A particle undergoes uniform circular motion. The angular momentum of the particle remains conserved about, (a

I. Multiple Choice Questions · Q7

Q. A particle undergoes uniform circular motion. The angular momentum of the particle remains conserved about,

(a) the center point of the circle

(b) the point on the circumference of the circle

(d) any point outside the circle

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Concept understanding — Angular Momentum

Angular Momentum: The Rotational Cousin of Momentum

You already know linear momentum — how hard it is to stop a moving object. A truck moving at 20 m/s has more momentum than a bicycle at the same speed. Now imagine something spinning: a bicycle wheel, a spinning top, or a figure skater pulling their arms in. There is a similar "quantity of motion" for rotation, and that is angular momentum.

The core intuition is simple: angular momentum measures how much rotation an object has, and how hard it is to change that rotation. Just as a heavy truck is hard to stop, a heavy flywheel spinning fast is hard to stop spinning.

The Two Faces of Angular Momentum

Angular momentum appears in two forms, depending on what you are studying.

For a single particle moving in a straight line or a curve, angular momentum is defined relative to a chosen point (usually the centre of rotation). It is the cross product of the position vector and the linear momentum:

\mathbf{L} = \mathbf{r} \times \mathbf{p}

Here $\mathbf{r}$ is the vector from the reference point to the particle, and $\mathbf{p} = m\mathbf{v}$ is its linear momentum. The magnitude is $L = r p \sin\theta$ , where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{p}$ . This tells you: the farther the particle is from the point, and the faster it moves perpendicular to that line, the greater its angular momentum.

For a rigid body rotating about a fixed axis, the formula simplifies beautifully. Every particle in the body contributes, and when you add them all up, you get:

L = I \omega

where $I$ is the moment of inertia (the rotational analogue of mass) and $\omega$ is the angular velocity (how fast it spins). This is the direct parallel of $p = mv$ .

\text{Linear: } p = mv \quad \longleftrightarrow \quad \text{Rotational: } L = I\omega

Why Angular Momentum Matters

The real power of angular momentum is its conservation. In the absence of an external torque (the rotational analogue of force), angular momentum stays constant. This is why a figure skater spins faster when she pulls her arms in — her moment of inertia $I$ decreases, so $\omega$ must increase to keep $L$ constant.

Important

Conservation of Angular Momentum: If net external torque $\tau_{\text{ext}} = 0$ , then $\mathbf{L}$ is constant in both magnitude and direction.

This principle explains everything from why a bicycle stays upright to why neutron stars spin at incredible speeds after a supernova collapse.

Connecting the Two Definitions

The particle definition $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ is the fundamental one. The rigid-body formula $L = I\omega$ is derived from it by summing over all particles in the body. For a single particle moving in a circle of radius $r$ with speed $v$ , you get $L = rmv = mr^2\omega = I\omega$ , since $I = mr^2$ for that particle.

So the two definitions are not separate — they are the same idea at different levels of description.

A Quick Check on Direction …

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Step 1. Recall the condition for angular momentum to be conserved about a point.

\vec\tau = \dfrac{d\vec L}{dt},

so $\vec L$ about a given point stays constant precisely when the net torque about that same point is zero at every instant.

Step 2. Identify the force acting on the particle.

In uniform circular motion, the only (net) force on the particle is the centripetal force, which has constant magnitude and is always directed radially inward, from the particle's current position straight toward the center of the circle.

Step 3. Compute the torque about the center.

Torque about a point is $\vec\tau = \vec r \times \vec F$ , where $\vec r$ is measured from that point to where the force acts. About the center, $\vec r$ (from center to particle) and $\vec F$ (from particle back toward center) are exactly anti-parallel (both lie along the same radial line). Since the cross product of two parallel (or anti-parallel) vectors is zero,

\vec\tau_{center} = \vec r \times \vec F = 0 \quad\text{at every instant.}

So $\vec L$ about the center never changes — it is conserved.

