MHT-CET Physics · Rotational Dynamics

Torque, Angular Momentum and Its Conservation

Torque τ = r × F is the turning effect of a force and changes angular momentum L = Iω; with no external torque L stays fixed, so a body that pulls its mass in spins faster.

All Rotational Dynamics notes MHT-CET Physics strategy

Why this matters

22 PYQs, only one HARD — a page of steady marks. Three shapes: a torque from a force or from a change in spin, the links between L, ω, I and kinetic energy, and a second disc or ring dropped onto a spinning one (or a skater folding their arms), where L is conserved and kinetic energy is not.

Concept 1 of 3: Torque: From a Force, and From a Change in Spin

A force turns a body more the further from the axis it acts and the more squarely it pushes: τ = rF sin θ. That torque is the rotational version of force: it produces angular acceleration, τ = Iα, and spinning a body up or down in a known time tells you the torque that did it.

Definition

$\vec\tau = \vec r \times \vec F$ — ORDER matters: $\vec F \times \vec r$ has the opposite sign.
$\tau = I\alpha$ ; average torque $= \dfrac{I(\omega_2 - \omega_1)}{t}$ (use $\omega = 2\pi n$ ).
A force on the rim: $F = \dfrac{\tau}{R}$ . Door: the same torque at a third of the width needs three times the force.
Rod pivoted at an end, at angle $\theta$ to the vertical: $\tau = mg\dfrac{L}{2}\sin\theta$ , $\alpha = \dfrac{3g\sin\theta}{2L}$ .
Power: $P = \tau\omega = I\alpha\omega$ .
Vector forms: $\vec v = \vec\omega \times \vec r$ , $\vec L = \vec r \times \vec p$ , $\vec\tau = \vec r \times \vec F$ .

Torque

\vec\tau = \vec r \times \vec F, \qquad \tau = I\alpha

Worked example

A force

\vec F = 2\hat i + \hat k

acts at

\vec r = \hat i + 3\hat j

. Torque about the origin?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 21 April Shift I · Q12Moderate

Example 1 · Rotational Dynamics · Angular Momentum, Torque, and Conservation

A disc of mass 25 kg and radius 0.2 m is rotating at 240 r.p.m. A retarding torque brings it to rest in 20 second. If the torque is due to a force applied tangentially on the rim of the disc, then the magnitude of the force is

Writing F × r

Torque is

\vec r \times \vec F

and angular momentum

\vec r \times \vec p

. The options swap the order in one term to reverse its sign.

Drill 6 more on torque: from a force, and from a change in spin

Concept 2 of 3: Angular Momentum and Its Links to Energy and Force

L = Iω for a spinning body, and L = mvr for a particle on a circle. Rotational kinetic energy can be written through L as L²/2I, which is the form every 'equal kinetic energies' or 'equal angular momenta' question wants.

Definition

$L = I\omega$ ; particle on a circle $L = mvr = mr^2\omega$ .
$K = \dfrac{1}{2}I\omega^2 = \dfrac{L^2}{2I} = \dfrac{1}{2}L\omega$ , so $I = \dfrac{L^2}{2K}$ .
Equal $K$ : $L \propto \sqrt{I}$ . Equal $L$ : $K \propto \dfrac{1}{I}$ .
Particle on a circle: $K = \dfrac{L^2}{2mr^2}$ and centripetal force $F = \dfrac{L^2}{mr^3}$ .
A particle moving in a straight line keeps a constant $L$ about any point (its perpendicular distance never changes); on a circle at changing speed only the DIRECTION of $L$ stays fixed.
The Earth: $L = \dfrac{2}{5}MR^2\cdot\dfrac{2\pi}{T}$ .

Angular momentum and energy

L = I\omega, \qquad K = \frac{L^2}{2I}

Worked example

A wheel of I = 0.5 kg m² spins at 4 rad/s. Its angular momentum and kinetic energy?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q32Moderate

Example 2 · Rotational Dynamics · Angular Momentum, Torque, and Conservation

Two bodies have their moments of inertia I and 2I respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momenta will be in the ratio.

Equal energy means equal L

At equal kinetic energy,

L = \sqrt{2IK} \propto \sqrt{I}

: moments of inertia I and 2I give

1 : \sqrt{2}

, not 1 : 2.

Drill 9 more on angular momentum and its links to energy and force

Concept 3 of 3: Conservation of Angular Momentum

With no outside torque, I ω stays the same. Drop a second disc onto a spinning one and I rises, so ω falls; fold your arms on a turntable and I falls, so ω rises. Kinetic energy is not conserved in either: it is lost when the discs grab each other, and supplied by your muscles when you pull in.

Definition

$I_1\omega_1 = I_2\omega_2$ when no external torque acts.
Disc $M$ with a coaxial disc $\dfrac{M}{3}$ placed on it: $\omega' = \dfrac{3}{4}\omega$ ; with $\dfrac{M}{2}$ : $\dfrac{2}{3}\omega$ .
Two discs brought together: $\omega = \dfrac{I_1\omega_1 + I_2\omega_2}{I_1 + I_2}$ , $K = \dfrac{(I_1\omega_1 + I_2\omega_2)^2}{2(I_1 + I_2)}$ .
At fixed $L$ , $K = \dfrac{L^2}{2I}$ : I down to 75% ⇒ K up by a third (33.3%).

No external torque

I_1\omega_1 = I_2\omega_2, \qquad K \propto \frac{1}{I}\ \text{at fixed } L

Worked example

A disc of I = 0.2 kg m² spins at 30 rad/s. A second disc of I = 0.1 kg m² is dropped on it coaxially. New ω, and the kinetic energy lost?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 2 · Q14Moderate

Example 3 · Rotational Dynamics · Angular Momentum, Torque, and Conservation

A thin uniform circular disc of mass 'M' and radius 'R' is rotating with angular velocity '

\omega

' in a horizontal plane about an axis passing through its centre and perpendicular to its plane. Another disc of same radius but of mass

\frac{M}{3}

is placed gently on the first disc co-axially. The new angular velocity will be

Conserving kinetic energy

Setting

\frac{1}{2}I_1\omega_1^2 = \frac{1}{2}I_2\omega_2^2

gives

\omega

changing as

\sqrt{I}

, which matches a planted option. The conserved quantity is

I\omega

Drill 4 more on conservation of angular momentum

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Torque: From a Force, and From a Change in Spin
Torque
$\vec\tau = \vec r \times \vec F, \qquad \tau = I\alpha$
Angular Momentum and Its Links to Energy and Force
Angular momentum and energy
$L = I\omega, \qquad K = \frac{L^2}{2I}$
Conservation of Angular Momentum
No external torque
$I_1\omega_1 = I_2\omega_2, \qquad K \propto \frac{1}{I}\ \text{at fixed } L$

Watch out for (3)

Writing F × r
→ Torque: From a Force, and From a Change in Spin
Equal energy means equal L
→ Angular Momentum and Its Links to Energy and Force
Conserving kinetic energy
→ Conservation of Angular Momentum

Test yourself on Rotational Dynamics

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Rotational Dynamics chapter test Drill all 22 questions

Related notes

Moment of Inertia and Radius of Gyration →