Rotational form of Newton’s 2nd law

We said that torque is a measure of the rotational effect of the force. What does this ‘rotational effect’ mean?

The greater the torque, the greater the change in the rotational state of motion of the object. This change is reflected in the angular acceleration of the object. Without torque, there is no angular acceleration: the object is at rest or continues to rotate at a constant angular speed. A non-zero torque will cause proportional angular acceleration, that is, a speeding up or slowing down of the rotation at a rate proportional to the magnitude of the torque. Let us analyse this proportionality.

For a single particle of mass \(m\), which is moving in a circle of radius \(R\) under the effect of a net force \(\mathbf{F}\), the magnitude of the torque is

\[\tau = R F \sin\vartheta = R F_{\mathrm{t}},\]

wherein \(\vartheta\) is the angle between the position vector \(\mathbf{R}\) and the force and \(F_{\mathrm{t}}\) denotes the tangential component of the force. We have seen when discussing circular motion that the tangential component of the force and the tangential acceleration are related through

\[F_{\mathrm{t}} = m a_{\mathrm{t}}.\]

Substituting the relationship between the angular acceleration and the tangential acceleration into this equation, we can write

\[F_{\mathrm{t}} = m R \alpha.\]

This way, we can link torque to angular acceleration by

\[\tau = m R^2 \alpha.\]

The quantity \(m R^2\) is called the moment of inertia of the particle, and it is the rotational analogue of mass. The greater the moment of inertia, the less angular acceleration a given torque causes, that is, the less the change in the rotational state of motion of the object.

We can generalise what we have found for a particle to extended rigid objects. It can also be proved that it is also true in vector form.