Q.What is the relation between Torque and Angular Momentum?

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:

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