Law of Rotation And Work-Energy Theorem

Q: Why do torque and work have the same unit?

Both are computed as force times length. Dimensionally torque and work are M L squared T to the minus two, giving SI units of newton metre. However work is expressed in joules and torque is never called joules because they are physically different quantities: torque is a vector that turns things while work is a scalar that transfers energy.

Q: Is tau equals I alpha valid about any point?

It is valid without correction only about a fixed axis (a real hinge or pivot fixed in space) or an axis through the centre of mass. About any other accelerating point extra pseudo-torque terms appear and the simple equation fails. For most JEE problems choose the hinge or the CM.

Q: How does the work-energy theorem look when a body both translates and rotates?

Total work done by all forces equals the change in total kinetic energy, which is the sum of translational half M v squared and rotational half I omega squared parts. Internal forces such as tension in a rigid body do zero net work.

Q: What is the condition for a body to be in equilibrium?

Two conditions must hold simultaneously: the net external force must be zero so the centre of mass has no linear acceleration, and the net external torque about any point must be zero so the body has no angular acceleration. If either condition fails the body is not in equilibrium.

PhysicsSystem Of Particles And Rotational MotionFor NEET aspirants

Torque is the rotational analogue of force. About a point $O$ , the torque of a force $F$ acting at position $r$ is $τ = r \times F$ , and its magnitude is $τ = r F sin θ$ . For a rigid body rotating about a fixed axis, the law of rotation states $τ_{net} = I α$ , the exact analogue of Newton's second law. Rotational work is $W = \int τ d θ$ , rotational kinetic energy is $K = \frac{1}{2} I ω^{2}$ , and the rotational work-energy theorem reads $W_{net} = Δ K_{rot}$ . These four relations solve almost every JEE, NEET and Advanced problem on hinged rods, pulleys with mass, ladders and rigid-body equilibrium.

Key Formulas - Quick Reference

Torque about a point: $τ = r \times F$ , $∣ τ ∣ = r F sin θ$
Torque as force $\times$ perpendicular arm: $τ = F \cdot d_{⊥}$
Rigid-body rotation about a fixed axis: $τ_{net} = I α$
Equilibrium of a rigid body: $\sum F = 0$ and $\sum τ = 0$
Rotational work: $W = \int_{θ_{1}}^{θ_{2}} τ d θ$ ; if $τ$ constant, $W = τ Δ θ$
Rotational kinetic energy: $K = \frac{1}{2} I ω^{2}$
Rotational power: $P = τ ω$
Work-energy theorem (rotation): $W_{net} = Δ K_{rot} = \frac{1}{2} I ω_{f}^{2} - \frac{1}{2} I ω_{i}^{2}$
Kinematic equations (constant $α$ ): $ω = ω_{0} + α t$ ; $θ = ω_{0} t + \frac{1}{2} α t^{2}$ ; $ω^{2} = ω_{0}^{2} + 2 α θ$

1. Torque About a Point

Figure 1: Torque

τ = r \times F

. Its magnitude is

r F sin θ

or equivalently

F \times

(perpendicular arm

d

The torque of a force about a chosen point $O$ measures the turning effect of that force. If the force $F$ acts at position $r$ from $O$ ,

$τ = r \times F, ∣ τ ∣ = r F sin θ$

where $θ$ is the angle between $r$ and $F$ . Two equivalent readings help:

$τ = (r sin θ) F = d_{⊥} F$ : force times perpendicular arm, the shortest distance from $O$ to the line of action of $F$ .
$τ = r (F sin θ) = r F_{⊥}$ : distance times the component of $F$ perpendicular to $r$ .

Direction is given by the right-hand rule. SI unit is $N m$ (dimensionally the same as work, but never called joule when it means torque).

Sign convention. For planar problems, treat anticlockwise torques as positive and clockwise as negative (or the opposite: be consistent within one problem). The vector form handles signs automatically through the cross product.

Solved Example 1

A particle of mass

m

is projected from origin

O

with speed

u

at angle

θ

above the horizontal. Find the torque of gravity about

O

when the particle is at the top of its trajectory.

Solution:

At the top, the horizontal distance from $O$ is $x = \frac{u ^{2} sin 2 θ}{2 g}$ (half the range) and the height is the maximum height $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ . Gravity acts downward with magnitude $m g$ .

