Question

moment of inertia of a ring is 3 kgm^2. It is rotated for 20s from its rest position by a torque of 6 Nm. Calculate the work done​

Answer 1

Explanation:

I=3kgm

2

initial angular velocity = 0

t = 20 seconds

Now, Torque =Iα

6=3×α

∴α=2rad/s

2

Θ=w

0

t+

2

1

αt

2

Θ=0×t+

2

1

×2×20

2

Θ=400rad

Hope this helps!