Question

The moment of inertia of a flywheel making 300 revolutions per minute is 0.3 kg m². Find the torque required to bring it to rest in 20 s.

Answer 1

Given,
moment of inertia = 0.3 kgm²

angular velocity = 300 rev/min = 300 × 2π/60 rad/s = 10π rad/s

Use formula, $\omega=\omega_0+\alpha t$
0 = 10π + $\alpha$ × 20

[ Final angular velocity is taken zero because body becomes rest after 20sec.]

Angular acceleration, $\alpha$ = -π/2 rad/s

We know, Torque = moment of inertia × angular acceleration.

so, torque = 0.3 × -π/2 = -0.15π Nm.

hence, magnitude of torque is 0.15π Nm.

Answer 2

Answer:

Given,

moment of inertia = 0.3 kgm²

angular velocity = 300 rev/min = 300 × 2π/60 rad/s = 10π rad/s

Use formula, \omega=\omega_0+\alpha tω=ω

0

+αt

0 = 10π + \alphaα × 20

[ Final angular velocity is taken zero because body becomes rest after 20sec.]

Angular acceleration, \alphaα = -π/2 rad/s

We know, Torque = moment of inertia × angular acceleration.

so, torque = 0.3 × -π/2 = -0.15π Nm.

hence, magnitude of torque is 0.15π Nm.