How much of JEE Main comes from Rotational Motion?

JEEGenie estimates about 5% of JEE Main marks come from Rotational Motion, based on recent papers. The share varies from year to year.

Which formulas matter most in Rotational Motion?

The Rotational Motion sheet on this page has 22 formula entries, starting with Angular Displacement, Velocity & Acceleration, Rotational Kinematics Equations, Moment of Inertia (I), Parallel Axis Theorem, Perpendicular Axis Theorem. Each one comes with when to use it and the trap to avoid.

What topics does Rotational Motion cover for JEE?

Rotational Motion covers Centre of Mass, Torque and Angular Momentum, Moment of Inertia, Rotational Kinematics, Rolling Motion, Conservation of Angular Momentum.

How should I revise Rotational Motion before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Rotational Motion questions on JEEGenie.

How many previous-year questions are there on Rotational Motion?

JEEGenie has 623 previous-year questions on Rotational Motion, and this page shows a few with worked explanations.

Is practice on Rotational Motion free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Rotational Motion for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 5% of JEE Main marks come from Rotational Motion. This page covers 22 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Confusing rigid body with point mass; rigid body has distributed mass.

Key points to know

Rigid Body & Rotational Motion. A rigid body has a fixed shape; its particles maintain fixed distances. Rotational motion occurs when all particles move in circles about a common axis.
Angular Displacement, Velocity & Acceleration. Angular displacement $θ$ is angle rotated; angular velocity $ω = d θ / d t$ ; angular acceleration $α = d ω / d t$ . Angular velocity and angular acceleration are vectors along the axis (right-hand rule); a finite angular displacement is not a vector (only an infinitesimal one is).
Moment of Inertia (I). Rotational inertia: $I = \sum m_{i} r_{i}^{2}$ for discrete masses, $I = \int r^{2} d m$ for continuous bodies. Depends on mass distribution and axis.
Standard Moments of Inertia. Common bodies: Ring (axis through center): $M R^{2}$ ; Disc (axis through center): $\frac{1}{2} M R^{2}$ ; Solid sphere: $\frac{2}{5} M R^{2}$ ; Rod (center): $\frac{1}{12} M L^{2}$ ; Rod (end): $\frac{1}{3} M L^{2}$ .
Torque ( $τ$ ). Torque is the rotational analogue of force: $τ = r \times F$ (cross product). Magnitude: $τ = r F \sin θ$ (lever arm times force).
Angular Momentum ( $L$ ). Angular momentum of a particle: $L = r \times p$ ; for rigid body: $L = I ω$ . Conserved if net external torque is zero.
Pulley Problems with Rotation. For a pulley with mass $M$ and radius $R$ , torque $τ = (T_{1} - T_{2}) R = I α$ , and $a = R α$ .
Radius of Gyration ( $k$ ). Distance from axis where all mass could be concentrated to give same $I$ : $I = M k^{2}$ .
Angular Impulse. Change in angular momentum equals angular impulse: $\int τ d t = Δ L$ .
Torque as a Vector. Torque is a pseudovector given by cross product $\vec{τ} = \vec{r} \times \vec{F}$ . Direction by right-hand rule.
Rolling on Inclined Plane. Acceleration $a = \frac{g \sin θ}{1 + I / (M R^{2})}$ . Time to reach bottom independent of mass but depends on shape.