Torque about $O$ = force $\times$ perpendicular distance from $O$ to the vertical line of action of gravity. That perpendicular distance is $x$ . So

$τ = m g \cdot x = m g \cdot \frac{u ^{2} sin 2 θ}{2 g} = \frac{m u ^{2} sin 2 θ}{2}$

Direction is into the page (clockwise as seen from the standard orientation).

2. Torque About an Axis

When a rigid body can rotate only about a fixed axis (a hinged door, a pulley on its shaft), only the component of torque along that axis produces angular acceleration. All other torque components are absorbed by the constraint that holds the axis in place.

Practical shortcut: for planar motion where all forces lie in a plane perpendicular to the axis, the torque of each force about the axis is simply $\pm F \cdot d_{⊥}$ , where $d_{⊥}$ is the perpendicular distance from the axis to the line of action, and the sign follows your chosen anticlockwise-positive convention.

Solved Example 2

A force of

20 N

is applied to a door at a distance

0.8 m

from the hinge. Find the torque about the hinge axis when the force is applied (i) perpendicular to the door, (ii) at

30°

to the door, (iii) along the door pointing toward the hinge.

Solution:

Torque about the hinge axis is $τ = r F sin ϕ$ , where $ϕ$ is the angle between $r$ (from hinge to point of application) and $F$ .

(i) $ϕ = 90°$ : $τ = (0.8) (20) (1) = 16 N m$ . This is the maximum for a given $F$ and $r$ .

(ii) $ϕ = 30°$ : $τ = (0.8) (20) sin 30° = (0.8) (20) (0.5) = 8 N m$ .

(iii) $ϕ = 180°$ : $sin ϕ = 0$ , so $τ = 0$ . A push directed along the door has no turning effect — this is why pushing near the hinge, or at a shallow angle to the door, is inefficient.

3. Equilibrium of a Rigid Body

A rigid body is in complete equilibrium when it is in translational and rotational equilibrium simultaneously:

$\sum F_{ext} = 0 and \sum τ_{ext} = 0$

Useful theorem. If the net external force on a body is zero, then the net torque about any point is the same. So for a body in translational equilibrium you may compute torques about whichever point makes the algebra cleanest, usually a point where an unknown force acts (so it drops out).

Solved Example 3

A uniform rod of mass

m

and length

L

is hinged at one end and held horizontal by a string tied to its other end, making an angle

θ

with the rod. Find the tension in the string.

Solution:

Take torques about the hinge (this eliminates the unknown hinge reaction).

Weight $m g$ acts vertically downward at the centre of the rod (distance $L /2$ from the hinge). Its torque about the hinge is $m g \cdot (L /2)$ , clockwise.

String tension $T$ acts at the free end. Its perpendicular component to the rod is $T sin θ$ , and its perpendicular arm from the hinge is the full length $L$ . Its torque about the hinge is $T sin θ \cdot L$ , anticlockwise.

For rotational equilibrium,

$T sin θ \cdot L = m g \cdot \frac{L}{2} \Rightarrow T = \frac{m g}{2 sin θ}$

3.1 The Ladder Problem

Figure 2: Ladder against a smooth wall on a rough floor. Weight

m g

acts at the centre;

N_{1}

is wall reaction;

N_{2}

and

f

are floor reaction and friction.

A uniform ladder of length $L$ and mass $m$ leans against a smooth vertical wall at angle $θ$ with the rough horizontal floor. Wall reaction $N_{1}$ is horizontal; floor reaction is $N_{2}$ upward with friction $f$ acting horizontally at the base.

Translational equilibrium gives $N_{1} = f$ and $N_{2} = m g$ . For rotational equilibrium, take torques about the base (so $N_{2}$ and $f$ drop out). Weight acts at the mid-point:

$N_{1} \cdot L sin θ = m g \cdot \frac{L}{2} cos θ \Rightarrow N_{1} = \frac{m g cot θ}{2}$

Therefore $f = N_{1} = \frac{m g cot θ}{2}$ . For the ladder not to slip, $f \leq μ_{s} N_{2} = μ_{s} m g$ , which requires

$tan θ \geq \frac{1}{2 μ _{s}}$

4. Law of Rotation: $τ = I α$

When a rigid body rotates about a fixed axis with angular acceleration $α$ under net external torque $τ_{net}$ about that axis,

$τ_{net} = I α$

This is the exact rotational counterpart of Newton's second law $F = ma$ . The proportionality constant is the moment of inertia about the axis.

Solved Example 4

A uniform rod of mass

m

and length

L

is hinged at one end and released from rest in the horizontal position. Find the angular acceleration of the rod immediately after release, and the tangential acceleration of its free end.

