Question

A 20 kg slender rod pivoted at its center has a length of 6 m. A motor applied a torque of 2400 Nm on the rod to turn it from rest.
a) Find the the average power by the motor when the rod rotates an angle of 500 radians.
b) Find the instantaneous power of the motor at the moment that the rod rotates an angle of 500 radians.

Answer 1

First compute the moment of inertia, I. It is something like
I = M L^2/12,
but don't trust me. Look it up.

(a) Compute the angular acceleration rate (neglecting friction):
alpha = (Torque)/I (radians/sec^2)

Compute the time T required to rotate 500 radians:

(Angle) = (1/2)* (alpha) * T^2

Average Power = (work done)/T
= (Torque)*(Angle)/T
= (1/2)(Torque)*(alpha)*T
b) Instantaneous power + (Torque)*(angular velocity) = (Torque)*w
= (Torque)*(alpha)*T