Formula sheet

Name	Formula	When to use
Angular Displacement, Velocity & Acceleration	$ω = \frac{d θ}{d t}$ , $α = \frac{d ω}{d t}$	Describing rotational kinematics of a rigid body.
Rotational Kinematics Equations	$ω = ω_{0} + α t$ ; $θ = ω_{0} t + \frac{1}{2} α t^{2}$ ; $ω^{2} = ω_{0}^{2} + 2 α θ$	Problems with constant torque or angular acceleration.
Moment of Inertia (I)	$I = \sum m_{i} r_{i}^{2}$ or $I = \int r^{2} d m$	Calculating rotational kinetic energy, torque, angular momentum.
Parallel Axis Theorem	$I = I_{c m} + M d^{2}$	Find $I$ about an axis offset from center of mass.
Perpendicular Axis Theorem	$I_{z} = I_{x} + I_{y}$	Find $I$ of a lamina about an axis perpendicular to its plane.
Standard Moments of Inertia	$I_{ring} = M R^{2}$ , $I_{disc} = \frac{1}{2} M R^{2}$ , $I_{sphere} = \frac{2}{5} M R^{2}$ , $I_{rod,center} = \frac{1}{12} M L^{2}$ , $I_{rod,end} = \frac{1}{3} M L^{2}$	Direct substitution in rotational dynamics problems.
Torque ( $τ$ )	$\vec{τ} = \vec{r} \times \vec{F}$ , $∣ τ ∣ = r F \sin θ$	Calculating rotational effect of a force.
Newton's Second Law for Rotation	$τ_{net} = I α$	Problems with angular acceleration due to torques.
Angular Momentum ( $L$ )	$\vec{L} = \vec{r} \times \vec{p}$ (particle), $L = I ω$ (rigid body)	Conservation of angular momentum in collisions or when torque is zero.
Conservation of Angular Momentum	$I_{1} ω_{1} = I_{2} ω_{2}$ (if $τ_{ext} = 0$ )	Problems where shape or axis changes without external torque.
Rotational Kinetic Energy	$K_{rot} = \frac{1}{2} I ω^{2}$	Energy conservation in rotational systems.
Rolling Without Slipping	$v_{cm} = R ω$ , $K = \frac{1}{2} M v_{cm}^{2} + \frac{1}{2} I_{cm} ω^{2}$	Problems with wheels, balls, cylinders rolling on surfaces.
Work-Energy Theorem for Rotation	$W = \int τ d θ = \frac{1}{2} I ω_{f}^{2} - \frac{1}{2} I ω_{i}^{2}$	Problems with variable torque or angular displacement.
Pulley Problems with Rotation	$τ = (T_{1} - T_{2}) R = I α$ , $a = R α$	Problems with massive pulleys and hanging masses.
Radius of Gyration ( $k$ )	$k = \sqrt{\frac{I}{M}}$	Comparing rotational inertia of different bodies.
Equilibrium of Rigid Bodies	$\sum \vec{F} = 0$ , $\sum \vec{τ} = 0$	Problems with ladders, beams, or objects in balance.
Angular Impulse	$\int τ d t = Δ L = I Δ ω$	Problems with collisions or sudden torques.
Rotational Collisions	$L_{initial} = L_{final}$	Problems where a particle hits a rotating body or two rotating bodies collide.
Torque as a Vector	$\vec{τ} = \vec{r} \times \vec{F}$	Calculating torque in 3D or when force is not perpendicular.
Relation Between Torque and Angular Momentum	$\vec{τ} = \frac{d \vec{L}}{d t}$	Deriving conservation or analyzing time-varying rotation.
Moment of Inertia of Composite Bodies	$I_{total} = \sum I_{i}$	Problems with multiple masses or shapes attached.
Rolling on Inclined Plane	$a = \frac{g \sin θ}{1 + \frac{I}{M R^{2}}}$	Comparing rolling times or speeds of different objects.

