JEE Mains Physics · System of Particles and Rotational Motion

Torque, Angular Acceleration and Rotational Energy

About a fixed axis, torque gives angular acceleration through τ = Iα, and a spinning body stores kinetic energy ½Iω², which energy conservation trades with height and with the motion of attached blocks.

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Why this matters

Twenty PYQs, eleven of them asking for a number, and seven from 2026. Eleven use τ = Iα: a string pulled off a wheel, a heavy pulley between two blocks, a torque that brings a spinning body to rest. Nine use energy: a falling block spins a flywheel, a rod falls about a pivot, two wheels share a belt.

Concept 1 of 2: Torque and angular acceleration about a fixed axis

τ = Iα is Newton's second law for rotation: torque plays the part of force and moment of inertia the part of mass. When a string runs over a wheel, the wheel and the string are tied together: the string's acceleration is Rα. Write one equation for each body, then link them with a = Rα.

Definition

$\tau = I\alpha$ about a fixed axis. A string pulled off a rim with force F gives $\tau = FR$ .
A string that does not slip: $a = R\alpha$ .
Block m hanging from a string wound on a disc pulley M: $mg - T = ma$ and $TR = \tfrac{1}{2}MR^{2}\dfrac{a}{R}$ , so $T = \tfrac{1}{2}Ma$ and $a = \dfrac{mg}{m + M/2}$ .
Two blocks over a pulley with moment of inertia I: $a = \dfrac{(m_1 - m_2)g}{m_1 + m_2 + I/R^{2}}$ . The tensions on the two sides are NOT equal; their difference turns the pulley.
Rod hung level from strings at its ends, one string cut: about the other end, $\alpha = \dfrac{mg(L/2)}{mL^{2}/3} = \dfrac{3g}{2L}$ , so $a_{cm} = \tfrac{3}{4}g$ and the remaining tension is $mg/4$ .
Stopping a spinning body in time t: $\tau = I\omega_0/t$ . Work done by a torque $= \tau\theta$ ; power $P = \tau\omega$ .

Rotational second law, heavy pulley, and power

\tau = I\alpha \qquad a = \frac{(m_1 - m_2)g}{m_1 + m_2 + I/R^{2}} \qquad P = \tau\omega

Worked example

A 1 kg block hangs from a light string wound round a uniform disc pulley of mass 6 kg and radius 0.2 m, free to turn on a fixed axle. Find the block's acceleration, the tension, and the pulley's angular acceleration (g = 10 m/s²).

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q6Moderate

Example 1 · System of Particles and Rotational Motion · Torque, Angular Acceleration and Rotational Energy

A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the center of the sphere) with an angular velocity of 1200 rpm . It is brought to rest in 10 s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are

\_\_\_\_

and

\_\_\_\_

respectively.

T = mg for an accelerating block

If the block accelerates downward, the string pulls with less than its weight: T = m(g − a). Using T = mg makes the pulley's acceleration too large.

Equal tensions over a heavy pulley

Equal tensions on both sides give zero net torque, so a pulley with mass could never start turning. With a massive pulley the tensions differ, and I/R² joins the masses in the denominator.

Angular speed in rpm

τ = Iω₀/t needs ω₀ in rad/s. 1200 rpm is 40π rad/s; leaving it as 1200, or as 20 revolutions a second, gives a torque off by a factor of 2π/60 or of 2π.

Drill 10 more on torque and angular acceleration about a fixed axis

Concept 2 of 2: Rotational kinetic energy and energy conservation

A spinning body stores ½Iω², the rotational twin of ½mv². When a block falls and spins a wheel through a string, the height it loses is shared between its own kinetic energy and the wheel's. When a rod swings about a pivot, only its centre of mass rises or falls, and the energy goes into ½Iω² about the pivot.

Definition

$K_{\text{rot}} = \tfrac{1}{2}I\omega^{2}$ .
Block on a string wound on a wheel: $mgh = \tfrac{1}{2}mv^{2} + \tfrac{1}{2}I\omega^{2}$ , with $v = \omega R$ . Count EVERY moving body.
A force F pulling a cord a length l off a wheel does work $Fl = \tfrac{1}{2}I\omega^{2}$ .
A rod about a pivot: $mg \times (\text{drop of the centre of mass}) = \tfrac{1}{2}I_{\text{pivot}}\omega^{2}$ . Hinged at its foot and falling from upright to flat: $\omega = \sqrt{3g/L}$ . Clamped at its foot and falling all the way to hang straight down, the centre of mass drops L: the free end moves at $\sqrt{6gL}$ .
Wheels joined by a belt share the same rim speed, $\omega_1R_1 = \omega_2R_2$ . Equal kinetic energies then need $I_1/I_2 = (R_1/R_2)^{2}$ .

Rotational kinetic energy, and energy for a block on a wheel

K_{\text{rot}} = \tfrac{1}{2}I\omega^{2} \qquad mgh = \tfrac{1}{2}mv^{2} + \tfrac{1}{2}I\omega^{2},\ v = \omega R

Worked example

A 1 kg block hangs from a string wound on a flywheel that is a uniform disc of mass 2 kg and radius 0.5 m. The block falls 1.8 m from rest. Find its speed and the flywheel's kinetic energy (g = 10 m/s²).

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q25Moderate

Example 2 · System of Particles and Rotational Motion · Torque, Angular Acceleration and Rotational Energy

A fly wheel having mass 3 kg and radius 5 m is free to rotate about a horizontal axis. A string having negligible mass is wound around the wheel and the loose end of the string is connected to a 3 kg mass. The mass is kept at rest initially and released. Kinetic energy of the wheel when the mass descends by 3 m is

\_\_\_\_

J.\left( g = 10\text{ }m/s^{2} \right)

Leaving out the block's kinetic energy

The falling block is moving too. mgh = ½Iω² alone gives the wheel all the energy; the block's ½mv² must be on the same side.

A rod's centre of mass falls half as far

A rod falling from upright to flat about its foot lowers its centre of mass by L/2, not L. Using L gives √(6g/L) in place of √(3g/L).

Moment of inertia about the pivot

A rod turning about its end has I = mL²/3 about that end. Using mL²/12, the value about its centre, ignores the motion of the centre of mass.

Drill 8 more on rotational kinetic energy and energy conservation

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 (2)

Torque and angular acceleration about a fixed axis
Rotational second law, heavy pulley, and power
$\tau = I\alpha \qquad a = \frac{(m_1 - m_2)g}{m_1 + m_2 + I/R^{2}} \qquad P = \tau\omega$
Rotational kinetic energy and energy conservation
Rotational kinetic energy, and energy for a block on a wheel
$K_{\text{rot}} = \tfrac{1}{2}I\omega^{2} \qquad mgh = \tfrac{1}{2}mv^{2} + \tfrac{1}{2}I\omega^{2},\ v = \omega R$

Watch out for (6)

T = mg for an accelerating block
→ Torque and angular acceleration about a fixed axis
Equal tensions over a heavy pulley
→ Torque and angular acceleration about a fixed axis
Angular speed in rpm
→ Torque and angular acceleration about a fixed axis
Leaving out the block's kinetic energy
→ Rotational kinetic energy and energy conservation
A rod's centre of mass falls half as far
→ Rotational kinetic energy and energy conservation
Moment of inertia about the pivot
→ Rotational kinetic energy and energy conservation

Test yourself on System of Particles and Rotational Motion

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