Question

A horizontal disc rotating freely about a vertical axis through its center makes $90$ revolutions per minute. A small piece of wax of mass $m$ falls vertically on the disc and sticks to it at a distance $\;r\;$ from the axis. If the number of revolutions per minute reduces to$60\;$, then the moment of inertia of the disc is?

Answer 1

The quantity that expresses a body’s tendency to resist angular acceleration is called the moment of inertia. It is the sum of mass times the square of the perpendicular distance to the axis of rotation. It can also be defined as the amount of torque needed for a specific angular acceleration around the rotational axis. Moment of inertia’s SI unit is given as $kg/{m^2}$.

Complete step by step solution:
Given that the horizontal axis is rotating freely. Let the moment of inertia of the disc initially be $I$. The disc is rotating in the vertical axis through the center making $90$revolutions per minute.
Now a piece of mask falls on the disc vertically. The mass of the wax is $m$. This wax sticks to the disk at a distance of $\;r\;$ from the axis.
Therefore the net moment of inertia is $I + m{r^2}$
After wax fell on the disc the no of rotations of the disc per minute reduces to $60\;$
There is no external torque acting on the system. Therefore the angular momentum remains constant.
The angular momentum formula is given as,
$L = I\omega$
Here $L$ is the angular momentum.
Equating both angular momenta of the disc and disc with wax
$I\omega$=$\omega (I + m{r^2})$
$1.5I = I + m{r^2}$
$1.5I - I = m{r^2}$
$0.5I = m{r^2}$
$I = \dfrac{1}{{0.5}}m{r^2}$
$I = 2m{r^2}$
Therefore the amount of inertia of the disc is $2m{r^2}$

Note:
The role of the moment of inertia in rotational motion is the same as the role of mass in linear motion. The moment of inertia depends upon the density of the material, shape, and size of the material and the axis of rotation. The rotating bodies can be categorized as discrete, that is a system of particles and continuous, which refers to the rigid body.