Open the Rotational Motion formula page

What JEE asks

Rigid Body & Rotational Motion. JEE often tests if you recognize that a rigid body's angular acceleration is same for all points.
Angular Displacement, Velocity & Acceleration. JEE asks to relate linear and angular quantities: $v = r ω$ , $a_{t} = r α$ .
Rotational Kinematics Equations. JEE often gives initial angular velocity and asks for time to stop or number of rotations.
Moment of Inertia (I). JEE asks to compute $I$ for composite bodies or using parallel/perpendicular axis theorems.
Parallel Axis Theorem. JEE often uses this for rods, discs, rings about edge or tangent.
Perpendicular Axis Theorem. JEE uses this for discs, rings, squares, etc. to find $I$ about diameter or perpendicular axis.
Standard Moments of Inertia. JEE expects you to recall these instantly for composite bodies or energy conservation.
Torque ( $τ$ ). JEE often asks net torque about a point, especially in equilibrium or dynamics.
Newton's Second Law for Rotation. JEE combines with linear dynamics for rolling or pulley problems.
Angular Momentum ( $L$ ). JEE tests conservation in problems like a person on rotating platform changing arms.
Conservation of Angular Momentum. JEE often uses this for ice skater, collapsing star, or rotating stool.
Rotational Kinetic Energy. JEE combines with work-energy theorem for rolling down incline.
Rolling Without Slipping. JEE asks to find acceleration down incline using $a = \frac{g \sin θ}{1 + I / (M R^{2})}$ .
Work-Energy Theorem for Rotation. JEE often uses this for rotating systems with constant torque or energy conservation.
Pulley Problems with Rotation. JEE asks to find acceleration of masses or tension difference.
Radius of Gyration ( $k$ ). JEE may ask to find $k$ for composite bodies or use in rolling motion.
Equilibrium of Rigid Bodies. JEE often asks to find forces or angles in equilibrium with multiple supports.
Angular Impulse. JEE tests this in rotational collisions (e.g., a disc hitting a rod).
Rotational Collisions. JEE often asks for final angular velocity after a particle sticks to a disc or rod.
Torque as a Vector. JEE may ask to find torque vector given force and position vectors.
Relation Between Torque and Angular Momentum. JEE uses this to show conservation when $τ = 0$ .
Moment of Inertia of Composite Bodies. JEE often gives a rod with masses at ends or a disc with holes.
Rolling on Inclined Plane. JEE asks which object reaches bottom first (lowest $I / (M R^{2})$ wins).

Common traps

Confusing rigid body with point mass; rigid body has distributed mass. Rigid Body & Rotational Motion
Forgetting direction: use right-hand rule for vector nature. Angular Displacement, Velocity & Acceleration
Using these only when $α$ is constant; not for variable acceleration. Rotational Kinematics Equations
Assuming $I$ is same for all axes; it changes with axis. Moment of Inertia (I)
Using it for axes not parallel; only for parallel axes. Parallel Axis Theorem
Only applicable for planar bodies (2D). Perpendicular Axis Theorem
Mixing up disc and ring formulas. Standard Moments of Inertia
Using $r$ as distance from axis, not perpendicular distance. Torque ( $τ$ )
Forgetting to take net torque, not just one torque. Newton's Second Law for Rotation
Using $L = I ω$ for particles; only for rigid bodies about axis. Angular Momentum ( $L$ )
Applying when external torque is present (e.g., friction). Conservation of Angular Momentum
Forgetting to add translational KE for rolling objects. Rotational Kinetic Energy
Assuming friction is always present; pure rolling can occur without friction on horizontal surface. Rolling Without Slipping
Forgetting that torque must be about the axis of rotation. Work-Energy Theorem for Rotation

Topics in Rotational Motion

Centre of Mass
Torque and Angular Momentum
Moment of Inertia
Rotational Kinematics
Rolling Motion
Conservation of Angular Momentum

Previous-year questions on Rotational Motion

JEEGenie has 623 previous-year questions on Rotational Motion. A few worked examples follow.

