it has neither linear nor angular acceleration.

Translational Equilibrium: Vector sum of all external forces is zero ( $\sum \vec{F}_i = 0$ ).
Rotational Equilibrium: Vector sum of all external torques is zero ( $\sum \vec{\tau}_i = 0$ $\sum τ_{i} = 0$ ).
- If $\sum \vec{F} = 0$ , the net torque is completely independent of the choice of origin.

Partial Equilibrium: A body can have zero net force but non-zero net torque (e.g., a couple), causing pure rotation. Or it can have zero net torque but non-zero net force, causing pure translation.

Couple

A pair of equal and opposite forces with different lines of action. It produces pure rotation.

Proof of Invariance: The moment of a couple is $(\vec{r}_2 - \vec{r}_1) \times \vec{F} = \vec{AB} \times \vec{F}$ . Since $\vec{AB}$ is the relative position vector between the points, torque depends only on their separation, independent of the origin.

Principle of Moments (Lever): $\text{Load Arm} \times \text{Load} = \text{Effort Arm} \times \text{Effort}$ . Mechanical Advantage (M.A.) = $\frac{\text{Load}}{\text{Effort}} = \frac{\text{Effort Arm}}{\text{Load Arm}}$ .

Centre of Gravity (CG)

The point where the total gravitational torque on the body is strictly zero. If $g$ is uniform across the body, CG and CM exactly coincide.

Experimental Determination: Suspend an irregular planar body (like cardboard) from a point; a vertical plumb line passes through the CG. Intersecting two such lines from different points locates the CG.

08Moment of Inertia & Radius of Gyration

Moment of Inertia (

I

)

The rotational analogue of mass. $I = \sum m_i r_i^2$ , where $r_i$ is the perpendicular distance to the axis of rotation.

Depends on: body mass, shape/size, distribution of mass, and position/orientation of the axis.

Radius of Gyration ( $k$ ): The distance from the axis where the entire mass $M$ could be concentrated to give the same $I$ . $I = Mk^2$ .

Standard Moments of Inertia (Table 6.1)

Thin circular ring (radius R, mass M): Axis through center $\perp$ to plane $\rightarrow I = MR^2$ . Diameter $\rightarrow I = MR^2/2$ .
Thin rod (length L): Axis through midpoint $\perp$ to rod $\rightarrow I = ML^2/12$ . Radius of gyration $k = L/\sqrt{12}$ .
Circular disc (radius R): Axis through center $\perp$ to disc $\rightarrow I = MR^2/2$ . Diameter $\rightarrow I = MR^2/4$ .
Hollow cylinder (radius R): Axis of cylinder $\rightarrow I = MR^2$ .
Solid cylinder (radius R): Axis of cylinder $\rightarrow I = MR^2/2$ .
Solid sphere (radius R): Diameter $\rightarrow I = 2MR^2/5$ .

Flywheel: Used in engines because its large $I$ prevents jerky motions by resisting sudden changes in angular speed.

09Dynamics of Rotational Motion

Derivation of Rotational Work and Power: A force acting tangentially undergoes displacement $ds = r d\theta$ . Work done $dW = F_{\perp} r d\theta = \tau d\theta$ . Dividing by $dt$ gives power $P = \tau \omega$ .
Work Done: $W = \int \tau d\theta$ .
Power: $P = \tau \omega$ .
Kinetic Energy: $K = \frac{1}{2}I\omega^2$ .
Work-Energy Theorem for Rotation: The work done by external torques equals the increase in rotational kinetic energy.
Newton's Second Law for Fixed Axis: Equating power to rate of change of KE ( $dK/dt = I\omega \alpha$ ) yields $\tau = I\alpha$ .

