Rotational Motion — Concepts, Formulas & Solved Numericals

Rotational Motion — The Physics Behind Every Spinning Thing

When a cricket ball spins in the air or a spinning top refuses to fall, rotational motion is at work. This chapter is essentially “Newton’s Laws, but for spinning objects” — and once you see that analogy, everything clicks.

Most Class 11 students treat this chapter as a memory exercise. That’s a mistake. Rotational motion has a beautiful one-to-one correspondence with linear motion: mass ↔ moment of inertia, force ↔ torque, linear velocity ↔ angular velocity. Learn the correspondence, and you automatically know the formulas.

This chapter carries high weightage in JEE Main (1-2 questions every session) and is a favourite for CBSE 5-mark numericals. The concepts also form the foundation for angular momentum conservation problems that appear in JEE Advanced.

Key Terms and Definitions

Rigid body — An object where the distance between any two particles remains constant during motion. Real objects aren’t perfectly rigid, but we treat them as such in this chapter. A disc, rod, ring, or sphere — all rigid bodies.

Angular displacement ( $\theta$ ) — The angle swept by any line fixed in the body. Measured in radians. One full revolution = $2\pi$ radians.

Angular velocity ( $\omega$ ) — Rate of change of angular displacement.

\omega = \frac{d\theta}{dt}

Unit: rad/s. For uniform circular motion, $\omega = 2\pi f$ where $f$ is frequency in Hz.

Angular acceleration ( $\alpha$ ) — Rate of change of angular velocity.

\alpha = \frac{d\omega}{dt}

Unit: rad/s². Positive $\alpha$ means the body is speeding up (in the direction of rotation), negative means slowing down.

Torque ( $\tau$ ) — The rotational equivalent of force. A force applied at a perpendicular distance from the axis creates torque.

\tau = r \times F = rF\sin\theta

Where $\theta$ is the angle between the position vector $\vec{r}$ and force $\vec{F}$ .

Moment of Inertia ( $I$ ) — The rotational equivalent of mass. Unlike mass, it depends on both the mass distribution AND the choice of axis.

I = \sum m_i r_i^2

For a continuous body: $I = \int r^2 \, dm$

The Linear-Rotational Correspondence

This is the single most useful table in all of Class 11 mechanics. Memorise this once and you’ll never forget the rotational formulas.

Linear Quantity	Symbol	Rotational Equivalent	Symbol
Displacement	$x$	Angular displacement	$\theta$
Velocity	$v$	Angular velocity	$\omega$
Acceleration	$a$	Angular acceleration	$\alpha$
Mass	$m$	Moment of inertia	$I$
Force	$F$	Torque	$\tau$
Momentum	$p = mv$	Angular momentum	$L = I\omega$
Kinetic energy	$\frac{1}{2}mv^2$	Rotational KE	$\frac{1}{2}I\omega^2$
Newton’s 2nd law	$F = ma$	Rotational 2nd law	$\tau = I\alpha$

Whenever you’re stuck on a rotational formula, think: “what’s the linear version of this?” Then replace $m \to I$ , $v \to \omega$ , $a \to \alpha$ , $F \to \tau$ . Works almost every time.

Moment of Inertia — Standard Results

You cannot derive all of these in an exam. Learn the standard results and the conditions (which axis, which body).

Thin ring (axis through centre, perpendicular to plane):

I = MR^2

Disc / solid cylinder (axis through centre, perpendicular to plane):

I = \frac{1}{2}MR^2

Solid sphere (axis through centre):

I = \frac{2}{5}MR^2

Hollow sphere (axis through centre):

I = \frac{2}{3}MR^2

Thin rod (axis through centre, perpendicular to length):

I = \frac{ML^2}{12}

Thin rod (axis through one end, perpendicular to length):

I = \frac{ML^2}{3}

Parallel Axis Theorem

When you know $I_{cm}$ (moment of inertia about the axis through the centre of mass), the moment of inertia about any parallel axis at distance $d$ is:

I = I_{cm} + Md^2

Perpendicular Axis Theorem

Only for flat (planar) bodies. If $x$ and $y$ axes lie in the plane of the body, and $z$ is perpendicular to the plane:

I_z = I_x + I_y

The perpendicular axis theorem only works for 2D planar objects (disc, ring, lamina). Never apply it to a sphere, cylinder, or rod. This is a very common error in CBSE board exams.