Solution:

About the hinge, the only torque is due to gravity acting at the centre of mass at perpendicular arm $L /2$ :

$τ = m g \cdot \frac{L}{2}$

Moment of inertia of the rod about the hinge (one end) is $I = \frac{m L ^{2}}{3}$ . From $τ = I α$ :

$α = \frac{τ}{I} = \frac{m g \cdot L /2}{m L ^{2} /3} = \frac{3 g}{2 L}$

The tangential acceleration of the free end (at distance $L$ from the hinge) is

$a_{t} = Lα = \frac{3 g}{2}$

which is greater than $g$ . Every point on the rod farther than $2 L /3$ from the hinge falls with tangential acceleration greater than $g$ .

4.1 Pulley With Mass (Rotational Motion of the Pulley)

When a pulley has non-negligible moment of inertia $I$ , the tensions on the two sides of the string are different. The difference in tension supplies the torque that angularly accelerates the pulley.

Figure 3: Massive pulley. Because

I \neq = 0

, the two tensions differ, and it is the torque

(T_{1} - T_{2}) R

that angularly accelerates the pulley.

Solved Example 5

A pulley of moment of inertia

I

and radius

R

carries a light inextensible string with masses

M

and

m

(

M > m

) hanging from its two ends. If the string does not slip on the pulley, find the acceleration of the masses.

Solution:

Let $a$ be the linear acceleration of the masses ( $M$ down, $m$ up). Since the string does not slip, the tangential acceleration of the rim equals $a$ , so the angular acceleration of the pulley is $α = a / R$ .

Newton's second law for each mass:

$M g - T_{1} = M a (1)$

$T_{2} - m g = ma (2)$

Rotational equation for the pulley (net torque about its axis):

$T_{1} R - T_{2} R = I α = \frac{I a}{R} (3)$

Adding $(1)$ and $(2)$ gives $M g - m g - (T_{1} - T_{2}) = (M + m) a$ . Using $(3)$ : $T_{1} - T_{2} = I a / R^{2}$ . Substituting,

$a = \frac{( M - m ) g}{M + m + I / R ^{2}}$

When $I = 0$ (massless pulley) this reduces to the familiar Atwood result.

5. Rotational Work and Kinetic Energy

When a torque $τ$ turns a body through an infinitesimal angle $d θ$ , the work done is $d W = τ d θ$ . Total work done by a torque as the body rotates from $θ_{1}$ to $θ_{2}$ is

$W_{rot} = \int_{θ_{1}}^{θ_{2}} τ d θ$

If $τ$ is constant, $W_{rot} = τ \cdot Δ θ$ . Power delivered by a torque rotating with angular speed $ω$ is $P = τ ω$ .

The rotational kinetic energy of a rigid body spinning about a fixed axis with angular speed $ω$ is

$K_{rot} = \frac{1}{2} I ω^{2}$

Derivation: each mass element $d m$ at distance $r$ from the axis moves with speed $v = r ω$ , so its kinetic energy is $\frac{1}{2} (d m) r^{2} ω^{2}$ . Summing over the body and pulling out $\frac{1}{2} ω^{2}$ leaves $\int r^{2} d m = I$ .

6. Work-Energy Theorem for Rotation

The net rotational work done on a rigid body equals its change in rotational kinetic energy:

$W_{net} = Δ K_{rot} = \frac{1}{2} I ω_{f}^{2} - \frac{1}{2} I ω_{i}^{2}$

This is the direct analogue of the linear work-energy theorem and is the fastest route to problems asking for final angular speed after a body has rotated through a given angle.

6.1 Conservation of Mechanical Energy

If only conservative forces (gravity, ideal springs) do work, mechanical energy is conserved. For rotational problems this reads

$K_{i} + U_{i} = K_{f} + U_{f}$

where the kinetic energy term includes both translational $\frac{1}{2} M v_{cm}^{2}$ (if the CM moves) and rotational $\frac{1}{2} I ω^{2}$ pieces.

Solved Example 6

A uniform rod of mass

m

and length

L

is hinged at one end and held vertically upward. It is released from rest and swings down under gravity. Find its angular speed as it passes through the horizontal position, and the speed of its free end at that instant.

Figure 4: Rod released from rest in the vertical-upward position (dashed) swings down to horizontal (solid). Centre of mass falls a height

L /2

; hinge does no work.