10Formulae, Equations & Units

Quantity	Symbol / Formula	SI Unit	Dimensions
Position of CM	$\vec{R} = \frac{\sum m_i \vec{r}_i}{M} = \frac{1}{M}\int \vec{r} dm$	$\text{m}$	$[L]$
Velocity of CM	$\vec{V}_{cm} = \frac{d\vec{R}}{dt} = \frac{\sum m_i \vec{v}_i}{M}$	$\text{m s}^{-1}$	$[LT^{-1}]$
Acceleration of CM	$\vec{A}_{cm} = \frac{d\vec{V}_{cm}}{dt} = \frac{\vec{F}_{ext}}{M}$	$\text{m s}^{-2}$	$[LT^{-2}]$
Linear Momentum	$\vec{P} = M\vec{V}_{cm}$	$\text{kg m s}^{-1}$	$[MLT^{-1}]$
Vector Product	$\vec{c} = \vec{a} \times \vec{b} = (ab \sin\theta)\hat{n}$	(Depends on $a,b$ )	(Depends on $a,b$ )
Linear/Angular Vel.	$\vec{v} = \vec{\omega} \times \vec{r}$	$\text{m s}^{-1}$	$[LT^{-1}]$
Torque	$\vec{\tau} = \vec{r} \times \vec{F}$	$\text{N m}$	$[ML^2T^{-2}]$
Angular Momentum	$\vec{l} = \vec{r} \times \vec{p}$ (Particle)	$\text{kg m}^2 \text{ s}^{-1}$	$[ML^2T^{-1}]$
Moment of Inertia	$I = \sum m_i r_i^2$	$\text{kg m}^2$	$[ML^2]$
Radius of Gyration	$k = \sqrt{I / M}$	$\text{m}$	$[L]$
Rotational KE	$K = \frac{1}{2}I\omega^2$	$\text{J}$	$[ML^2T^{-2}]$
Rotational Work	$W = \int \tau d\theta$	$\text{J}$	$[ML^2T^{-2}]$
Rotational Power	$P = \tau \omega$	$\text{W}$	$[ML^2T^{-3}]$
Newton's 2nd Law	$\vec{\tau}_{ext} = \frac{d\vec{L}}{dt} = I\vec{\alpha}$ (Fixed Axis)	$\text{N m}$	$[ML^2T^{-2}]$
Mech. Advantage	$M.A. = \frac{F_1}{F_2} = \frac{d_2}{d_1}$	Dimensionless	N/A

11Conditions & Limitations

Rigid Body Assumption: Equations like $v = \omega r$ specifically apply to objects assumed to undergo zero deformation ( $r$ doesn't change).
Kinematic Equations: $\omega = \omega_0 + \alpha t$ and related rotational kinematic equations are ONLY valid if the angular acceleration ( $\alpha$ ) is constant.
$\vec{L} = I \vec{\omega}$ Vector Form: This scalar-to-vector parallel form strictly applies ONLY if the axis of rotation is an axis of symmetry for the rigid body. Otherwise, the component of angular momentum perpendicular to the axis is non-zero, making $\vec{L}$ non-parallel to $\vec{\omega}$ .
$\tau = I \alpha$ : This scalar equation assumes fixed axis rotation where perpendicular components of torque are entirely cancelled out by the bearings/supports (forces of constraint).
CM = CG: The Center of Mass and Center of Gravity coincide ONLY in a uniform gravitational field (i.e., $g$ is constant over the extent of the body).

12COMMON MISCONCEPTIONS & SIGN CONVENTIONS

Vector Order is Crucial

$\vec{v} = \vec{\omega} \times \vec{r}$ (Correct) vs $\vec{v} = \vec{r} \times \vec{\omega}$ (Incorrect: gives opposite sign).

$\vec{\tau} = \vec{r} \times \vec{F}$ (Correct) vs $\vec{\tau} = \vec{F} \times \vec{r}$ (Incorrect).

$\vec{l} = \vec{r} \times \vec{p}$ (Correct).

Angle constraint for Cross Products: When applying the right-hand rule or the magnitude formula $ab \sin\theta$ , you must strictly use the smaller angle between the two vectors (i.e., $\theta < 180^\circ$ ).
Torque Reference Point: A body can have zero net torque about its Center of Mass, but non-zero net torque about a different origin unless the net external force on the body is also zero ( $\sum \vec{F}_{ext} = 0$ ). If net force is zero, torque is invariant of origin.
Central Forces Assumption in Rigid Bodies: We assume that internal torques sum to zero ( $\sum \vec{\tau}_{int} = 0$ ). This critically requires the strong form of Newton's Third Law—that internal action-reaction forces act exactly along the straight line joining the two interacting particles.
Variable Gravity Edge Case: If an object is so massive/extended that $g$ varies from one part to the other, the CM (mass distribution) and the CG (gravitational torque balance) will decouple and exist at different locations.
Choice of $\vec{r}$ in $\vec{v} = \vec{\omega} \times \vec{r}$ : $\vec{r}$ does NOT have to be the perpendicular distance from the axis. It can be the position vector originating from any arbitrary point chosen on the axis of rotation.
Internal Forces Illusion: JEE Tip Students often think internal explosions alter the total momentum or CM trajectory. They do NOT. Sum of internal forces is mathematically constrained to zero.
Lever Sign Conventions: When evaluating the Principle of Moments ( $\sum \tau = 0$ ), the standard convention is that anticlockwise moments are positive (+) and clockwise moments are negative (-).
Torque constraints for Fixed-Axis Problems: When solving fixed-axis dynamics, any forces that act parallel to the rotation axis, or forces whose line of action passes directly through the axis, generate zero effective torque along the rotation axis and do no rotational work.