Equations of Rotational Kinematics

Since $\alpha$ (angular acceleration) plays the same role as $a$ (linear acceleration), the kinematic equations are identical in form:

\omega = \omega_0 + \alpha t

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

\omega^2 = \omega_0^2 + 2\alpha\theta

\theta = \frac{(\omega_0 + \omega)}{2} \cdot t

These are directly analogous to $v = u + at$ , $s = ut + \frac{1}{2}at^2$ , etc.

Angular Momentum and Its Conservation

Angular momentum $L = I\omega$ for a rigid body rotating about a fixed axis.

For a particle: $\vec{L} = \vec{r} \times \vec{p}$

The most important result for JEE:

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

When $\tau_{net} = 0$ (no external torque), angular momentum is conserved:

I_1\omega_1 = I_2\omega_2

This is why a figure skater spins faster when they pull their arms in — they reduce $I$ , so $\omega$ must increase to keep $L$ constant.

Angular momentum conservation is a JEE Advanced favourite. It appears in problems involving a person walking on a rotating platform, a ball hitting a rod, or a satellite changing orbit. In JEE Main 2023 April Shift 2, a direct numerical on $I_1\omega_1 = I_2\omega_2$ was asked — straightforward if you identify the system correctly.

Rolling Motion — Pure Rolling Condition

A body rolls without slipping when the contact point has zero velocity. This gives us:

v_{cm} = R\omega

Where $v_{cm}$ is the velocity of the centre of mass and $R$ is the radius.

The total kinetic energy during rolling:

KE_{total} = \frac{1}{2}Mv_{cm}^2 + \frac{1}{2}I\omega^2

Substituting $v_{cm} = R\omega$ :

KE_{total} = \frac{1}{2}Mv_{cm}^2\left(1 + \frac{I}{MR^2}\right)

The term $\frac{I}{MR^2}$ is the key. For a ring it’s 1, for a disc it’s $\frac{1}{2}$ , for a solid sphere it’s $\frac{2}{5}$ .

Race on an inclined plane: When different shapes roll down from rest, the one with smaller $\frac{I}{MR^2}$ reaches the bottom first (more energy goes to translation). Order: solid sphere > disc > ring.

Solved Examples

Example 1 — Easy (CBSE Level)

A flywheel of moment of inertia 5 kg·m² is rotating at 120 rpm. Calculate its angular velocity and rotational kinetic energy.

Step 1: Convert rpm to rad/s.

\omega = \frac{2\pi \times 120}{60} = 4\pi \text{ rad/s} \approx 12.57 \text{ rad/s}

Step 2: Calculate KE.

KE = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 5 \times (4\pi)^2 = \frac{1}{2} \times 5 \times 16\pi^2

= 40\pi^2 \approx 394.8 \text{ J}

Example 2 — Medium (JEE Main Level)

A solid cylinder of mass 3 kg and radius 0.4 m rolls without slipping on a horizontal surface with velocity 5 m/s. Find the total kinetic energy.

For a solid cylinder, $I_{cm} = \frac{1}{2}MR^2$ .

Pure rolling: $\omega = \frac{v}{R} = \frac{5}{0.4} = 12.5$ rad/s

KE_{trans} = \frac{1}{2}Mv^2 = \frac{1}{2}(3)(25) = 37.5 \text{ J}

KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{1}{2}MR^2 \cdot \frac{v^2}{R^2} = \frac{1}{4}Mv^2 = \frac{1}{4}(3)(25) = 18.75 \text{ J}

KE_{total} = 37.5 + 18.75 = 56.25 \text{ J}

Notice the ratio: translational KE : rotational KE = 2:1 for any solid cylinder. For a disc (same formula), same ratio. Memorise these ratios for quick MCQ solving.

Example 3 — Hard (JEE Advanced Level)

A thin uniform rod of mass M and length L can rotate freely about a vertical axis through one end. A particle of mass m moving horizontally with velocity $v_0$ strikes the other end of the rod and sticks to it. Find the angular velocity just after collision.

This is an angular momentum conservation problem. External torques about the vertical axis are zero during collision (impulsive forces only act at the contact point).

Before collision:

Angular momentum of particle about the pivot = $m \cdot v_0 \cdot L$ (particle moves perpendicular to rod)
Angular momentum of rod = 0 (at rest)

After collision:

System (rod + particle) rotates together
$I_{system} = \frac{ML^2}{3} + mL^2 = L^2\left(\frac{M}{3} + m\right)$

Applying conservation:

mv_0 L = I_{system} \cdot \omega

\omega = \frac{mv_0 L}{L^2\left(\frac{M}{3} + m\right)} = \frac{mv_0}{\left(\frac{M}{3} + m\right)L} = \frac{3mv_0}{(M + 3m)L}

Exam-Specific Tips

CBSE Board Exam Strategy

CBSE typically asks:

1 numerical on moment of inertia (using parallel axis or perpendicular axis theorem) — 3 marks
1 numerical on rolling motion or angular momentum conservation — 5 marks
Short answer on torque and angular momentum relationship — 2 marks

Marking scheme tip: In CBSE, writing the formula first, then substituting values, then getting the answer earns full step marks even if the final calculation is wrong. Never skip writing the formula.

