A flywheel in a motor is spinning at 540 rpm when a power failure suddenly occur

Question

A flywheel in a motor is spinning at 540 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 37.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 230 complete revolutions. At what rate is the flywheel spinning when the power comes back on? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time? N=?

Explanation / Answer

Since the information given and the required answers are in terms of revolutions it is easier NOT to convert to radians - just work in revolutions, seconds, and revs/sec (rps)

540 rpm = 540/60 rps = 9 rps

s = ut + 0.5at^2
230 revs = 9 rps * 37s + 0.5*a*37^2
a = - 0.1504 revs/sec^2 (note -ve sign indicates slowing down)

v = u + at
v = 9 + (-0.1504)*37 = 9 - 5.5676 rps = 3.4352 rps

So when the power comes on the flywheel is doing 3.4356 revs/sec = 206.136 rpm

Time to stop = u / a = 9 rps / 0.1504 rpsps = 59.84 seconds

Number of revs in 59.84 seconds s = 9*59.84 + 0.5*(-0.1504)*(59.84)^2
s = 538.56 - 269.28 = 269 revolutions

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