Why is tau equals dL by dt more general than tau equals I alpha?

Because tau equals dL by dt holds even when the moment of inertia I changes with time or when the axis of rotation is not fixed. Expanding L equals I omega gives dL by dt equals I times d omega by dt plus omega times dI by dt, which is I alpha plus omega times dI by dt. Only when I is constant does this reduce to I alpha.

Angular Momentum, Its Conservation and Angular Impulse

Q: In a collision between a particle and a hinged rod, is linear momentum conserved?

No. During the collision the hinge exerts a large impulsive force on the rod, so linear momentum of the particle-plus-rod system is not conserved. Only angular momentum about the hinge is conserved, because the hinge force acts at the hinge and contributes zero torque about it.

Q: What is angular impulse and when is it useful?

Angular impulse is the time integral of torque, and it equals the change in angular momentum. It is useful whenever a torque acts over a short interval and you care only about the net change in rotational state, not the detailed instantaneous angular acceleration. Bat hitting a ball, brake stopping a spinning wheel, and impulsive forces on hinged bodies all fit here.

Q: Is angular momentum a vector?

Yes. It is defined as L equals r cross p, a cross product, so it has both magnitude and direction. Its direction follows the right-hand rule and is perpendicular to the plane of r and p. For rotation about a fixed axis L lies along that axis; its sense is given by curling the fingers along the direction of rotation with the thumb showing L.

PhysicsSystem Of Particles And Rotational MotionFor NEET aspirants

Angular momentum is the rotational analogue of linear momentum. For a particle it is $L = r \times p$ ; for a rigid body rotating about a fixed axis with angular speed $ω$ it reduces to $L = I ω$ . The net external torque on a system equals the rate of change of its angular momentum: $τ_{ext} = d L / d t$ . When $τ_{ext} = 0$ , angular momentum is conserved. Angular impulse is $\int τ d t$ and equals the change in angular momentum. Together these tools crack JEE and NEET problems on spinning skaters, hinged rods struck by particles, jumping off rotating platforms, and any situation where a sudden collision changes rotational state.

Key Formulas - Quick Reference

Angular momentum of a particle about a point: $L = r \times p = m (r \times v)$
Magnitude: $L = m v r sin θ = m v r_{⊥} = p \cdot d_{⊥}$
Rigid body about a fixed axis: $L = I ω$
Rigid body in general motion: $L = L_{cm} + M (r_{cm} \times v_{cm})$
Newton's second law for rotation: $τ_{ext} = \frac{d L}{d t}$
Conservation: if $τ_{ext} = 0$ , then $L = constant$
For a rigid body: $I_{1} ω_{1} = I_{2} ω_{2}$ (when net external torque is zero and $I$ can change)
Angular impulse: $J_{ang} = \int_{t_{1}}^{t_{2}} τ d t = L_{2} - L_{1}$
SI unit of $L$ : $kg m^{2} s^{- 1}$ ; dimensions $[M L^{2} T^{- 1}]$

1. Angular Momentum of a Particle About a Point

Figure 1: Angular momentum

L = r \times p

. Magnitude

L = m v r sin θ = p \cdot r_{⊥}

For a particle of mass $m$ moving with velocity $v$ at position $r$ relative to a chosen point $O$ , its angular momentum about $O$ is

$L = r \times p = m (r \times v)$

Magnitude: $L = m v r sin θ$ , where $θ$ is the angle between $r$ and $v$ . Two equivalent readings:

$L = (r sin θ) p = r_{⊥} p$ : momentum times perpendicular distance from $O$ to the line of motion.
$L = r (m v sin θ) = r p_{⊥}$ : distance times the component of momentum perpendicular to $r$ .

Direction is given by the right-hand rule for $r \times v$ . SI unit is $kg m^{2} s^{- 1}$ , dimensions $[M L^{2} T^{- 1}]$ .

Key insight. A particle moving in a straight line has a nonzero angular momentum about any point not on that line. Its magnitude

L = m v r_{⊥}

stays constant along the entire straight-line motion because

v

is constant and the perpendicular distance

r_{⊥}

from any fixed external point to the fixed line of motion is constant. Angular momentum is not just about circular motion.

Solved Example 1

A particle of mass

m

is projected from origin with speed

u

at angle

θ

above the horizontal. Find its angular momentum about the point of projection when it is at the highest point of its trajectory.

