Question

A horizontal disc rotating freely about a vertical axis through its centre 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 $60$, the moment of inertia of the disc is

• A. $m{r}^2$
• B. $\frac{3}{2}m{r}^2$
• C. $2m{r}^2$
• D. $3m{r}^2$

Answer 1

The correct option is C $2m{r}^2$

Given,
Initial angular velocity of the disc $\left({\omega }_1\right)=90\text{rpm}$
${\omega }_1=90\times \frac{2\pi }{60}=3\pi \text{rad/s}$ (revolutions per second)
Initial moment of inertia $I_1=I$
Here, $I=$ MOI of the disc


After piece of wax sticks to the stick:


Final angular velocity of the dic $\left({\omega }_2\right)$
$=60\text{rpm}$
$=60\times \frac{2\pi }{60}=2\pi \text{rad/s}$
Moment of inertia of the system (disc + wax)
$I_2=(I+m{r}^2)$
As we see, there is no external torque on the system. Hence angular momentum of the system will be conserved.
$L_i=L_f$
$I_1{\omega }_1=I_2{\omega }_2$
$I\left(3\pi \right)=(I+m{r}^2)\left(2\pi \right)$

$\frac{3}{2}I=I+m{r}^2$

$\frac{3}{2}I-I=m{r}^2$

$\frac{3I-2I}{2}=m{r}^2$
i.e $I=2m{r}^2$