What is the difference between torque and work, since both have units of N·m?

Their dimensions are identical (M L² T⁻²), but torque is a vector while work is a scalar. NCERT states this explicitly: 'Moment of a force is a vector, while work is a scalar.' To avoid confusion the SI unit of torque is written as newton metre (N m) and never as the joule, which is reserved for energy.

When is the torque on a particle zero even though a force acts on it?

Torque τ = rF sinθ vanishes if r = 0 (force applied at the origin), F = 0, or θ = 0° or 180° (the line of action of the force passes through the origin, so the moment arm r⊥ = r sinθ is zero). A force whose line of action passes through the axis produces no turning effect — which is why pushing a door at the hinge line does not rotate it.

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

No external torque acts about her vertical spin axis, so her angular momentum L = Iω is conserved. Pulling her arms in brings mass closer to the axis, lowering her moment of inertia I. Because Iω must stay constant, ω rises. The same principle is used by divers, acrobats and ballet dancers performing a pirouette.

How is τ = dL/dt related to Newton's second law?

It is the exact rotational analogue. For translation, F = dp/dt; for rotation, τ = dL/dt. Force changes linear momentum; torque changes angular momentum. Setting the right-hand side to zero gives the two conservation laws: zero net force conserves p, zero net external torque conserves L.

Do internal forces inside a system contribute to its total torque?

No. Assuming Newton's third law and that the force between any two particles acts along the line joining them, the internal torques cancel in pairs, so τinternal = 0. Only external torques change the system's angular momentum: dL/dt = τexternal. This is why a skater's muscular (internal) forces cannot change her total angular momentum.

Why does a planet move faster when it is nearer the Sun?

The Sun's gravity on a planet always points toward the Sun, i.e. along the position vector r, so it exerts zero torque about the Sun (r × F = 0 when F is along r). The planet's angular momentum about the Sun is therefore conserved. As r shrinks near perihelion, the speed must rise to keep L = mvr⊥ constant — the dynamical basis of Kepler's law of equal areas.

Torque & Angular Momentum — NEET Physics

Q: Are angular momentum L and angular velocity ω always parallel?

Not in general. NCERT notes that L and ω are not necessarily parallel vectors. The simple relation L = Iω holds only when rotation is about a fixed axis that is also an axis of symmetry of the rigid body. NEET problems almost always use this symmetric case, so L = Iω applies, but the qualification is worth remembering.

Angular velocity and the rotation axis

In rotation about a fixed axis, every particle of a rigid body moves in a circle whose plane is perpendicular to the axis and whose centre lies on it. NCERT §6.6 defines the instantaneous angular velocity $\omega = d\theta/dt$, and crucially notes that the same $\omega$ describes every particle — so we call it the angular velocity of the whole body. Particles on the axis have $r_\perp = 0$ and so $v = \omega r_\perp = 0$, confirming the axis is stationary.

Angular velocity is in fact a vector. It lies along the axis of rotation, pointing in the direction a right-handed screw advances when its head turns with the body. The linear velocity of any particle then follows from a vector product, $\mathbf{v} = \boldsymbol{\omega}\times\mathbf{r}$, which is perpendicular to both $\boldsymbol{\omega}$ and $\mathbf{r}$ and tangent to the particle's circular path. The angular acceleration is $\boldsymbol{\alpha} = d\boldsymbol{\omega}/dt$; for a fixed axis it reduces to the scalar $\alpha = d\omega/dt$.

Torque — the moment of a force

NCERT §6.7.1 asks for the rotational analogue of force using a familiar object: a door. A force applied along the hinge line cannot rotate the door at all, while the same force applied at right angles to the outer edge is the most effective. It is not the force alone, "but how and where the force is applied" that matters in rotation. That combination is the moment of force, used interchangeably with torque.

