Question

A flywheel of mass 182 kg has an effective radius of 0.62 m (assume the mass is concentrated along a circumference located at the effective radius of the flywheel).
(a) How much work is done to bring this wheel from rest to a speed of 120 rev/min in a time. interval of 30.0 s?
(b) What is the applied torque on the fly-wheel (assumed constant)?

Answer 1

Answer:

A)5524J,

B) 29.2Nm

Explanation:

This question can be treated using work- energy theorem

Work= change in Kinectic energy

W= Δ KE

Work= difference between the final Kinectic energy and intial Kinectic energy.

We know that

Kinectic energy= 1/2 mv^2 .............eqn(1)

This can be written in term of angular velocity, as

KE= 1/2 I