Year	Questions
2026	34
2025	34
2024	32
2023	38
2022	38
2021	59
2020	49
2019	79
2018	17
2017	21
2016	12
2015	11
2014	8
2013	11
2012	15
2011	12
2010	6
2009	7
2008	14
2007	13
2006	11
2005	10
2004	6
2003	8
2002	10
2001	3
2000	5
1999	5
1998	5
1997	8
1996	1
1995	2
1994	3
1993	3
1992	4
1991	3
1990	3
1989	3
1988	2
1987	2
1986	3
1985	2
1984	3
1983	1
1982	2
1981	1
1980	1
1979	2
1978	1

Question 1 medium

The position of an object having mass 0.1 kg as a function of time

t

is given as

\vec{r} = (10 t^{2} \hat{i} + 5 t^{3} \hat{j})

m. At

t = 1

s, which of the following statements are correct? A. The linear momentum

\vec{p} = (2 \hat{i} + 1.5 \hat{j})

kg·m/s. B. The force acting on the object

\vec{F} = (2 \hat{i} + 3 \hat{j})

N. C. The angular momentum of the object about its origin

\vec{L} = 15 \hat{k}

J·s. D. The torque acting on the object about its origin

\vec{τ} = 20 \hat{k}

N·m. Choose the correct answer from the options given below:

A, B and C only
B, C and D only
A, C and D only
A, B and D only

Answer: D

Explanation.

\vec{r} = (10 t^{2}) \hat{i} + (5 t^{3}) \hat{j}

\vec{v} = \frac{d \vec{r}}{d t} = 20 t \hat{i} + 15 t^{2} \hat{j}

t = 1

\vec{v} = 20 \hat{i} + 15 \hat{j}

Mass

m = 0.1

kg, so

\vec{p} = m \vec{v} = 0.1 (20 \hat{i} + 15 \hat{j}) = (2 \hat{i} + 1.5 \hat{j}) kg \cdot m/s

Statement A is correct. Acceleration:

\vec{a} = \frac{d \vec{v}}{d t} = 20 \hat{i} + 30 t \hat{j}

t = 1

\vec{a} = 20 \hat{i} + 30 \hat{j}

\vec{F} = m \vec{a} = 0.1 (20 \hat{i} + 30 \hat{j}) = (2 \hat{i} + 3 \hat{j}) N

Statement B is correct. Angular momentum about origin:

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

t = 1

\vec{r} = 10 \hat{i} + 5 \hat{j}

\vec{p} = 2 \hat{i} + 1.5 \hat{j}

\vec{L} = (10 \hat{i} + 5 \hat{j}) \times (2 \hat{i} + 1.5 \hat{j})

= 10 (1.5) \hat{k} - 5 (2) \hat{k} = 15 \hat{k} - 10 \hat{k} = 5 \hat{k} J \cdot s

Statement C is incorrect. Torque:

\vec{τ} = \vec{r} \times \vec{F}

t = 1

\vec{F} = 2 \hat{i} + 3 \hat{j}

\vec{τ} = (10 \hat{i} + 5 \hat{j}) \times (2 \hat{i} + 3 \hat{j})

= 10 (3) \hat{k} - 5 (2) \hat{k} = 30 \hat{k} - 10 \hat{k} = 20 \hat{k} N \cdot m

Statement D is correct. Correct statements: A, B, D.

D

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 easy

Which of the following are correct expression for torque acting on a body? A.

\vec{τ} = \vec{r} \times \vec{L}

\vec{τ} = \frac{d}{d t} (\vec{r} \times \vec{p})

\vec{τ} = \vec{r} \times \frac{d \vec{p}}{d t}

\vec{τ} = I \vec{α}

\vec{τ} = \vec{r} \times \vec{F}

(

\vec{r} =

position vector;

\vec{p} =

linear momentum;

\vec{L} =

angular momentum;

\vec{α} =

angular acceleration;

I =

moment of inertia;

\vec{F} =

force;

t =

time) Choose the correct answer from the options given below:

A, B, D and E Only
C and D Only
B, C, D and E Only
B, D and E Only

Answer: C

Explanation. Torque is defined as

\vec{τ} = \vec{r} \times \vec{F}

(E). Since

\vec{F} = \frac{d \vec{p}}{d t}

, we also have

\vec{τ} = \vec{r} \times \frac{d \vec{p}}{d t}

(C). From

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

, differentiating:

\frac{d \vec{L}}{d t} = \frac{d}{d t} (\vec{r} \times \vec{p}) = \vec{τ}

(B). For rigid bodies,

\vec{τ} = I \vec{α}

(D). Option A (

\vec{r} \times \vec{L}

) is incorrect because

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

, so

\vec{r} \times \vec{L} = \vec{r} \times (\vec{r} \times \vec{p}) \neq \vec{τ}

. Thus correct are B, C, D, E. The answer is option C.

From NTA/IIT papers: JEE Main 2025 · 4 Apr Shift 1

Question 3 medium

A solid sphere and a hollow cylinder roll up without slipping on same inclined plane with same initial speed

v

. The sphere and the cylinder reaches upto maximum heights

h_{1}

and

h_{2}

respectively, above the initial level. The ratio

h_{1} : h_{2}

\frac{n}{10}

. The value of

n

is __________.

Answer: 7

Explanation. For rolling without slipping, total kinetic energy is

K = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}

Since

v = r ω

K = \frac{1}{2} m v^{2} + \frac{1}{2} I \frac{v^{2}}{r^{2}} = \frac{1}{2} m v^{2} (1 + \frac{I}{m r^{2}})

This kinetic energy is converted into gravitational potential energy at maximum height

h

m g h = \frac{1}{2} m v^{2} (1 + \frac{I}{m r^{2}})

So,

h = \frac{v^{2}}{2 g} (1 + \frac{I}{m r^{2}})

For a solid sphere,

I = \frac{2}{5} m r^{2}

Thus,

h_{1} = \frac{v^{2}}{2 g} (1 + \frac{2}{5}) = \frac{v^{2}}{2 g} \cdot \frac{7}{5}

For a hollow cylinder,

I = m r^{2}

Thus,

h_{2} = \frac{v^{2}}{2 g} (1 + 1) = \frac{v^{2}}{2 g} \cdot 2

Therefore,

\frac{h_{1}}{h_{2}} = \frac{\frac{7}{5}}{2} = \frac{7}{10}

Given that

\frac{h_{1}}{h_{2}} = \frac{n}{10}

we get

\frac{n}{10} = \frac{7}{10}

n = 7

Final answer:

7.0

From NTA/IIT papers: JEE Main 2024 · 4 Apr Shift 1

Question 4 hard

A thin circular coin of mass

5

gm and radius

4 / 3

cm is initially in a horizontal

x y

-plane. The coin is tossed vertically up (

+ z

direction) by applying an impulse of

\sqrt{\frac{π}{2}} \times 10^{- 2}

N-s at a distance

2 / 3

cm from its center. The coin spins about its diameter and moves along the

+ z

direction. By the time the coin reaches back to its initial position, it completes

n

rotations. The value of

n

is ____. [Given: The acceleration due to gravity

g = 10

m s

^{- 2}

]

Answer: 30

Explanation. We treat the coin as a disc. 1. Mass:

m = 5 \times 10^{- 3} kg, R = \frac{4}{3} \times 10^{- 2} m

2. Moment of inertia about a diameter:

I = \frac{1}{4} m R^{2} = \frac{1}{4} (5 \times 10^{- 3}) {(\frac{4}{3} \times 10^{- 2})}^{2}

= \frac{5 \times 10^{- 3}}{4} \cdot \frac{16}{9} \times 10^{- 4} = \frac{20}{9} \times 10^{- 7} {kg m}^{2}

3. Impulse

J = \sqrt{\frac{π}{2}} \times 10^{- 2}

N-s applied at distance

d = \frac{2}{3} \times 10^{- 2} m

from center gives angular impulse:

J d = I ω

ω = \frac{J d}{I} = \frac{\sqrt{π / 2} \times 10^{- 2} \cdot \frac{2}{3} \times 10^{- 2}}{\frac{20}{9} \times 10^{- 7}}