Solution:

At the highest point the velocity is horizontal, $v_{x} = u cos θ$ , and the height is $H = \frac{u ^{2} sin ^{2} θ}{2 g}$ . The perpendicular distance from the origin to this horizontal velocity vector is exactly $H$ .

$L = m v_{x} H = m (u cos θ) \cdot \frac{u ^{2} sin ^{2} θ}{2 g} = \frac{m u ^{3} sin ^{2} θ cos θ}{2 g}$

Direction: perpendicular to the plane of motion (into the page for a projectile going up and to the right).

Solved Example 2

A particle of mass

2 kg

moves with constant velocity

5 m/s

along a straight line whose perpendicular distance from a fixed point

O

3 m

. Show that its angular momentum about

O

is constant, and find its value.

Solution:

Because the particle moves along a fixed straight line with constant speed, the perpendicular distance $r_{⊥}$ from $O$ to that line never changes, and neither does $v$ . Hence

$L = m v r_{⊥} = 2 \times 5 \times 3 = 30 kg m^{2} / s$

is constant throughout the motion. This is consistent with $τ = d L / d t$ : no force acts on the particle, so no torque acts about any point, and $L$ about every point is separately conserved. Direction: perpendicular to the plane containing $O$ and the line of motion (right-hand rule).

2. Angular Momentum of a Rigid Body About a Fixed Axis

For a rigid body rotating about a fixed axis with angular speed $ω$ , every particle at perpendicular distance $r_{i}$ from the axis moves in a circle with speed $v_{i} = r_{i} ω$ . Its contribution to angular momentum about the axis is $m_{i} v_{i} r_{i} = m_{i} r_{i}^{2} ω$ . Summing over the body,

$L = i \sum m_{i} r_{i}^{2} ω = ω i \sum m_{i} r_{i}^{2} = I ω$

So along the fixed axis,

$L = I ω$

This is the direct rotational analogue of $p = m v$ . Direction of $L$ is along the axis, given by the right-hand rule wrapped around the sense of rotation.

2.1 Angular Momentum in Combined Translation and Rotation

When a rigid body both translates (its centre of mass moves with velocity $v_{cm}$ ) and rotates about an axis through its centre of mass, its total angular momentum about a fixed point $O$ splits into two clean pieces:

$L = L_{cm} + M (r_{cm} \times v_{cm})$

Here $L_{cm} = I_{cm} ω$ is the spin angular momentum (about the CM), and the second term is the orbital angular momentum of the CM about $O$ . A rolling wheel is the classic example.

3. Relation Between Torque and Angular Momentum

Differentiating $L = r \times p$ with respect to time,

$\frac{d L}{d t} = \frac{d r}{d t} \times p + r \times \frac{d p}{d t} = v \times m v + r \times F$

The first term is zero (a vector crossed with itself), leaving

$\frac{d L}{d t} = τ_{ext}$

This is the rotational Newton's second law in its most general form. Note the analogy with $F_{ext} = d p / d t$ . If we specialize to a rigid body about a fixed axis with $L = I ω$ and constant $I$ , we recover $τ = I α$ .

4. Conservation of Angular Momentum

If the net external torque on a system is zero, the total angular momentum of the system is conserved:

$τ_{ext} = 0 ⟹ L = constant$

For a rigid body whose moment of inertia can change (a skater pulling in arms, a person on a rotating stool),

$I_{1} ω_{1} = I_{2} ω_{2}$

Pull the mass closer to the axis ( $I$ decreases) and $ω$ rises. Push it out ( $I$ increases) and $ω$ falls. This is why a spinning skater speeds up dramatically by folding arms in.

Choice of reference point matters. Angular momentum is conserved only about a point (or axis) where the net external torque is zero. In a hinged-rod-plus-particle collision, gravity acts and the hinge exerts an impulsive force. About the hinge itself, the hinge reaction contributes zero torque and gravity's impulse during the very short collision is negligible, so angular momentum about the hinge is conserved. About any other point, it is not.

4.1 Hinged Rod Struck by a Particle

Solved Example 3

A uniform rod of mass

m

and length

L

is hinged at one end

O

and free to rotate in a horizontal plane about a vertical axis through

O

. A point particle of the same mass

m

, moving horizontally with speed

u

perpendicular to the rod, strikes the free end and sticks to it. Find the angular speed of the rod immediately after the collision.