For a force $\mathbf{F}$ acting at a point P with position vector $\mathbf{r}$ about the origin O, the torque is the vector product

$$ \boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}, \qquad \tau = rF\sin\theta $$

where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{F}$. Torque is a vector; its direction is perpendicular to the plane containing $\mathbf{r}$ and $\mathbf{F}$, fixed by the right-handed screw rule. Its SI unit is the newton metre (N m), and its dimensions are $[M\,L^2\,T^{-2}]$ — the same as energy, yet it is a completely different quantity.

Torque $\boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}$. Its magnitude $rF\sin\theta$ equals the force times the moment arm $r_\perp = r\sin\theta$, the perpendicular distance from O to the line of action of $\mathbf{F}$. The torque vector points out of the plane (right-hand rule). Based on NCERT Fig. 6.18.

Moment arm and when torque vanishes

The magnitude can be regrouped in two equivalent ways that NCERT highlights, and both carry physical meaning:

$$ \tau = r\,(F\sin\theta) = r\,F_\perp \qquad\text{and}\qquad \tau = (r\sin\theta)\,F = r_\perp F $$

Here $F_\perp = F\sin\theta$ is the component of force perpendicular to $\mathbf{r}$, and $r_\perp = r\sin\theta$ is the moment arm — the perpendicular distance of the line of action of the force from the origin. The second form is the door insight made exact: only the part of the force "off the line" turns the body.

Angular momentum of a particle

Angular momentum is the rotational analogue of linear momentum — the "moment of (linear) momentum". For a particle of mass $m$ and momentum $\mathbf{p} = m\mathbf{v}$ at position $\mathbf{r}$ about the origin O, NCERT §6.7.2 defines

$$ \mathbf{l} = \mathbf{r}\times\mathbf{p}, \qquad l = r\,p\sin\theta = r_\perp\, p = r\, p_\perp $$

Just as torque mirrors the structure of $\mathbf{r}\times\mathbf{F}$, angular momentum mirrors $\mathbf{r}\times\mathbf{p}$. It is zero if $p = 0$, if the particle sits at the origin ($r = 0$), or if the line of $\mathbf{p}$ passes through the origin ($\theta = 0^\circ$ or $180^\circ$). The vector $\mathbf{l}$ is perpendicular to the plane of $\mathbf{r}$ and $\mathbf{p}$.

Angular momentum $\mathbf{l} = \mathbf{r}\times\mathbf{p}$ of a particle about O. Magnitude $l = r\,p\sin\theta = r_\perp\,p$. A particle moving with constant velocity has constant $\mathbf{l}$ about any point, because $r_\perp$ (here $=OM$) stays fixed — NCERT Example 6.6.

The master relation τ = dL/dt

Differentiating $\mathbf{l} = \mathbf{r}\times\mathbf{p}$ with respect to time and using $\mathbf{v} = d\mathbf{r}/dt$, $\mathbf{p} = m\mathbf{v}$ and $\mathbf{F} = d\mathbf{p}/dt$:

$$ \frac{d\mathbf{l}}{dt} = \frac{d\mathbf{r}}{dt}\times\mathbf{p} + \mathbf{r}\times\frac{d\mathbf{p}}{dt} = (\mathbf{v}\times m\mathbf{v}) + (\mathbf{r}\times\mathbf{F}) = \boldsymbol{\tau} $$

The first term vanishes because the cross product of parallel vectors ($\mathbf{v}$ and $m\mathbf{v}$) is zero. What remains is the rotational analogue of Newton's second law:

$$ \boldsymbol{\tau} = \frac{d\mathbf{l}}{dt} \quad\longleftrightarrow\quad \mathbf{F} = \frac{d\mathbf{p}}{dt} $$

Torque and angular momentum of a system

For a system of $n$ particles, the total angular momentum about a point is the vector sum of the individual angular momenta, and the total torque is the vector sum of the individual torques:

$$ \mathbf{L} = \sum_i \mathbf{l}_i = \sum_i \mathbf{r}_i\times\mathbf{p}_i, \qquad \boldsymbol{\tau} = \sum_i \mathbf{r}_i\times\mathbf{F}_i $$