Step 4. Check what happens about any other point. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 18 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. Write S. I. unit of Angular Momentum.$
›Reveal solutionSolution

$L=I\omega$ has SI unit $\text{kg}\,\text{m}^2\,\text{s}^{-1}$ , same as J·s.

Angular momentum is $L=\vec r \times \vec p = I\omega$ , where moment of inertia $I$ has SI unit $\text{kg}\,\text{m}^2$ and angular velocity $\omega$ has SI unit $\text{s}^{-1}$ (rad/s, but radian is dimensionless). Multiplying:

$[L] = \text{kg}\,\text{m}^2 \times \text{s}^{-1} = \text{kg}\,\text{m}^2\,\text{s}^{-1}$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. A particle is moving with a constant velocity along a straight line parallel to positive x-axis. The magnitude of its angular momentum with respect to origin is:$
(a) decreasing with x
(b) zero
(c) remaining constant
(d) increasing with x
›Reveal solutionSolution

Angular momentum about the origin depends on the perpendicular distance from the origin to the particle's line of motion; for straight-line motion parallel to the x-axis, this perpendicular distance (the particle's fixed y-coordinate) never changes, so L is constant.

The angular momentum of a particle about the origin is

L = r x p = m(r x v)

For a particle moving along a straight line parallel to the positive x-axis at a fixed height y = y0, with constant velocity v = vx-hat, the position vector is r = xx-hat + y0*y-hat.

L = m(r x v) = m[(xx-hat + y0y-hat) x (vx-hat)] = m[xv*(x-hat x x-hat) + y0v(y-hat x x-hat)] = -mvy0*z-hat

…

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CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: Moment of linear momentum is called ____.$
›Reveal solutionSolution

The moment of linear momentum about a point is called angular momentum.

Just as the moment of a force about a point is called torque (τ = r × F), the moment of linear momentum p about a point is defined as angular momentum:

L = r × p …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. A particle is moving with constant velocity parallel to x-axis. Its angular momentum relative to origin point (A) is zero (B) remains constant (C) goes on increasing (D) goes on decreasing$
›Reveal solutionSolution

A particle in straight-line uniform motion has constant angular momentum about any fixed point (unless it passes through that point along the line).

Angular momentum about the origin is $\vec L=\vec r\times\vec p$ , with magnitude $L=p\cdot d$ , where $d$ is the perpendicular distance from the origin to the line along which the particle moves. Since the particle moves with constant velocity, $\vec p=m\vec v$ is constant in both m …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. A body of mass M is moving with uniform angular velocity ω about its axis of rotation. I is its moment of inertia about this axis. Its angular momentum will be (A) ½Iω^2 (B) MIω^2 (C) I^2ω (D) Iω$
›Reveal solutionSolution

Angular momentum of a rotating body is $L=I\omega$ .

Just as linear momentum is $p=mv$ , angular momentum about the rotation axis is defined as $L=I\omega$ , where $I$ is the moment of inertia about that axis and $\omega$ is the a …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The product of moment of inertia and angular velocity is called$
(a) Torque
(b) Impulse
(c) Linear momentum
(d) Angular momentum
›Reveal solutionSolution

Angular momentum L is defined as the product of moment of inertia I and angular velocity ω: L = Iω.

This is the rotational analogue of linear momentum p = mv, with mass replaced by moment of inertia and linear velocity replaced by angular velocity. Torque is the rotational analogue of force, impulse is force …

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CBSE 2024Set sz1 markMCQ
$Q. Angular momentum is a: (A) Polar vector (B) Axial vector (C) Scalar (D) None of these$
›Reveal solutionSolution

Angular momentum is an axial (pseudo) vector because it is defined as the cross product of two polar vectors, position and linear momentum.

Angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$ .