Figure 2: Top view. Particle of speed

u

strikes the free end perpendicular to the rod. Hinge reaction has zero torque about

O

, so angular momentum about

O

is conserved.

Solution:

During the collision, gravity acts vertically (perpendicular to the horizontal plane of motion), and the hinge exerts a horizontal impulsive force. About the hinge $O$ , both these forces have zero torque during the collision, so angular momentum about $O$ is conserved.

Before: only the particle carries angular momentum: $L_{i} = m uL$ .

After: rod plus stuck particle rotates about $O$ with angular speed $ω$ . Moment of inertia:

$I = rod \frac{m L ^{2}}{3} + particle at end m L^{2} = \frac{4 m L ^{2}}{3}$

Angular momentum after: $L_{f} = I ω = \frac{4 m L ^{2}}{3} ω$ .

Setting $L_{i} = L_{f}$ :

$m uL = \frac{4 m L ^{2}}{3} ω \Rightarrow ω = \frac{3 u}{4 L}$

Solved Example 4

A bullet of mass

m

moving horizontally with speed

u

hits a uniform rod of mass

M

and length

L

that is hinged at its upper end and hanging vertically. The bullet embeds itself at a distance

d

from the hinge

(d \leq L)

. Find the angular speed of the rod immediately after the collision.

Solution:

The hinge exerts an impulsive reaction, so linear momentum is not conserved. During the very brief collision, gravity's angular impulse is negligible, and the hinge force has zero torque about the hinge. So angular momentum about the hinge is conserved.

Before: only the bullet contributes. Its perpendicular distance from the hinge is $d$ , so $L_{i} = m u d$ .

After: rod plus embedded bullet rotates about the hinge with angular speed $ω$ . Moment of inertia about the hinge:

$I_{f} = rod about end \frac{M L ^{2}}{3} + bullet at distance d m d^{2}$

Conservation gives $m u d = I_{f} ω$ , so

$ω = \frac{3 m u d}{M L ^{2} + 3 m d ^{2}}$

Checks: (i) $d = L$ and $M = m$ recovers $ω = 3 u / (4 L)$ from Example 3. (ii) $d \to 0$ (hit at hinge) gives $ω \to 0$ , correct because the bullet delivers no angular impulse about the hinge.

Solved Example 5

A uniform disc of mass

M

and radius

R

rotates freely about a vertical axis through its centre with angular speed

ω_{0}

. A stationary particle of mass

m

gently lands on and sticks to the rim. Find the new angular speed.

Figure 3: Top view. Particle lands gently on the rim and sticks. External torque about the spin axis is zero, so

L = I ω

is conserved.

Solution:

No external torque about the axis, so angular momentum about the axis is conserved.

Before: $L_{i} = I_{disc} ω_{0} = \frac{1}{2} M R^{2} ω_{0}$ .

After: the particle sits at distance $R$ from the axis, adding $m R^{2}$ to the moment of inertia. New $I_{f} = \frac{1}{2} M R^{2} + m R^{2}$ .

$ω_{f} = \frac{I _{i} ω _{0}}{I _{f}} = \frac{\frac{1}{2} M R ^{2} ω _{0}}{\frac{1}{2} M R ^{2} + m R ^{2}} = \frac{M ω _{0}}{M + 2 m}$

Solved Example 6

A ballerina spins on a frictionless point with her arms outstretched. In this position her moment of inertia about the vertical spin axis is

I_{1} = 6 kg m^{2}

and her angular speed is

ω_{1} = 1 rev/s

. She then pulls her arms in tightly, reducing her moment of inertia to

I_{2} = 2 kg m^{2}

. Find (a) her new angular speed, and (b) the ratio of final to initial rotational kinetic energy. Where does the extra kinetic energy come from?

Solution:

(a) Friction at the point contact and gravity have no torque about the vertical spin axis, so angular momentum about that axis is conserved:

$I_{1} ω_{1} = I_{2} ω_{2} \Rightarrow ω_{2} = \frac{I _{1}}{I _{2}} ω_{1} = \frac{6}{2} \times 1 = 3 rev/s$

(b) Ratio of kinetic energies:

$\frac{K _{2}}{K _{1}} = \frac{\frac{1}{2} I _{2} ω _{2}^{2}}{\frac{1}{2} I _{1} ω _{1}^{2}} = \frac{I _{2}}{I _{1}} \cdot (\frac{ω _{2}}{ω _{1}})^{2} = \frac{2}{6} \times 9 = 3$

So the KE triples. Since $L$ is fixed and $K = L^{2} / (2 I)$ , halving/tripling $I$ inversely scales $K$ . The extra kinetic energy comes from work done by the ballerina's muscles as she pulls her arms inward against the outward centrifugal reaction — internal work, not from any external torque.

