Question

A ring of radius 20 cm and mass 4 kg is rotating about an axis passing through its centre and perpendicular to its plane. If the angular acceleration of the ring is 2 $\frac{rad}{s^2}$, What is its rate of change in angular momentum?

• A. 1.6 N-m
• B. 3.2 N-m
• C. 0.16 N-m
• D. 0.32 N-m

Answer 1

The correct option is D 0.32 N-m

Magnitude of Net torque on the ring = I$\alpha$
$=M{R}^2\alpha$
$=4(0.2{)}^2.2N-m$
$=0.32N-m$
$\left|\overrightarrow{\tau }\right|=\frac{\left|d\overrightarrow{L}\right|}{dt}=0.32N-m$