Question

The moment of inertia of wheel about the axis of rotation is $3Kg,m^2$. If its $KE$ is $600J$, its period of rotation:

• A. $20s$
• B. $3.14s$
• C. $0.05s$
• D. $0.314s$

Answer 1

The correct option is D $0.314s$

$KE=\frac{1}{2}I{\omega }^2$

$600=\frac{1}{2}3{\omega }^2$

$\omega =20rad/sec=\frac{2\pi }{T}$

$T=0.314secs$