4.2 Linear vs Angular Momentum: When to Use Which

Situation	Linear momentum	Angular momentum
Particle strikes free rod on smooth surface (no hinge)	Conserved	Conserved about any point
Particle strikes hinged rod	Not conserved (hinge impulse)	Conserved about the hinge only
Skater pulls in arms	Trivially conserved (no external horizontal force)	Conserved (no external vertical torque)
Sphere rolling on smooth incline	Not conserved (gravity)	Not conserved about arbitrary points

5. Angular Impulse

The angular impulse of a torque acting over a time interval is the time integral of the torque:

$J_{ang} = \int_{t_{1}}^{t_{2}} τ d t$

Combined with $τ = d L / d t$ , this gives the angular impulse-momentum theorem:

$J_{ang} = Δ L = L_{2} - L_{1}$

The angular impulse of the net torque equals the change in angular momentum. This is the rotational counterpart of $F Δ t = Δ p$ . Useful whenever a torque acts for a short but non-zero time and you want a change in angular speed without computing the detailed time-varying angular acceleration.

Solved Example 7

A wheel of moment of inertia

I = 0.5 kg m^{2}

is initially rotating with angular speed

10 rad/s

. A tangential braking force applies a constant torque of

2 N m

opposing the rotation for

2 s

. Find the angular speed at the end of the interval.

Solution:

Angular impulse (magnitude) delivered by the brake: $J = τ Δ t = 2 \times 2 = 4 N m s$ , opposing the rotation.

Change in angular momentum: $Δ L = - J = - 4 kg m^{2} / s$ .

Initial angular momentum: $L_{i} = I ω_{i} = 0.5 \times 10 = 5 kg m^{2} / s$ .

Final: $L_{f} = 5 - 4 = 1 kg m^{2} / s$ , so $ω_{f} = L_{f} / I = 1/0.5 = 2 rad/s$ .

6. Person Jumping Off a Rotating Platform

Solved Example 8

A circular platform of moment of inertia

I

and radius

R

rotates freely about its vertical axis with angular speed

ω_{0}

. A person of mass

m

stands at the rim. The person then jumps off tangentially with speed

u

relative to the ground, in the direction opposite to the platform's motion at that point. Find the new angular speed of the platform.

Figure 4: Top view. Person jumps tangentially opposite to rim velocity. External torque about spin axis is zero throughout, so total angular momentum about the axis is conserved.

Solution:

No external torque about the platform's axis, so angular momentum about that axis is conserved.

Before: person plus platform rotate together. $L_{i} = (I + m R^{2}) ω_{0}$ .

After the jump: person moves with speed $u$ in the direction opposite to the platform's rim velocity at that instant. The person's angular momentum about the axis is $- m u R$ (negative because it opposes the platform's rotation).

Platform continues to rotate with angular speed $ω$ ; its angular momentum is $I ω$ .

Conservation:

$(I + m R^{2}) ω_{0} = I ω - m u R$

$ω = \frac{( I + m R ^{2} ) ω _{0} + m u R}{I}$

The platform speeds up, as expected: the person's negative angular momentum forces the platform's angular momentum to grow to keep the total unchanged.

Common Mistakes to Avoid

Watch out

Applying conservation of angular momentum about an arbitrary point without checking that the net external torque about that point is zero.
Assuming linear momentum is conserved in a collision with a hinged body. The hinge always delivers an impulsive reaction, so only angular momentum about the hinge is guaranteed to be conserved.
Forgetting the second term $M (r_{cm} \times v_{cm})$ when computing angular momentum of a body that both translates and rotates.
Writing $L = I ω$ for a rigid body about a point that is not on the axis of rotation. That formula holds only about the axis itself.
Confusing angular momentum $L = m v r sin θ$ with angular momentum in circular motion $L = m v r$ . The general formula includes $sin θ$ ; only when $v$ is perpendicular to $r$ does the sine equal one.
Assuming a particle moving in a straight line has zero angular momentum. It has zero angular momentum only about points on that line.
Neglecting the sign (direction) of angular momentum when combining contributions of different objects that rotate the opposite way about the same axis.
Using $I_{i} ω_{i} = I_{f} ω_{f}$ when external torques are present. This shortcut applies only when $τ_{ext} = 0$ about the axis in question.

