Question

A recording disc rotates steadily at $45rpm$ on a table. When a small mass of $0.02kg$ is dropped gently on the disc at a distance of $0.04m$ from its axis, it gets stuck to the disc and the rate of revolution falls to $36rpm$. The moment of inertia of the disc about its centre is

• A. $1.28\times {10}^{-4}\text{kg}-{\text{m}}^2$
• B. $1.28\times {10}^{-5}\text{kg}-{\text{m}}^2$
• C. $1.28\times {10}^{-3}\text{kg}-{\text{m}}^2$
• D. $1.28\times {10}^{-2}\text{kg}-{\text{m}}^2$

Answer 1

The correct option is A $1.28\times {10}^{-4}\text{kg}-{\text{m}}^2$

By conservation of angular momentum,
$I_1{\omega }_1=I_2{\omega }_2$ .....(i)
where,
$I_1$= Moment of Inertia of disc before mass gets stuck
$I_2$= Moment of Inertia of disc after mass gets stuck
${\omega }_1$= Angular velocity of disc before mass gets stuck
${\omega }_2$ =Angular velocity of disc after mass gets stuck

using eq. (i),
$I\times (2\pi \times \frac{45}{60})=[I+0.02\times (0.04{)}^2]\times (2\pi \times \frac{36}{60})$
$I=1.28\times {10}^{-4}\text{kg}-{\text{m}}^2$