The angular momentum of a planet is given by L = 5 – Class 11 Physics

Question 85 of 87

Q. The angular momentum of a planet is given by L = 5t^2 i - 6t j + 3k (i, j, k are unit vectors). Calculate the torque experienced by the planet. Will the torque be in the same direction as that of the angular momentum?

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2026Subjective· 3mImportance★★★★★

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Concept understanding — Torque

What is Torque? The Intuition

Think about opening a door. You push near the handle, not near the hinges. Why? Because the same push is far more effective when applied farther from the hinge. That "turning effectiveness" of a force is exactly what torque measures.

If you push a door with 10 N of force right at the hinge, the door barely moves. Push with the same 10 N at the handle, and it swings open easily. The force is identical — what changed is the distance from the axis of rotation. Torque combines force and distance into one quantity that tells you how much a force will make something spin.

Note

Torque is to rotation what force is to translation. Force makes things move in a straight line; torque makes things rotate.

The Precise Definition

Torque $\vec{\tau}$ about a point (or axis) is defined as the cross product of the position vector $\vec{r}$ (from the axis to the point where force acts) and the force vector $\vec{F}$ :

\vec{\tau} = \vec{r} \times \vec{F}

The magnitude is:

\tau = rF\sin\theta

where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$ .

The direction of torque is given by the right-hand rule: curl your fingers from $\vec{r}$ toward $\vec{F}$ , and your thumb points along $\vec{\tau}$ . This direction is perpendicular to both $\vec{r}$ and $\vec{F}$ .

Tip

For maximum torque, push perpendicular to the lever arm ( $\theta = 90^\circ$ , so $\sin\theta = 1$ ). Pushing directly toward or away from the axis ( $\theta = 0^\circ$ or $180^\circ$ ) produces zero torque — that's why you can't open a door by pushing straight into it.

Why Torque Equals Rate of Change of Angular Momentum

This is the rotational analogue of Newton's second law: $\vec{F} = d\vec{p}/dt$ .

For a particle of mass $m$ at position $\vec{r}$ with velocity $\vec{v}$ , its linear momentum is $\vec{p} = m\vec{v}$ . Angular momentum about the same point is:

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

Now differentiate with respect to time:

\frac{d\vec{L}}{dt} = \frac{d\vec{r}}{dt} \times \vec{p} + \vec{r} \times \frac{d\vec{p}}{dt}

Since $d\vec{r}/dt = \vec{v}$ and $\vec{p} = m\vec{v}$ , the first term is $\vec{v} \times m\vec{v} = 0$ (cross product of parallel vectors). The second term uses $d\vec{p}/dt = \vec{F}$ , giving:

$\frac{d\vec{L}}{dt} = \vec{r} \times \vec{F} = \vec{\tau}$ …

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Differentiating L with respect to time gives tau = 10t i - 6j; this is not parallel to L = 5t^2 i - 6t j + 3k, so the torque and angular momentum do not point in the same direction.

Torque and angular momentum are related by

tau = dL/dt

Given:

L = 5t^2 i - 6t j + 3k

Differentiating each component with respect to time t:

d(5t^2)/dt = 10t

d(-6t)/dt = -6

d(3)/dt = 0 (since 3 is a constant)

So:

tau = 10t i - 6j + 0k = 10t i - 6j

Now let's check whether tau is parallel to L (i.e., whether tau = lambda*L for some scalar lambda at any time t):

L = (5t^2, -6t, 3), tau = (10t, -6, 0)

For tau to be parallel to L, we would need tau = lambdaL, i.e.: 10t = lambda(5t^2) so lambda = 2/t (for t not equal to 0)

-6 = lambda*(-6t) so lambda = 1/t

0 = lambda*(3) so lambda = 0

…

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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 16 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. What is the unit of Torque?$
(a) N
(b) Nm
(c) Nm^2
(d) None of these
›Reveal solutionSolution

Torque (moment of force) tau = r x F has units of (distance) x (force) = metre x newton = Nm.

Torque is defined as tau = r F sin(theta), the product of the force applied and the perpendicular distance (lever arm) from the axis of rotation.

SI unit of force = newton (N)

SI unit of distance = metre (m)

So SI unit of torque = N x m = Nm.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If the rate of change in angular momentum is zero, then the net torque acting on the body will be -$
(a) Constant
(b) Minimum
(c) Maximum
(d) Zero
›Reveal solutionSolution

Net torque τ = dL/dt; if the rate of change of angular momentum is zero, the net torque is zero.

Just as force is the rate of change of linear momentum (F = dp/dt), torque is the rotational analogue: it is the rate of change of angular momentum, τ = dL/dt. If dL/dt = 0 (angular momentum is constant, i.e. conserved), it directly follows that the net exter …

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CBSE 2024Set SET-NDP60001 markMCQ
$Q. Which of the following relations is not correct:$
(a) F = r × τ
(b) L = r × p
(c) τ = I α
(d) τ = dL/dt
›Reveal solutionSolution
$F = r \times \tau$ is not a valid relation; torque is defined as $\tau = r \times F$ , not the other way round.

Check each option against the standard definitions used in rotational mechanics:

(a) The defining relation for torque is $\vec{\tau} = \vec{r} \times \vec{F}$ (torque = position vector crossed with force). Writing it as $\vec{F} = \vec{r} \times \vec{\tau}$ reverses this relationship and is not a valid physical identity — force cannot generally be recovered this way from torque and position.

(b) $\vec{L} = \vec{r} \times \vec{p}$ is the correct definition of angular momentum of a particle about a point.

