Question

8. A torque of magnitude \( 400 N - m \), acting on a body of mass \( 40 kg \), produces an angular acceleration of \( 20 rad / s ^{2} \). Calculate the M.I. and radius of gyration of the body.

Answer 1

\(\tau\) = 400 N^-m 

M = 40 kg

α = 20 rad/s² 

I = ?

k = ?

We know that

\(\tau\) = Iα

∵ I = Mk² 

\(\tau\) = Mk²α

k = \(\sqrt{\frac{\tau}{M\alpha}}\) where k = radius of gyration.

k = \(\sqrt{\frac{400}{40 \times 20}}\)

k = \(\sqrt{\frac{1}{2}}\)

Radius of gyration k = 0.70 m

Moment of inertia I = Mk² 

I = 40 × \(\frac{1}{2}\)

I = 20 kg m²