4. Simplify:

ω = \frac{\sqrt{π / 2} \cdot \frac{2}{3} \times 10^{- 4}}{\frac{20}{9} \times 10^{- 7}} = \frac{\sqrt{π / 2} \cdot \frac{2}{3}}{\frac{20}{9}} \times 10^{3}

= \sqrt{\frac{π}{2}} \cdot \frac{2}{3} \cdot \frac{9}{20} \times 10^{3} = \sqrt{\frac{π}{2}} \cdot \frac{3}{10} \times 10^{3} = 300 \sqrt{\frac{π}{2}} rad/s

5. Linear impulse gives upward velocity:

v = \frac{J}{m} = \frac{\sqrt{π / 2} \times 10^{- 2}}{5 \times 10^{- 3}} = 2 \sqrt{\frac{π}{2}} m/s

6. Time to go up and return to initial position:

T = \frac{2 v}{g} = \frac{2 \cdot 2 \sqrt{π / 2}}{10} = \frac{4 \sqrt{π / 2}}{10} = \frac{2}{5} \sqrt{\frac{π}{2}} s

7. Number of rotations:

n = \frac{ω T}{2 π} = \frac{300 \sqrt{π / 2} \cdot \frac{2}{5} \sqrt{π / 2}}{2 π}

= \frac{300 \cdot \frac{2}{5} \cdot \frac{π}{2}}{2 π} = \frac{300 \cdot \frac{π}{5}}{2 π} = \frac{60 π}{2 π} = 30

30

From NTA/IIT papers: JEE Advanced 2023

Question 5 hard

At time

t = 0

, a disk of radius

1 m

starts to roll without slipping on a horizontal plane with an angular acceleration of

α = \frac{2}{3} r a d s^{- 2}

. A small stone is stuck to the disk. At

t = 0

, it is at the contact point of the disk and the plane. Later, at time

t = \sqrt{π} s

, the stone detaches itself and flies off tangentially from the disk. The maximum height (in

m

) reached by the stone measured from the plane is

\frac{1}{2} + \frac{x}{10}

. The value of

x

is ______. [Take

g = 10 m s^{- 2}

Answer: 0.52

Explanation. **What's really being asked** Find how high a stone flies after it leaves a spinning disk that is speeding up, when it detaches at a specific time — and then use that height to solve for the unknown number

x

. **The one concept you need** **Projectile motion** — after the stone leaves the disk, it moves only under gravity (no other forces), so its path is a parabola. Its initial velocity is exactly the tangential velocity of the disk at the instant of detachment, and its initial height is the height of that point on the disk above the ground. **Step-by-step** 1. **Find the angular speed at detachment.** The disk starts from rest (

ω_{0} = 0

) and has constant angular acceleration

α = \frac{2}{3} {rad/s}^{2}

. Angular speed after time

t

is:

ω = ω_{0} + α t = 0 + \frac{2}{3} \cdot \sqrt{π} = \frac{2 \sqrt{π}}{3} rad/s .

2. **Find the tangential speed of the stone.** For a point on a rotating disk, tangential speed

v = ω R

. Here

R = 1 m

, so:

v = ω \cdot 1 = \frac{2 \sqrt{π}}{3} m/s .

3. **Find the position of the stone at detachment.** The stone starts at the bottom contact point at

t = 0

. The disk rolls without slipping, so the angle rotated by time

t

is:

θ = \frac{1}{2} α t^{2} = \frac{1}{2} \cdot \frac{2}{3} \cdot (\sqrt{π})^{2} = \frac{1}{2} \cdot \frac{2}{3} \cdot π = \frac{π}{3} rad .