The force on each particle splits into external and internal parts. NCERT assumes Newton's third law and that the forces between any two particles act along the line joining them. Under these assumptions the internal torques cancel in action–reaction pairs, so $\boldsymbol{\tau}_{\text{internal}} = 0$. What survives is the system law:

$$ \frac{d\mathbf{L}}{dt} = \boldsymbol{\tau}_{\text{external}} $$

This is the rotational analogue of $d\mathbf{P}/dt = \mathbf{F}_{\text{external}}$, and it holds for any system of particles — rigid or not. NCERT also flags an independence worth keeping: the vanishing of total external force and the vanishing of total external torque are separate conditions. A couple has zero net force but non-zero net torque.

Foundation

Both $\boldsymbol{\tau}$ and $\mathbf{L}$ are cross products — revisit the rules at vector product of two vectors if the right-hand rule feels shaky.

The relation L = Iω

For a rigid body rotating about a fixed axis that is also an axis of symmetry, the component of angular momentum along the axis reduces to a simple product of the moment of inertia and the angular velocity:

$$ L = I\omega $$

This sits neatly in the analogy table: $\mathbf{p} = m\mathbf{v}$ for translation becomes $L = I\omega$ for rotation, with mass replaced by moment of inertia and linear velocity by angular velocity. Combined with $\tau = dL/dt$, a constant $I$ gives $\tau = I\,d\omega/dt = I\alpha$.

Conservation of angular momentum

The payoff of $d\mathbf{L}/dt = \boldsymbol{\tau}_{\text{ext}}$ appears when the right-hand side is zero. If the total external torque on a system is zero,

$$ \boldsymbol{\tau}_{\text{ext}} = 0 \;\Rightarrow\; \mathbf{L} = \text{constant}, \qquad\text{i.e. } I\omega = \text{constant} $$

This is the rotational counterpart of conservation of linear momentum, and it is among the highest-yield ideas in this chapter for NEET. NCERT's swivel-chair demonstration captures it: a person spinning on a frictionless chair with folded arms speeds up; stretching the arms out increases $I$ about the axis and so reduces $\omega$, while drawing them in does the reverse. Because friction at the pivot is neglected there is no external torque about the axis, so $I\omega$ stays constant.

With no external torque about the spin axis, $I\omega$ is conserved. Pulling the arms in lowers $I$, so $\omega$ rises. NCERT §6.12.1 cites this for skaters, divers, acrobats and dancers performing a pirouette.

The same law governs planetary motion. The Sun's gravitational pull on a planet always points toward the Sun — along the position vector $\mathbf{r}$ drawn from the Sun. A force parallel to $\mathbf{r}$ gives $\mathbf{r}\times\mathbf{F} = 0$, so the torque about the Sun is zero and the planet's angular momentum is conserved. As the planet swings closer to the Sun, $r_\perp$ shrinks, so its speed must rise to keep $L = mvr_\perp$ constant. This is the dynamical root of Kepler's law of equal areas.

Worked examples

Example 1 · Torque from a force

Find the torque about the origin when a force of $3\hat{\mathbf{j}}$ N acts on a particle whose position vector is $2\hat{\mathbf{k}}$ m. (NEET 2020)

Set up. $\mathbf{r} = 2\hat{\mathbf{k}}$ m, $\mathbf{F} = 3\hat{\mathbf{j}}$ N. Apply $\boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}$.

Cross product. $\boldsymbol{\tau} = 2\hat{\mathbf{k}}\times 3\hat{\mathbf{j}} = 6\,(\hat{\mathbf{k}}\times\hat{\mathbf{j}})$. Since $\hat{\mathbf{k}}\times\hat{\mathbf{j}} = -\hat{\mathbf{i}}$,

$$ \boldsymbol{\tau} = -6\,\hat{\mathbf{i}}\ \text{N m}. $$

Takeaway. Order matters in a cross product. Reversing the factors would flip the sign — a frequent NEET slip.

