Torque

How hard does gravity twist the rod about the pivot?

A rod pinned at one end can’t slide anywhere. It can only turn, so what matters is not the force on it but how strongly that force twists it about the pivot. That twist is the torque.

Think of pushing a door. The same push does far more near the handle than near the hinge. Torque captures both pieces: how hard you push, and how far from the pivot you push.

\[\tau = F_{\perp}\, d\]

Here \(F_{\perp}\) is the part of the force that points across the rod, and \(d\) is how far from the pivot it acts (see physics).

From the force decomposition, the part of gravity that swings the rod is \(-mg\sin\theta\). It acts at the centre of mass, which for a uniform rod sits halfway down, at \(d = L/2\):

\[\tau = -\,mg\,\frac{L}{2}\,\sin\theta\]

The part along the rod, \(mg\cos\theta\), points straight through the pivot. Its distance across from the pivot is zero, so it contributes no torque at all, which confirms that it plays no part in the swing.

There is a second way to see the same result. Keep the whole weight \(mg\), pointing straight down, and ask how far the centre of mass sits sideways from the vertical line through the pivot. That sideways distance, the lever arm, is \(\frac{L}{2}\sin\theta\), and multiplying gives the same torque. Either view works, because both count only the part of gravity that misses the pivot.

The checks carry over from the force:

• At \(\theta = 0\) the centre of mass is directly below the pivot, so there is no lever arm and no torque.
• At \(\theta = 90^\circ\) the rod is horizontal, so the lever arm is the full \(L/2\) and the torque is at its largest.

Torque tells us how hard gravity twists the rod. How much the rod actually speeds up depends on how hard it is to spin, which is the next step: Moment of Inertia.