a flywheel has an angular speed of 1200rev/min when its motor is turned off.the wheel attains constant decelerations of 1.5rad/s2 due to friction in its bearing.determine the time required for the wheel to come to rest and the numer of revolutions the wheel makes before it comes to rest.

Answer 1

Wi= 1200 rev/min = 20 rev/sec
= 20*2*pi rad/sec
Wf = 0; alpha= -1.5 rad/s^2

Wf = Wi + alpha*t
0 = 40*pi - 1.5*t
t = 40*pi/1.5 = 84 sec
Now,
theta = Wi*t + (1/2)*alpha*t^2

Plug in the values in this eqn. to get theta - the rotation of the wheel (in radians)before it comes to halt. Divide the result by 2*pi to convert into no. of revolutions.

Answer 2

Please help me to elaborate the answer as i tried but couldn't