Both $\vec{r}$ (position) and $\vec{p}$ (linear momentum) are polar (true) vectors — their direction is along an actual physical displacement or motion. The cross product of two polar vectors, however, produces an axial vector (also called a pseudovector): its direction is assigned by convention (the right-hand rule) perpendicular to the plane cont …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. A rigid body rotates with an angular momentum L. If its kinetic energy is halved, the angular momentum becomes :$
(a) 2L
(b) L
(c) L/√2
(d) L/2
›Reveal solutionSolution

Since KE = L^2/(2I), halving KE while keeping I fixed means L is scaled by 1/sqrt(2).

The rotational kinetic energy of a rigid body is

KE = (1/2) I omega^2

Angular momentum is L = I omega, so omega = L/I. Substituting:

KE = (1/2) I (L/I)^2 = L^2 / (2I)

So L = sqrt(2 x I x KE), i.e. L is proportional to sqrt(KE) for a fixed I (no external torque changes I here).

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. SI unit of angular momentum is:$
(a) Joule x Second
(b) Newton x Meter
(c) kg x m^2
(d) Newton x m / Sec
›Reveal solutionSolution
Angular momentum is measured in Joule-second (equivalently kg m^2/s), NOT in kg x m^2 alone or in Newton x metre.

Angular momentum L = I*omega, where moment of inertia I has units kg m^2 and angular velocity omega has units rad/s (dimensionless rad, so effectively 1/s). Hence:

[L] = kg m^2 x (1/s) = kg m^2/s

Now check the given options:

Joule x Second = (kg m^2/s^2) x s = kg m^2/s -- matches

Newton x Metre = (kg m/s^2) x m = kg m^2/s^2 -- this is the unit of TORQUE, not angular momentum …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. Match the column: Angular momentum L — match with the correct expression.$
(a) sqrt(2gR)
(b) sqrt(T/m)
(c) GMm/r^2
(d) I*omega
(e) 2pisqrt(l/g)
(f) sqrt(gR)
(g) m*R^2
›Reveal solutionSolution

Angular momentum is L = Iomega, the rotational counterpart of linear momentum p = mv, matching option (d).

Just as linear momentum p = mv combines mass (inertia for straight-line motion) with linear velocity, angular momentum L combines moment of inertia I (inertia for rotational motion) with angular velocity omega: L = Iomega

…

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CBSE 2022Set TERM11 markMCQ
$Q. Moment of linear momentum is$
(1) Couple
(2) Torque
(3) Impulse
(4) Angular momentum
›Reveal solutionSolution
'Moment of linear momentum' is the literal definition of angular momentum, L = r x p (position vector crossed with linear momentum), just as torque is defined as the moment of force.

In rotational mechanics, taking the 'moment' of a vector quantity about a point means crossing the position vector r (from that point to where the quantity acts) with the quantity itself.

Moment of FORCE = r x F = Torque

Moment of LINEAR MOMENTUM = r x p = Angular Momentum, denoted L …
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CBSE 2022Set TERM11 markMCQ
$Q. A particle performing uniform circular motion has angular momentum L. If its angular frequency is doubled and its kinetic energy halved, then the new angular momentum is$
(1) L/2
(2) L/4
(3) 2L
(4) 4L
›Reveal solutionSolution

For rotational motion, L and KE are related through L = 2(KE)/omega (from KE = (1/2)L.omega). Plugging in the new KE (halved) and new omega (doubled) gives the new angular momentum as one-quarter of the original.

For a particle in uniform circular motion (with moment of inertia I about the axis):

Angular momentum: L = I omega

Kinetic energy: KE = (1/2) I omega^2 = (1/2) (I omega) omega = (1/2) L omega

So: L = 2 (KE) / omega ... (*)

Original state: L1 = 2 KE1 / omega1

New state: omega2 = 2 omega1, KE2 = KE1 / 2

Using (*) for the new state: …

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Q.A particle undergoes uniform circular motion. The angular momentum of the particle remains conserved about,

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