The net external torque on a system of particles about an axis is zero. Which of the following are compatible

NCERT Exemplar · Q11

Q. The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it? (Note: more than one of the given options may be correct.)

(a) The forces may be acting radially from a point on the axis.

(b) The forces may be acting on the axis of rotation.

(d) The torque caused by some forces may be equal and opposite to that caused by other forces.

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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 about an axis means the component of $\vec{r}\times\vec{F}$ ALONG that axis. Options (A), (B), and (C) each describe forces whose line of action gives zero torque individually, and (D) describes the general way non-zero torques can still cancel in the sum. All four options — (A), (B), (C), and (D) — are compatible with zero net external torque.

Setting up the axial torque

Take the axis of rotation as the $z$ -axis. For a force $\vec{F}=(F_x,F_y,F_z)$ applied at a point with position $(x,y,z)$ relative to the axis, the torque about the axis is the $z$ -component of $\vec{r}\times\vec{F}$ :

\tau_z = xF_y - yF_x

Notice this depends only on $x$ , $y$ , $F_x$ , $F_y$ — not at all on $z$ or $F_z$ .

Checking each option

(A) Forces acting radially from a point on the axis. If a force's line of action passes through a point on the axis, then $\vec{F}$ is parallel to $\vec{r}$ (measured from that point), so $\vec{r}\times\vec{F}=\vec{0}$ entirely — including its $z$ -component. $\tau_z=0$ . Compatible.

(B) Forces acting on the axis of rotation. If the force is applied at a point on the axis itself, $x=y=0$ , so $\tau_z = (0)F_y-(0)F_x=0$ regardless of $\vec{F}$ . Compatible.

(C) Forces acting parallel to the axis of rotation. Here $F_x=F_y=0$ (all of $\vec{F}$ is along $z$ ), so $\tau_z = x(0)-y(0)=0$ identically, no matter where the force is applied. The full 3-D torque vector $\vec{r}\times\vec{F}$ is generally non-zero and points perpendicular to the axis — but that perpendicular part is not "torque about the axis"; only the axial ( $z$ ) component is. Since that axial component is exactly zero, this option is also compatible. …

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Q.The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it? (Note: more than one of the given options may be correct.)

Concept understanding — Torque

What is Torque? The Intuition

The Precise Definition

Why Torque Equals Rate of Change of Angular Momentum

Setting up the axial torque

Checking each option

Concept: Torque About an Axis Is the Component of $\vec r \times \vec F$ Along That Axis

Q.The net external torque on a system of particles about an axis is zero. Which of the following are compatible with it? (Note: more than one of the given options may be correct.)

Setting up the axial torque

Checking each option

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