Example 2 · Moment of a force about a point

The moment of the force $\mathbf{F} = 4\hat{\mathbf{i}} + 5\hat{\mathbf{j}} - 6\hat{\mathbf{k}}$ at the point $(2,0,-3)$, about the point $(2,-2,-2)$, is what? (NEET 2018)

Position vector relative to the pivot. $\mathbf{r} = (2-2)\hat{\mathbf{i}} + (0-(-2))\hat{\mathbf{j}} + (-3-(-2))\hat{\mathbf{k}} = 2\hat{\mathbf{j}} - \hat{\mathbf{k}}$.

Determinant. $$ \boldsymbol{\tau} = \mathbf{r}\times\mathbf{F} = \begin{vmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ 0 & 2 & -1\\ 4 & 5 & -6\end{vmatrix} $$

Expand. $\hat{\mathbf{i}}(2\cdot(-6) - (-1)\cdot 5) - \hat{\mathbf{j}}(0\cdot(-6) - (-1)\cdot 4) + \hat{\mathbf{k}}(0\cdot 5 - 2\cdot 4)$

$$ = (-12+5)\hat{\mathbf{i}} - (4)\hat{\mathbf{j}} + (-8)\hat{\mathbf{k}} = -7\hat{\mathbf{i}} - 4\hat{\mathbf{j}} - 8\hat{\mathbf{k}}\ \text{N m}. $$

Takeaway. When torque is taken about a point other than the origin, first build $\mathbf{r}$ as (point of application) − (pivot), then take the cross product.

Example 3 · Turntable Iω conservation

A child stands at the centre of a turntable with arms outstretched. The turntable is set rotating at $40$ rev/min. The child folds the arms back, reducing the moment of inertia to $2/5$ of the initial value. Find the new angular speed. The turntable rotates without friction. (NCERT Exercise 6.12a)

Identify the conserved quantity. No friction means no external torque about the vertical axis, so $L = I\omega$ is conserved: $I_1\omega_1 = I_2\omega_2$.

Substitute. With $I_2 = \tfrac{2}{5}I_1$ and $\omega_1 = 40$ rev/min,

$$ \omega_2 = \frac{I_1\omega_1}{I_2} = \frac{I_1\,(40)}{\tfrac{2}{5}I_1} = \frac{5}{2}\times 40 = 100\ \text{rev/min}. $$

Takeaway. Lowering $I$ raises $\omega$ in exact inverse proportion. The rotational kinetic energy $\tfrac{1}{2}I\omega^2$ actually increases — the extra energy comes from the muscular work the child does in pulling the arms in against the outward pull of the rotating masses.

Quick recap

Torque and angular momentum in one breath

Torque: $\boldsymbol{\tau} = \mathbf{r}\times\mathbf{F}$, magnitude $rF\sin\theta = r_\perp F$. Vector, unit N m, dimensions of energy but not energy.
$\tau = 0$ when the force's line of action passes through the axis ($\theta = 0^\circ$ or $180^\circ$).
Angular momentum of a particle: $\mathbf{l} = \mathbf{r}\times\mathbf{p}$; of a system: $\mathbf{L} = \sum_i \mathbf{l}_i$.
Master relation: $\boldsymbol{\tau} = d\mathbf{L}/dt$, the rotational analogue of $\mathbf{F} = d\mathbf{p}/dt$. Internal torques cancel, so only external torque counts.
For a symmetric axis, $L = I\omega$; with constant $I$, $\tau = I\alpha$.
Zero net external torque conserves $\mathbf{L}$: skater pulling arms in (small $I$, large $\omega$), and planets (central force) speeding near the Sun.

Torque & Angular Momentum