Question

A horizontal disc rotating freely about a vertical axis makes $200\text{rpm}$. A piece of wax of mass $20\text{g}$ falls verticaly on the disc and adheres to it at a distance of $8\text{cm}$ from the axis. If the number of revolutions per minute is thereby reduced to $120\text{rpm}$, then calculate the moment of inertia of disc.

• A. $3.8\times {10}^{-4}{\text{kg-m}}^2$
• B. $1.92\times {10}^{-4}{\text{kg-m}}^2$
• C. $1.28\times {10}^{-4}{\text{kg-m}}^2$
• D. $2.56\times {10}^{-4}{\text{kg-m}}^2$

Answer 1

The correct option is B $1.92\times {10}^{-4}{\text{kg-m}}^2$

Let $I_1$ is the moment of inertia of disc.
Here, system is disc and a piece of wax.
Since, there is no external torque acting on the system. Therefore, the angular momentum of the system will be conserved.
$L_i=L_f$
$I_1{\omega }_1=I_2{\omega }_2$
where $I_2$ be the moment of inertia of disc after wax adheres to it.

$(\because \omega =\frac{2\pi N}{60})$
$N=$ no. of revolutions per minute.
$\Rightarrow I_1\left(\frac{2\pi \times 200}{60}\right)=(I_1+20\times 8^2)\left(\frac{2\pi \times 120}{60}\right)$
$\Rightarrow I_1\times 200=(I_1+1280)\times 120$
$\Rightarrow 80I_1=1280\times 120$
$\Rightarrow I_1=1920{\text{g-cm}}^2$
$\Rightarrow I_1=1.92\times {10}^{-4}{\text{kg-m}}^2$