Solution:

The hinge does no work. From vertical to horizontal, the centre of mass falls a height $h = L /2$ , so the loss in gravitational PE is $m g L /2$ . This equals the gain in rotational KE about the hinge:

$m g \cdot \frac{L}{2} = \frac{1}{2} I ω^{2} = \frac{1}{2} \cdot \frac{m L ^{2}}{3} \cdot ω^{2}$

Solving,

$ω = \frac{3 g}{L}$

Speed of the free end: $v = L ω = 3 g L$ .

Solved Example 7

A solid cylinder of mass

M

and radius

R

rolls without slipping from rest down an incline of vertical height

h

. Find its speed at the bottom.

Figure 5: Solid cylinder rolling down an incline; the centre of mass falls a height

h

and both translational and rotational KE grow.

Solution:

Rolling without slipping means $v = R ω$ , so $ω = v / R$ . For a solid cylinder about its central axis, $I_{cm} = \frac{1}{2} M R^{2}$ .

Friction does no work in pure rolling because the contact point is momentarily at rest, so mechanical energy is conserved:

$M g h = \frac{1}{2} M v^{2} + \frac{1}{2} I_{cm} ω^{2} = \frac{1}{2} M v^{2} + \frac{1}{2} (\frac{1}{2} M R^{2}) (\frac{v}{R})^{2} = \frac{3}{4} M v^{2}$

Solving,

$v = \frac{4 g h}{3}$

This is less than the $2 g h$ a frictionless sliding block would attain — some of the gravitational PE has gone into spinning the cylinder rather than sliding it faster.

Solved Example 8

A flywheel of moment of inertia

I = 5 kg m^{2}

is acted on by a constant torque of

10 N m

. Starting from rest, find (i) the angular speed after

4 s

, (ii) the work done by the torque in this interval, (iii) the kinetic energy of the flywheel at

t = 4 s

Solution:

(i) Angular acceleration $α = τ / I = 10/5 = 2 rad/s^{2}$ . After $4 s$ , $ω = α t = 8 rad/s$ .

(ii) Angle turned: $θ = \frac{1}{2} α t^{2} = \frac{1}{2} (2) (16) = 16 rad$ . Work done by torque: $W = τ θ = 10 \times 16 = 160 J$ .

(iii) Kinetic energy: $K = \frac{1}{2} I ω^{2} = \frac{1}{2} (5) (64) = 160 J$ .

The two answers match, verifying the rotational work-energy theorem.

7. Rotational Kinematics (Constant Angular Acceleration)

When $α$ is constant, the equations of angular motion mirror the linear kinematic equations, symbol-for-symbol:

Linear	Rotational
$v = u + a t$	$ω = ω_{0} + α t$
$s = u t + \frac{1}{2} a t^{2}$	$θ = ω_{0} t + \frac{1}{2} α t^{2}$
$v^{2} = u^{2} + 2 a s$	$ω^{2} = ω_{0}^{2} + 2 α θ$

Solved Example 9

A grindstone of moment of inertia

2 kg m^{2}

is spinning at

300 rpm

. It is brought to rest in

20 s

by a constant friction torque at its bearing. Find (i) the friction torque, and (ii) the number of revolutions the stone makes before stopping.

Solution:

Convert to SI units: $ω_{0} = 300 \times \frac{2 π}{60} = 10 π rad/s \approx 31.4 rad/s$ . Final angular speed $ω = 0$ .

(i) Angular deceleration from $ω = ω_{0} + α t$ :

$α = \frac{ω - ω _{0}}{t} = \frac{0 - 10 π}{20} = - \frac{π}{2} rad/s^{2}$

Friction torque magnitude: $∣ τ ∣ = I ∣ α ∣ = 2 \times \frac{π}{2} = π N m \approx 3.14 N m$ .

(ii) Total angle turned:

$θ = ω_{0} t + \frac{1}{2} α t^{2} = 10 π (20) + \frac{1}{2} (- \frac{π}{2}) (400) = 200 π - 100 π = 100 π rad$

Number of revolutions: $N = \frac{θ}{2 π} = \frac{100 π}{2 π} = 50$ revolutions.