JEE Main Pattern

JEE Main asks 1-2 questions on this chapter, mostly as numerical value type (NVT) or MCQ. High-frequency subtopics:

Rolling on inclined planes (energy method)
Angular momentum conservation in collision problems
Torque calculations with multiple forces

The chapter has appeared in every JEE Main session in the past 3 years — treat it as guaranteed marks.

JEE Advanced Strategy

JEE Advanced goes deeper:

Rolling with slipping (friction analysis)
Angular impulse-momentum theorem
Compound pendulum and physical pendulum
Combined rotation + translation with constraints

Spend time on constraint equations — these are what separate average from excellent scores here.

Common Mistakes to Avoid

Mistake 1: Using $I = MR^2$ for a disc. The ring has $I = MR^2$ . The disc has $I = \frac{1}{2}MR^2$ . Students confuse these under pressure because both are “circular.” Trick: the disc has mass spread from centre to rim, so average $r^2$ is smaller.

Mistake 2: Applying perpendicular axis theorem to 3D objects. If the problem involves a sphere or cylinder, this theorem doesn’t apply. It’s strictly for planar (2D) bodies only.

Mistake 3: Forgetting $v = R\omega$ in rolling problems. In pure rolling, $v_{cm} = R\omega$ always. Students sometimes apply energy conservation without using this relation to express $\omega$ in terms of $v$ , leaving two unknowns.

Mistake 4: Taking torque about a moving axis incorrectly. The equation $\tau = I\alpha$ is valid only for a fixed axis or for an axis passing through the centre of mass. For any other moving point, you need to be careful. CBSE doesn’t test this, but JEE Advanced does.

Mistake 5: Ignoring which axis to apply parallel axis theorem from. The theorem says $I = I_{cm} + Md^2$ . The $d$ must be the distance from the centre of mass axis, not from some other convenient axis. Check that you’re starting from the CM axis every time.

Practice Questions

Q1. A ring and a disc, both of mass M and radius R, start rolling from rest down an incline of height h. Which reaches the bottom first, and what is the speed of each at the bottom?

Using energy conservation: $Mgh = \frac{1}{2}Mv^2\left(1 + \frac{I}{MR^2}\right)$

Ring: $\frac{I}{MR^2} = 1$ , so $v_{ring} = \sqrt{gh}$

Disc: $\frac{I}{MR^2} = \frac{1}{2}$ , so $v_{disc} = \sqrt{\frac{4gh}{3}}$

Since $\sqrt{\frac{4gh}{3}} > \sqrt{gh}$ , the disc reaches the bottom first.

Q2. The angular velocity of a body changes from $\omega_1$ to $\omega_2$ without any external torque. Show that the ratio of radii of gyration is $k_2 : k_1 = \sqrt{\omega_1} : \sqrt{\omega_2}$ .

Radius of gyration $k$ is defined by $I = Mk^2$ , so $I_1 = Mk_1^2$ and $I_2 = Mk_2^2$ .

No external torque → angular momentum conserved: $I_1\omega_1 = I_2\omega_2$

$Mk_1^2 \omega_1 = Mk_2^2 \omega_2$

$\frac{k_2^2}{k_1^2} = \frac{\omega_1}{\omega_2}$

$\frac{k_2}{k_1} = \sqrt{\frac{\omega_1}{\omega_2}}$

Hence $k_2 : k_1 = \sqrt{\omega_1} : \sqrt{\omega_2}$ .

Q3. A thin rod of mass 2 kg and length 1 m is pivoted at one end. It is held horizontal and released. Find the angular acceleration and linear acceleration of the free end at the instant of release.

$I$ about pivot = $\frac{ML^2}{3} = \frac{2 \times 1}{3} = \frac{2}{3}$ kg·m²

Torque due to gravity (acting at CM, distance $L/2$ from pivot):

$\tau = Mg \cdot \frac{L}{2} = 2 \times 10 \times 0.5 = 10$ N·m

$\alpha = \frac{\tau}{I} = \frac{10}{2/3} = 15$ rad/s²

Linear acceleration of free end: $a = \alpha L = 15 \times 1 = 15$ m/s²

Note: $a = 15$ m/s² $> g = 10$ m/s² — the free end accelerates faster than free fall at the instant of release!