Frequently Asked Questions

Does a particle moving in a straight line have angular momentum?

Yes, about any point not on its line of motion. The magnitude is $L = m v r_{⊥}$ , where $r_{⊥}$ is the perpendicular distance from the chosen point to the straight-line path. It is zero only about points on the line itself. Angular momentum is not restricted to circular or curved motion.

Why does a spinning skater speed up when she pulls in her arms?

No external torque acts about the vertical axis (friction from the ice is negligible), so her angular momentum $L = I ω$ is conserved. Pulling in her arms brings mass closer to the axis, decreasing her moment of inertia $I$ . For $L$ to stay the same, $ω$ must increase in the same ratio.

In a collision between a particle and a hinged rod, is linear momentum conserved?

No. During the collision, the hinge exerts a large impulsive force on the rod (that is what stops the hinge end from flying off), so linear momentum of the particle-plus-rod system is not conserved. Only angular momentum about the hinge is conserved, because the hinge force acts at the hinge and contributes zero torque about it.

What is angular impulse and when is it useful?

Angular impulse is the time integral of torque, $\int τ d t$ , and it equals the change in angular momentum. It is useful whenever a torque acts over a short interval and you care only about the net change in rotational state, not the detailed instantaneous angular acceleration. Bat hitting a ball, brake stopping a spinning wheel, and impulsive forces on hinged bodies all fit here.

Is angular momentum a vector?

Yes. It is defined as $L = r \times p$ , a cross product, so it has both magnitude and direction. Its direction follows the right-hand rule and is perpendicular to the plane of $r$ and $p$ . For rotation about a fixed axis, $L$ lies along that axis; its sense is given by curling the fingers along the direction of rotation with the thumb showing $L$ .

Can angular momentum be conserved even when linear momentum is not?

Yes, and this is very common in hinged-body problems. The hinge exerts an external force (so linear momentum changes) but that force passes through the hinge, so its torque about the hinge is zero and angular momentum about the hinge is conserved. Similarly, gravity acts on a rotating platform but its torque about the vertical spin axis is zero, so spin angular momentum survives even though linear momentum does not.

Why is $τ = d L / d t$ more general than $τ = I α$ ?

Because $τ = d L / d t$ holds even when the moment of inertia $I$ changes with time (a rotating body reshaping itself) or when the axis of rotation is not fixed. Expanding $L = I ω$ gives $d L / d t = I (d ω / d t) + ω (d I / d t) = I α + ω (d I / d t)$ . Only when $I$ is constant does this reduce to $I α$ .

In problems where a mass sticks to a rotating body, why is kinetic energy not conserved but angular momentum is?

Because the sticking process is a perfectly inelastic collision, and inelastic collisions always dissipate kinetic energy (into heat, sound, deformation). But if no external torque acts about the axis, angular momentum is still exactly conserved. Use angular-momentum conservation to find the new $ω$ ; then compute the KE loss as the difference between initial and final kinetic energies.

Previous year questions on Angular Momentum, Its Conservation and Angular Impulse

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

Show all 23 questions

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1. Angular Momentum of a Particle About a Point

2. Angular Momentum of a Rigid Body About a Fixed Axis

2.1 Angular Momentum in Combined Translation and Rotation

3. Relation Between Torque and Angular Momentum

4. Conservation of Angular Momentum

4.1 Hinged Rod Struck by a Particle

4.2 Linear vs Angular Momentum: When to Use Which

5. Angular Impulse

6. Person Jumping Off a Rotating Platform

Common Mistakes to Avoid

Frequently Asked Questions

Does a particle moving in a straight line have angular momentum?

Why does a spinning skater speed up when she pulls in her arms?

In a collision between a particle and a hinged rod, is linear momentum conserved?

What is angular impulse and when is it useful?

Is angular momentum a vector?

Can angular momentum be conserved even when linear momentum is not?

Why is τ=dL/dt more general than τ=Iα?

In problems where a mass sticks to a rotating body, why is kinetic energy not conserved but angular momentum is?

Previous year questions on Angular Momentum, Its Conservation and Angular Impulse

Ready to master System Of Particles And Rotational Motion?

Why is $τ = d L / d t$ more general than $τ = I α$ ?