Q.A door is hinged along one vertical edge and is free to rotate about that vertical axis. The weight of the door acts vertically downward at its centre of mass. Does the weight of the door produce any torque about the vertical hinge axis? Give the reason for your answer.

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.

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 $\vec{F}$:

The magnitude is:

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 $\vec{F}$.

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:

Now differentiate with respect to time:

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:

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