Q4. A solid sphere of mass 5 kg and radius 0.1 m rolls up an incline of angle 30°. If its initial velocity is 4 m/s, how far along the incline does it travel?

Total KE initially for solid sphere rolling:

$KE = \frac{1}{2}Mv^2\left(1 + \frac{2}{5}\right) = \frac{7}{10}Mv^2 = \frac{7}{10}(5)(16) = 56$ J

This converts to PE: $Mgh = 56$

$h = \frac{56}{5 \times 10} = 1.12$ m

Distance along incline: $s = \frac{h}{\sin 30°} = \frac{1.12}{0.5} = 2.24$ m

Q5. Calculate the moment of inertia of a rectangular plate of mass M, length $a$ , breadth $b$ , about an axis through its centre and perpendicular to its plane.

Using the perpendicular axis theorem (valid here — plate is a planar body):

$I_z = I_x + I_y$

For a rectangular plate:

$I_x = \frac{Mb^2}{12}$ (about axis parallel to breadth through centre)
$I_y = \frac{Ma^2}{12}$ (about axis parallel to length through centre)

$I_z = \frac{M(a^2 + b^2)}{12}$

Q6 — Q8. (Short practice)

Q6. Torque on a body is zero. What can you say about its angular momentum?

$\frac{dL}{dt} = \tau = 0$ , so $L$ = constant. Angular momentum is conserved. The body either stays at rest or rotates with constant angular velocity.

Q7. A disc of moment of inertia $I$ is rotating with angular velocity $\omega$ . Another disc of moment of inertia $I/2$ , initially at rest, is gently placed coaxially on it. Find the final angular velocity.

No external torque → $L$ conserved.

$I\omega = \left(I + \frac{I}{2}\right)\omega'$

$\omega' = \frac{I\omega}{\frac{3I}{2}} = \frac{2\omega}{3}$

Q8. For a solid sphere rolling without slipping, what fraction of total KE is rotational?

$KE_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2} \cdot \frac{2}{5}MR^2 \cdot \frac{v^2}{R^2} = \frac{1}{5}Mv^2$

$KE_{total} = \frac{7}{10}Mv^2$

Fraction $= \frac{\frac{1}{5}Mv^2}{\frac{7}{10}Mv^2} = \frac{2}{7}$

So $\frac{2}{7}$ of total energy is rotational — about 28.6%.

FAQs

What is the difference between moment of inertia and mass?

Mass resists linear acceleration; moment of inertia resists angular acceleration. Unlike mass, moment of inertia is not a fixed property of an object — it changes depending on which axis you choose. The same disc has different $I$ values for an axis through its centre versus an axis at its rim.

Why does a hollow sphere have more moment of inertia than a solid sphere?

In a hollow sphere, all the mass is concentrated at the rim (maximum distance from centre). In a solid sphere, most mass is closer to the centre. Since $I = \sum mr^2$ , larger $r$ means larger $I$ . This is why hollow sphere ( $\frac{2}{3}MR^2$ ) > solid sphere ( $\frac{2}{5}MR^2$ ) despite same $M$ and $R$ .

What does it mean when angular momentum is conserved?

It means the total “spinning tendency” of a system stays constant when no external torque acts. The classic example: a planet moves faster when closer to the sun (its $r$ decreases, so $\omega$ must increase to keep $L = mr^2\omega$ constant).

How is torque different from force?

Force causes linear acceleration. Torque causes angular acceleration. Torque depends not just on the magnitude of force but also on where it’s applied — same force applied farther from the axis creates more torque. This is why door handles are placed far from the hinges.

Can a body have angular momentum without rotating?

Yes. A particle moving in a straight line has angular momentum about any point not on that line. Angular momentum of a particle is $L = mvr_\perp$ where $r_\perp$ is the perpendicular distance from the point to the line of motion.

What is the condition for pure rolling? Why is it important?

Pure rolling requires $v_{cm} = R\omega$ — the contact point’s velocity is zero. This condition lets us link translation and rotation, reducing two unknowns to one. Without it, we’d need friction force equations for each problem individually.

Why do we use radians instead of degrees in rotational motion?

Because the relation $v = r\omega$ (or $a = r\alpha$ ) only holds when $\omega$ is in rad/s. These are not arbitrary unit choices — they come from the definition of arc length $s = r\theta$ , which is valid only when $\theta$ is in radians. Always convert rpm or degrees to radians before substituting in any formula.