Answer to Question #19420 in Mechanics | Relativity for Celesta
if the braking time is T, then final speed = 0 = omega + acc*T = 10 rad/s -
T = (10 rad/s) / (0.2 rad/s^2) = 50 s.
The number of revolutions will be N = angle/(2*Pi) = (omega/2)*T = ( (10
rad/s)*(50 s)/2 ) / (2*Pi) = 39.81;
Floored downwards, it's 39 revolutions.
Answer: (a) 50 seconds, (b) 39 revolutions
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