Posted by sanjeev kumar on Wednesday, August 1, 2012 at 3:34am.
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.

Dynamics  Ajayb, Wednesday, August 1, 2012 at 6:47am
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.

Dynamics  Rajan, Saturday, August 11, 2012 at 5:48pm
Please help me to elaborate the answer as i tried but couldn't
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