13Previous Year JEE Topics

Conservation of Angular Momentum ( $L_i = L_f$ ): Frequently tested via collisions involving rotating bodies, or objects dropping onto rotating discs (e.g., bug walking on a turntable, clay dropping on a spinning wheel).
Explosion & Center of Mass Path: Calculating the fragment trajectories of a projectile exploiting the fact that $X_{cm}$ and $Y_{cm}$ follow standard projectile equations.
Translational + Rotational Equilibrium: Ladder leaning on a wall problems (determining normal reaction and friction forces via $\sum \tau = 0$ and $\sum F_x = \sum F_y = 0$ ).
Angular Momentum of a particle in straight-line uniform motion: Using $l = mvr_\perp$ to find constant $L$ about a specified origin.
Torque Vector Evaluation: Calculating $\vec{\tau} = \vec{r} \times \vec{F}$ using determinants. Ensure $\vec{r}$ is precisely defined as $\vec{r}_{final} - \vec{r}_{origin}$ .

14JEE Traps

Angular Momentum Vector Alignment

✕Misconception

The total angular momentum vector $\vec{L}$ of a rotating rigid body is always parallel to its angular velocity vector $\vec{\omega}$ .

✓Reality

$\vec{L}$ is only parallel to $\vec{\omega}$ if the axis of rotation is an axis of symmetry of the body. For asymmetric bodies, $\vec{L}$ has a non-zero perpendicular component ( $\vec{L}_\perp$ ).

Torque of a Couple

✕Misconception

The net torque produced by a couple depends on the reference point or origin chosen to calculate it.

✓Reality

The moment of a couple is an invariant vector. It is absolutely independent of the choice of origin or reference point.

Particle in Straight Line Motion

✕Misconception

A particle moving in a straight line with constant velocity has zero angular momentum.

✓Reality

It has a constant, non-zero angular momentum about any origin that does not lie directly on its line of motion. $\vec{l} = mvr_\perp$ , and this perpendicular distance remains constant.

Internal Forces and Total Kinetic Energy

✕Misconception

Because internal forces strictly sum to zero, they can never change the kinetic energy of a system.

✓Reality

Internal forces (like an explosion) strictly cannot change linear momentum or CM path. However, they do work and absolutely can change the kinetic energy of the system.

Centre of Gravity vs. Centre of Mass

✕Misconception

Centre of Mass (CM) and Centre of Gravity (CG) are two names for the exact same physical point.

✓Reality

CM depends only on mass distribution. CG depends on gravitational torque. They coincide only in a uniform gravitational field.

The Radius of Gyration

✕Misconception

The radius of gyration ( $k$ ) is a fixed geometric constant for a given rigid body.

✓Reality

Radius of gyration is entirely dependent on the choice and orientation of the axis of rotation. If you change the axis, $k$ changes.

Equilibrium Constraints

✕Misconception

If the net external force on a rigid body is zero ( $\sum \vec{F} = 0$ ), the body must be in absolute mechanical equilibrium.

✓Reality

The body is only in translational equilibrium. It could still have a net non-zero torque (like a couple) acting on it, placing it in partial equilibrium where it undergoes pure angular acceleration.

Cross Product Mirror Reflection

✕Misconception

Because normal vectors reverse their sign in a mirror image ( $\vec{r} \to -\vec{r}$ ), their cross product must also reverse sign.

✓Reality

The vector product of two vectors does not change sign under reflection. Because both vectors flip, $(-\vec{a}) \times (-\vec{b}) = \vec{a} \times \vec{b}$ .

Constant Angular Acceleration Assumption

✕Misconception

You can always use $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ or $\omega^2 = \omega_0^2 + 2\alpha\theta$ to solve rotational kinematics.

✓Reality

These equations are strictly limited to cases where angular acceleration ( $\alpha$ ) is perfectly uniform/constant.

Conservation of Angular Momentum Trigger

✕Misconception

Angular momentum is conserved only if no forces are acting on the system.

✓Reality

Angular momentum is conserved as long as the net external torque is zero ( $\sum \vec{\tau}_{ext} = 0$ ). Tremendous forces can act, but as long as their line of action passes through the reference point/axis, $\vec{L}$ remains constant.