3.4 Moment of Inertia

3.4 Moment of Inertia

Consider a mass $m$ attached to the end of a massless rod as shown in Fig. 3.15. Assume that the bearing at the pivot point $O$ is frictionless. Let the system be in a horizontal plane. A force $F$ is acting on the mass perpendicular to the rod and hence, this will accelerate the mass according to:

In doing so, the force will cause the mass to rotate about $O$. Since tangential acceleration $a_{T}$ is related to angular acceleration $\alpha$ by the equation:

So

As turning effect is produced by torque $\tau$, it would, therefore, be better to write the equation for rotation in terms of torque. This can be done by multiplying both sides of the above equation by $r$. Thus:

which is rotational analogue of the Newton's second law of motion, $F = ma$.

Here $F$ is replaced by $\tau$, $a$ by $\alpha$ and $m$ by $mr^{2}$. The quantity $mr^{2}$ is known as the moment of inertia and is represented by $I$. The moment of inertia plays the same role in angular motion as the mass in linear motion. It may be noted that moment of inertia depends not only on mass $m$ but also on $r^{2}$.

Most rigid bodies have different mass concentration at different distances from the axis of rotation, which means the mass distribution is not uniform. As shown in Fig. 3.16(a), the rigid body is made up of $n$ small pieces of masses $m_{1}$, $m_{2}$, $\ldots$ at distances $r_{1}$, $r_{2}$, $\ldots$ from the axis of rotation $O$. Let the body be rotating with the angular acceleration $\alpha$, so the magnitude of the torque acting on $m_{i}$ is:

Similarly, the torque on $m_{2}$ is:

Since the body is rigid, so all the masses are rotating with the same angular acceleration $\alpha$.

Total torque $\tau$ is then given by:

or

where $I$ is the moment of inertia of the body and is expressed as:

For your Information — Moments of Inertia of various bodies about axis $AA'$: