Question

A rotating platform of moment of inertia $100$ kg-$m^2$ completes one rotation in $10$ seconds A person of mass is $50kg$ is standing at the center of the platform. If a person moves at a distance of $2m$ from the center along the radius then find the angular velocity of the platform.

• A. $\frac{\pi }{30}rad/sec$
• B. $\frac{\pi }{60}rad/sec$
• C. $\frac{\pi }{15}rad/sec$
• D. $\frac{\pi }{20}rad/sec$

Answer 1

Answer: (C)

$\frac{\pi }{15}rad/sec$

Given,

The platform completes one rotation in 10 sec.

$\omega =\frac{2\pi }{10}=\frac{\pi }{5}$

Initially $I=100kg-m^2$

Finally $I=100+50\times 2^2=300kg-m^2$

In this process of changing I, no external torque is involved,

Hence,

$I_1{\omega }_1=I_2{\omega }_2$

100\times \dfrac{\pi}{5}=300\omega_2$

${\omega }_2=\frac{\pi }{5\times 3}=\frac{\pi }{15}$

Option $\text{C}$ is the correct answer