(c) $\tau = I\alpha$ is the correct rotational analogue of Newton's second law ( $F = ma$ ), relating net torque, moment of inertia, and angular acceleration. …
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CBSE 2022Set TERM11 markMCQ
$Q. The product of moment of inertia and angular acceleration is called$
(1) angular momentum
(2) torque
(3) force
(4) work
›Reveal solutionSolution

Torque is the rotational analogue of force; the rotational form of Newton's second law is τ = I α, directly defining torque as the product of moment of inertia and angular acceleration.

In linear (translational) motion, Newton's second law states: F = m a (force = mass x linear acceleration).

The exact rotational analogue replaces force with torque, mass with moment of inertia, and linear acceleration with angular acceleration:

tau (torque) = I (moment of inertia) x alpha (angular acceleration)

…

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CBSE 2022Set ANNUAL1 mark
$Q. What is a couple?$
›Reveal solutionSolution
A couple is a pair of equal, opposite, non-collinear forces that produces rotation only.

A couple consists of two forces that are:

equal in magnitude,

opposite in direction, and

acting along two different (parallel but non-coincident) lines of action.

Because the two forces are equal and opposite, their vector sum is zero, so a couple produces no net translational force on the body — the centre of mass does not accelerate linearly due to a couple alone. However, since the two forces act along different lines separated by a perpendicular distance $d$ , each contributes a torque about any point, and these torques add up (they do not cancel, since both forces try to rotate the body in the same sense).

…
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CBSE 2022Set ANNUAL1 markMCQ
$Q. The moment of a couple of forces:$
(a) does not depend on the point around which we find the moment
(b) depends on the point around which we find the moment
(c) does not depend on the moment of the forces making the couple
(d) None of the above
›Reveal solutionSolution

The moment (torque) of a couple has the same magnitude about every point in the plane, so it does not depend on the reference point chosen.

A couple consists of two equal and opposite forces F, separated by a perpendicular distance d, whose lines of action do not coincide. If moments are taken about any arbitrary point O, the contributions from the perpendicular distances of the two forces to O always combine to give the same net moment F×d, because the point-dependent parts cancel out (this is a special property of a couple, unlike a single force whose moment does dep …

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CBSE 2021Set sz1 mark
$Q. Write relation between torque and force.$
›Reveal solutionSolution
Torque is the moment of a force about a point/axis: tau = r x F, its magnitude being r F sin(theta).

When a force F acts on a body at a point whose position vector (measured from the axis or pivot about which turning is considered) is r, the turning effect of the force is measured by the torque (moment of force):

tau = r x F

This is a vector (cross) product, so:

Magnitude: tau = r F sin(theta), where theta is the angle between r and F.

Direction: perpendicular to the plane containing r and F, given by the right-hand rule. …
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CBSE 2021Set ANNUAL1 markMCQ
$Q. A couple produces$
(a) purely rotational motion
(b) purely linear motion
(c) linear and rotational motion
(d) no motion
›Reveal solutionSolution

A couple has zero net force but a nonzero net torque, so it spins a body without moving its centre of mass.

A couple consists of two forces of equal magnitude, opposite direction, acting along two different parallel lines of action (e.g. the two forces used to turn a steering wheel or open a tap). Because the forces are equal and opposite, their vector sum is zero — so there is no net force to accelerate the body's centre of mass, and hence no translational motion. However, since the two forces act along different lines, their torques about any point …

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CBSE 2020Set ANN1 mark
$Q. The rotational analogue of force is$
(a) energy
(b) work
(c) inertia
(d) torque
›Reveal solutionSolution

Torque (moment of force) is the rotational counterpart of force in linear motion.

In translational (linear) motion, force F = dp/dt produces linear acceleration. In rotational motion, the corresponding quantity that produces angular acceleration is torque, τ = dL/dt (rate of change of angular momentum), also written τ = r × F for a force F applied at position vector …

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CBSE 2020Set ANNUAL1 markMCQ
$Q. The unit of torque is:$
(a) N-m
(b) kg m^2 s^-1
(c) kg m s^-1
(d) kg^2 m^2 s^-1
›Reveal solutionSolution

Torque = force x lever arm, so its unit is the same combination as force (N) and distance (m): N-m.

Torque (moment of force) is defined as tau = r x F, the vector (cross) product of the position vector and the force.

Units: [Force] x [Distance] = N x m = N-m.

…

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CBSE 2020Set ANNUAL1 markMCQ
$Q. Analogue of force in rotational motion is:$
(a) Moment of inertia
(b) Torque
(c) Radius of gyration
(d) Angular momentum
›Reveal solutionSolution

Rotational motion has direct analogues of every linear-motion quantity; the analogue of force (which produces linear acceleration) is torque (which produces angular acceleration).

In linear motion: F = m a.

In rotational motion: tau = I alpha (torque = moment of inertia x angular acceleration).

…

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CBSE 2020Set ANNUAL1 markMCQ
$Q. The product of moment of inertia and angular acceleration is called :$
(a) Angular momentum
(b) Torque
(c) Force
(d) Work
›Reveal solutionSolution

Torque $\tau = I\alpha$ is the rotational analogue of Newton's second law $F = ma$ .

Step 1 — Rotational analogue of Newton's second law.

For linear motion, $F = ma$ . For rotational motion about a fixed axis, mass $m$ is replaced by moment of inertia $I$ and linear acceleration $a$ by angular acceleration $\alpha$ :

$\tau = I\alpha$

Step 2 — Identify the quantity. …

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←Previous questionQ26 — Give any two differences between translational and rotational motions.→Next questionQ28 — Write note on any three types of equilibrium.

Q.The angular momentum of a planet is given by L = 5t^2 i - 6t j + 3k (i, j, k are unit vectors). Calculate the torque experienced by the planet. Will the torque be in the same direction as that of the angular momentum?

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