An electric motor is running at 1200rpm when its power is cut off for 40 seconds, then turned back on. If the rotor weighs 60lb, has a radius of gyration of 8in, and experiences a power-on net torque of 100lb/in and power-off friction torque of 8lb/in, the rotational speed in rpm of the motor 2 seconds after the power is turned back on is closest to which answer?
*Please show your work.
initial rotation speed wo = 1200 rpm= 1200/60 = 20 rps = 20*2*pi= 40pi rad/s
power cut duration t = 40 sec
mass of the rotor m = 60 lb
radius of gyration k = 8 in = 8/12 = 2/3 ft
power on net ...
This solution is provided in 168 words. It uses step by step calculations for the power off and power on phases for the system to find out angular acceleration and angular velocity.