The angular velocity of a flywheel decreases uniformly from 1000 rev/min to 500

Question

The angular velocity of a flywheel decreases uniformly from 1000 rev/min to 500 rev/min in 5 sec. Find: The angular acceleration in rad/sec^2, The number of revolutions made by the wheel in the 5-sec interval, and How many additional seconds are required for the wheel to come to rest? A grinding wheel whose angular acceleration is constant at 2 rad/sec^2 rotates through an angle of 100 radians in 5 sec. How long had it been in motion at the beginning of the 5-sec interval if at some seconds earlier it started from rest?

Explanation / Answer

5)

given

w1 = 1000 rev/min = 1000*2*pi/60 = 104.8 rad/s

w2 = 5000 rev/min = 500*2*pi/60 = 52.4 rad/s

time taken, t = 5 s

A) angular acceleration, alfa = (w2-w1)/t

= (52.4 - 104.8)/5

= -10.48 rad/s^2

B) theta = w1*t + 0.5*alf*t^2

= 104.8*5 + 0.5*(-10.48)*5^2

= 393 rad

= 393/(2*pi) revolutions

= 62.5 revolutions

C) let t is the time taken to stop.

w3 = w2 + alfa*t

0 = 52.4 + (-10.48)*t

==> t = 52.4/10.48

= 5 s

6) angular velsoity of the wheel, w = theta/time

= 100/5

= 20 rad/s

let t is the time taken to get this angular velocity

Apply, w = wo + alfa*t

==> t = (w - wo)/alfa

= (20 - 0 )/2

= 10 s

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