A particle of mass $m$ moves in the $yz$-plane along a straight line parallel to the $+y$-axis, staying at the constant height $z = a$ above the $y$-axis, with uniform speed $v$ directed along $+y$. It strikes a rigid wall that is perpendicular to the $y$-axis (at some fixed value of $y$) and rebounds elastically, so that it then travels back along the same line $z = a$ in the $-y$ direction with the same speed $v$. Taking $\hat{e}_x$ as the unit vector along the $x$-axis, find the change in the particle's angular momentum about the origin produced by the bounce. (a) $mva\,\hat{e}_x$ (b) $2mva\,\hat{e}_x$ (c) $ymv\,\hat{e}_x$ (d) $2ymv\,\hat{e}_x$

A particle of mass m moves in the yz-plane along a straight line parallel to the +y-axis, staying at the const

NCERT Exemplar · Q3

m

(a)

mva\,\hat{e}_x

(b)

2mva\,\hat{e}_x

(c)

ymv\,\hat{e}_x

(d)

2ymv\,\hat{e}_x

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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

Angular momentum is a vector. Its direction is given by the right-hand rule: curl your fingers in the direction of rotation, and your thumb points along $\mathbf{L}$ . This direction matters when you add or subtract angular momenta, or when torques change it.

Watch out

A common mistake is to treat angular momentum as a scalar. It is not — direction is crucial, especially in problems involving precession or collisions.

The Bottom Line

Angular momentum is the rotational twin of linear momentum. For a particle, it is $\mathbf{r} \times \mathbf{p}$ ; for a spinning rigid body, it is $I\omega$ . It is conserved when no external torque acts, and that conservation is one of the most powerful tools in physics — from explaining the spin of planets to the behaviour of gyroscopes.

Angular momentum: $L = I\omega$ (rigid body) or $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ (particle).

Angular momentum is a central concept in the NCERT Class 11 Physics chapter on System of Particles and Rotational Motion, and 'angular momentum formula L = Iω' or 'angular momentum important questions class 11 physics' are common board and JEE Main searches. This particle-versus-rigid-body distinction is also essential groundwork for the conservation-of-angular-momentum numericals that follow in the same chapter.

The particle's angular momentum about the origin is set by the fixed perpendicular distance $a$ from the $y$ -axis of motion, giving magnitude $mva$ along $x$ . Reversing the velocity in the bounce flips this vector's sign, so the change is twice the initial value: $2mva\,\hat{e}_x$ .

Concept

Angular momentum about the origin: $\vec{L} = \vec{r}\times \vec{p}$ , with $\vec{p}=m\vec{v}$ .

Set-up

The particle is at $\vec{r} = (0,\,y,\,a)$ (in the $yz$ -plane at height $z=a$ ). Its momentum before the bounce is $\vec{p}_i = (0,\,mv,\,0)$ .

Steps

Before the bounce:

\vec{L}_i = \vec{r}\times\vec{p}_i = \begin{vmatrix}\hat{e}_x & \hat{e}_y & \hat{e}_z\\ 0 & y & a\\ 0 & mv & 0\end{vmatrix} = (y\cdot 0 - a\cdot mv)\,\hat{e}_x = -mva\,\hat{e}_x.

After the elastic bounce the speed is unchanged but the direction reverses: $\vec{p}_f = (0,\,-mv,\,0)$ , so

\vec{L}_f = (y\cdot 0 - a\cdot(-mv))\,\hat{e}_x = +mva\,\hat{e}_x.

Change:

\Delta\vec{L} = \vec{L}_f - \vec{L}_i = mva\,\hat{e}_x - (-mva\,\hat{e}_x) = 2mva\,\hat{e}_x.

Note that the $y$ -coordinate cancels out; only the fixed perpendicular distance $a$ enters. Distractor (A) drops the factor of 2 (it is only $|\vec{L}_i|$ ); (C) and (D) wrongly use the variable $y$ instead of the true moment arm $a$ .

✓Final answer

2mva\,\hat{e}_x

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:

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

This is dimensionally identical to the joule-second (J·s), since $1\text{ J}=1\text{ kg}\,\text{m}^2\,\text{s}^{-2}$ , so $1\text{ J·s}=1\text{ kg}\,\text{m}^2\,\text{s}^{-1}$ .

✓Final answer
$\text{kg}\,\text{m}^2\,\text{s}^{-1}$
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

This magnitude, mvy0, depends only on the mass m, the speed v, and the perpendicular distance y0 from the origin to the line of motion — none of which change as the particle moves along the straight line. So the magnitude of L stays exactly constant, regardless of x.

This is a general result: for any particle in straight-line motion with constant velocity, the angular momentum about any fixed point is conserved, because no net torque acts about that point.

✓Final answer
$The correct option is (c) remaining constant — the perpendicular distance from the origin to the line of motion (y0) doesn't change, so L = m v y0 stays the same throughout the motion.$
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

where r is the position vector of the particle from the chosen point, and p = mv is its linear momentum. Angular momentum is the rotational counterpart of linear momentum and plays the same central role in rotational dynamics that p plays in translational dynamics (e.g., τ = dL/dt, analogous to F = dp/dt).

✓Final answer
$Moment of linear momentum is called angular momentum.$
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 magnitude and direction, and it keeps moving along the same straight line, so $d$ never changes either. Hence $L=pd$ stays constant throughout the motion.

✓Final answer
$(B) Remains constant.$
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 angular velocity. It is directly proportional to $\omega$ , not $\omega^2$ .

✓Final answer
$I\omega$
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 x time (or change in momentum), and linear momentum is mass x velocity — none of these match I x ω.

✓Final answer
$(d) Angular momentum.$
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 containing $\vec r$ and $\vec p$ , and it does not reverse sign under a mirror reflection the way a polar vector does. Torque, angular velocity, and magnetic field are other common axial vectors.

✓Final answer
$The correct option is (B) Axial vector.$
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).

If KE is halved (KE becomes KE/2), and I is unchanged,

L' = sqrt(2 I (KE/2)) = sqrt(I x KE) = sqrt(2 I KE) / sqrt(2) = L/sqrt(2)

✓Final answer
$The correct option is (c) L/sqrt(2).$
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

kg x m^2 -- missing the 1/s factor, incomplete

Newton x m / Sec = kg m^2/s^3 -- this is closer to a power-like unit Only option (a) has the correct dimensions.

✓Final answer
$The correct option is (a) Joule x Second.$
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

This is the standard definition used throughout rotational mechanics, appearing for instance in the rotational analogue of Newton's second law, torque = dL/dt. Among the 7 given options, only (d), I*omega, matches.

✓Final answer
$(d) I*omega.$
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

So by direct definition, the moment of linear momentum is called angular momentum. (Torque, by contrast, is the moment of force, not of momentum; a couple is a pair of equal and opposite forces; impulse is force x time.)

✓Final answer
$(4) Angular momentum.$
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:

L2 = 2 KE2 / omega2 = 2 (KE1/2) / (2 omega1) = KE1 / (2 omega1)

Compare to L1 = 2 KE1/omega1, so KE1/omega1 = L1/2.

L2 = (1/2) x (KE1/omega1) = (1/2) x (L1/2) = L1/4

So the new angular momentum is one-quarter of the original.

✓Final answer
$(2) L/4.$

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