Common Mistakes to Avoid

Watch out

Forgetting that torque depends on the choice of reference point. Always state which point (or axis) you are computing torque about.
Using $τ = r F$ without the $sin θ$ factor when $r$ and $F$ are not perpendicular.
Assuming equal tensions on both sides of a pulley when the pulley has significant moment of inertia. Different tensions supply the net torque that spins the pulley.
Applying $τ = I α$ about a point that is neither the centre of mass nor a fixed axis. In such a case additional pseudo-torque terms appear.
Confusing rotational work $W = τ θ$ (angle must be in radians) with linear work.
Adding the linear KE $\frac{1}{2} M v_{cm}^{2}$ twice when a body both translates and rotates. Use it once, plus $\frac{1}{2} I_{cm} ω^{2}$ once.
Applying rotational kinematic equations when $α$ varies with time. These equations require constant $α$ just as their linear counterparts require constant $a$ .
Forgetting that the hinge or pivot force does no work in a swinging-rod problem, so energy conservation involves only gravity plus rotational KE.

Frequently Asked Questions

Why do torque and work have the same unit?

Both are computed as force times length. Dimensionally, torque and work are $[M L^{2} T^{- 2}]$ , giving SI units of $N m$ . However, we express work in joules and never call torque joules, because torque and work are physically different quantities: torque is a vector that turns things, work is a scalar that transfers energy.

Can two forces produce zero net force but nonzero net torque?

Yes. Two equal and opposite forces acting along different lines form a couple. Their vector sum is zero, so no translational acceleration, but their torques add (both curl the body the same way), producing pure rotation. A wrench turning a nut is the everyday example.

Why is the tension different on the two sides of a massive pulley?

The pulley itself needs a net torque to gain angular acceleration. That torque comes from the difference $(T_{1} - T_{2}) R$ . If the two tensions were equal, the net torque on the pulley would be zero and it could not spin faster, contradicting the string not slipping over an accelerating rope. For a massless pulley $I = 0$ , so no torque is needed and the tensions match.

Is $τ = I α$ valid about any point?

It is valid without correction only about (i) a fixed axis (a real hinge or pivot fixed in space) or (ii) an axis through the centre of mass. About any other accelerating point, extra pseudo-torque terms appear and the simple equation fails. For most JEE problems, choose the hinge or the CM.

How does the work-energy theorem look when a body both translates and rotates?

Total work done by all forces equals the change in total kinetic energy, which is the sum of translational and rotational parts: $W_{net} = Δ [\frac{1}{2} M v_{cm}^{2} + \frac{1}{2} I_{cm} ω^{2}]$ . Internal forces (like tension in a rigid body) do zero net work.

Does the hinge reaction do work when a rod swings down?

No. The hinge reaction acts at a point that has zero velocity throughout the motion (the pin is fixed in space). Since work equals force dotted with the displacement of the point of application, and that displacement is zero, the hinge does no work. This is why energy conservation cleanly gives $ω$ in swinging-rod problems.

What is the condition for a body to be in equilibrium?

Two conditions must hold simultaneously: (i) the net external force must be zero, so the centre of mass has no linear acceleration, and (ii) the net external torque about any point must be zero, so the body has no angular acceleration. If either condition fails, the body is not in equilibrium.

In a ladder problem, why does friction act at the base but not at the smooth wall?

Friction acts wherever there is a rough contact that would otherwise slip. The wall is stated smooth so it exerts only a normal reaction. The floor is rough, so friction can act there to prevent the base from sliding outward. If both surfaces were rough, both would contribute friction; if both were smooth, the ladder could not stand at any angle other than vertical.

Previous year questions on Law of Rotation And Work-Energy Theorem

10 questions from past papers, each with a step-by-step solution.

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Practice questions →Take a mock test

1. Torque About a Point

2. Torque About an Axis

3. Equilibrium of a Rigid Body

3.1 The Ladder Problem

4. Law of Rotation: τ=Iα

4.1 Pulley With Mass (Rotational Motion of the Pulley)

5. Rotational Work and Kinetic Energy

6. Work-Energy Theorem for Rotation

6.1 Conservation of Mechanical Energy

7. Rotational Kinematics (Constant Angular Acceleration)

Common Mistakes to Avoid

Frequently Asked Questions

Why do torque and work have the same unit?

Can two forces produce zero net force but nonzero net torque?

Why is the tension different on the two sides of a massive pulley?

Is τ=Iα valid about any point?

How does the work-energy theorem look when a body both translates and rotates?

Does the hinge reaction do work when a rod swings down?

What is the condition for a body to be in equilibrium?

In a ladder problem, why does friction act at the base but not at the smooth wall?

Previous year questions on Law of Rotation And Work-Energy Theorem

Ready to master System Of Particles And Rotational Motion?

4. Law of Rotation: $τ = I α$

Is $τ = I α$ valid about any point?