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 reduce to $60$, then the moment of inertia of the disc is?

Answer 1

Step 1. Given data

1. A horizontal disc rotating freely about a vertical axis through its center makes $90$ revolutions per minute, i.e. ${\omega }_1=90rpm$
2. The mass of the wax =$m$
3. Wax sticks to the disk at a distance of $r$ from the axis
4. After wax fell on the disc the new angular velocity is ${\omega }_2=60rpm$
5. Let, the moment of inertia of the disc is $I$.

Step 2. Formula used

In this case, the total angular momentum of the system must be conserved.

$L=I\omega$is conserved.

Where $L$ is the angular momentum, $I$ moment of inertia, $\omega$is angular velocity

Step 3. Find the moment of inertia

The initial angular momentum is $L_1=I_1{\omega }_1=90I_1$

The wax is attached to the disc at a distance of $r$ from axis, so the new moment of inertia of the disc with the wax piece is $I+m{r}^2$.

The final angular momentum $L_2=$${\omega }_2\left(I+m{r}^2\right)=60\left(I+m{r}^2\right)$

From Law of conservation of angular momentum we have

$$\begin{gathered} L_1=L_2 \\ \Rightarrow I{\omega }_1=(I+m{r}^2){\omega }_2 \\ \Rightarrow 90I=60\left(I+m{r}^2\right) \\ \Rightarrow 30I=60m{r}^2 \\ \Rightarrow I=2m{r}^2 \end{gathered}$$

Hence, the moment of inertia of the disc is $2m{r}^2$.