So the stone has rotated

π / 3

radians (60

^{\circ}

) from the bottom point. 4. **Find the height of the stone above the ground at detachment.** The center of the disk is at height

R = 1 m

. The stone is on the rim, at an angle

θ = π / 3

measured from the downward vertical (because it started at the bottom). Its vertical position relative to the center is:

y_{rel} = - R \cos θ = - \cos (π / 3) = - \frac{1}{2} .

(Negative because it is below the center.) So absolute height above ground:

h_{0} = 1 - \frac{1}{2} = \frac{1}{2} m .

5. **Find the direction of the velocity at detachment.** The stone moves tangentially to the circle. At the point that is 60

^{\circ}

from the bottom (i.e., 60

^{\circ}

from the downward vertical), the tangent direction makes an angle with the horizontal. The radius vector from center to stone points downward and 60

^{\circ}

to the right (since the disk rolls to the right). The tangent is perpendicular to the radius. The angle of the velocity above the horizontal is

60^{\circ}

(because the radius is 60

^{\circ}

below the horizontal, so the tangent is 60

^{\circ}

above the horizontal). So:

v_{x} = v \cos 60^{\circ} = \frac{2 \sqrt{π}}{3} \cdot \frac{1}{2} = \frac{\sqrt{π}}{3},

v_{y} = v \sin 60^{\circ} = \frac{2 \sqrt{π}}{3} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3 π}}{3} .

6. **Find the maximum height reached above the detachment point.** For projectile motion, the extra height gained after launch is:

Δ h = \frac{v_{y}^{2}}{2 g} .

(This comes from energy: kinetic energy in vertical direction converts to potential energy.) Substitute:

Δ h = \frac{{(\frac{\sqrt{3 π}}{3})}^{2}}{2 \cdot 10} = \frac{\frac{3 π}{9}}{20} = \frac{π / 3}{20} = \frac{π}{60} .

7. **Add the initial height to get total maximum height from ground.**

H_{\max} = h_{0} + Δ h = \frac{1}{2} + \frac{π}{60} .

8. **Compare with given form.** The problem says

H_{\max} = \frac{1}{2} + \frac{x}{10}

. So:

\frac{x}{10} = \frac{π}{60} ⟹ x = \frac{10 π}{60} = \frac{π}{6} .

Using

π \approx 3.1416

x \approx \frac{3.1416}{6} \approx 0.5236 \approx 0.52.

**The trap** Many students forget that the stone starts at the *bottom* of the disk, not at the center or at the side. They incorrectly take the initial height as

R = 1

or as

R (1 + \cos θ)

. Also, they often take the velocity direction as purely horizontal, forgetting that at 60

^{\circ}

from the bottom, the tangent is inclined. Always draw the circle, mark the starting point at the bottom, rotate by the angle, and then find both the height and the velocity components from the geometry. **Remember** For a point on a rolling disk, first find

θ = \frac{1}{2} α t^{2}

, then use geometry to get height and velocity components — never guess the direction.

From NTA/IIT papers: JEE Advanced 2022

Frequently asked questions

How much of JEE Main comes from Rotational Motion?: JEEGenie estimates about 5% of JEE Main marks come from Rotational Motion, based on recent papers. The share varies from year to year.
Which formulas matter most in Rotational Motion?: The Rotational Motion sheet on this page has 22 formula entries, starting with Angular Displacement, Velocity & Acceleration, Rotational Kinematics Equations, Moment of Inertia (I), Parallel Axis Theorem, Perpendicular Axis Theorem. Each one comes with when to use it and the trap to avoid.
What topics does Rotational Motion cover for JEE?: Rotational Motion covers Centre of Mass, Torque and Angular Momentum, Moment of Inertia, Rotational Kinematics, Rolling Motion, Conservation of Angular Momentum.
How should I revise Rotational Motion before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Rotational Motion questions on JEEGenie.
How many previous-year questions are there on Rotational Motion?: JEEGenie has 623 previous-year questions on Rotational Motion, and this page shows a few with worked explanations.
Is practice on Rotational Motion free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.