A flat lamina lies in the $x$-$y$ plane. Two axes $z$ and $z'$ are both perpendicular to the lamina (parallel to each other) and pierce it at two different points. A force $\vec{F}$ acts on the lamina, in the plane of the lamina, at a point P that lies between the two axes and is closer to the $z'$-axis than to the $z$-axis. Taking $\hat{k}$ as the unit vector pointing out of the plane of the lamina, decide which of the following statements are correct; more than one may be correct. (a) The torque $\vec{\tau}$ produced by $\vec{F}$ about the $z$-axis is directed along $-\hat{k}$ (b) The torque $\vec{\tau}'$ produced by $\vec{F}$ about the $z'$-axis is directed along $-\hat{k}$ (c) The magnitude of the torque produced by $\vec{F}$ about the $z$-axis is greater than that about the $z'$-axis (d) The total torque is given by $\vec{\tau}_{total} = \vec{\tau} + \vec{\tau}'$

A flat lamina lies in the x-y plane. Two axes z and z' are both perpendicular to the lamina (parallel to each

NCERT Exemplar · Q12

x

(a) The torque

\vec{\tau}

produced by

\vec{F}

about the

z

-axis is directed along

-\hat{k}

(b) The torque

\vec{\tau}'

produced by

\vec{F}

about the

z'

-axis is directed along

-\hat{k}

\vec{F}

about the

z

-axis is greater than that about the

z'

-axis

(d) The total torque is given by

\vec{\tau}_{total} = \vec{\tau} + \vec{\tau}'

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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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Torque of an in-plane force about a perpendicular axis lies along $\pm\hat k$ . Because P sits between the two axes, the position vectors from $z$ and from $z'$ to P point in opposite in-plane senses, so the two torques have opposite signs; the torque about $z'$ is along $-\hat k$ . The moment arm about the farther axis $z$ is larger, so its torque magnitude is greater. That makes (B) and (C) correct.

Concept

Torque about an axis: $\vec{\tau} = \vec{r}\times\vec{F}$ , where $\vec r$ runs from the axis to the point of application. For an in-plane $\vec F$ and an in-plane $\vec r$ , $\vec\tau$ is perpendicular to the plane, i.e. along $\pm\hat k$ . Its magnitude is $F$ times the perpendicular distance (moment arm) from the axis to the line of action of $\vec F$ .

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