A flywheel in a motor is spinning at 550 when a power failure suddenly occurs. T

Question

A flywheel in a motor is spinning at 550 when a power failure suddenly occurs. The flywheel has mass 40.0 and diameter 75.0 . The power is off for 40.0 , 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 250 complete revolutions. A) At what rate is the flywheel spinning when the power comes back on? B) 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?

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) 550rpm = 550/60 rps = 9.167 rps s = ut + 0.5at^2 250 revs = 9.167rps * 40s + 0.5*a*40^2 a = -0.145 revs/sec/sec (note -ve sign indicates slowing down) a.)v = u + at v = 9.167 + (-0.145)*40 = 9.167 - 5.83 rps = 3.33 rps b.)Time to stop = u / a = 9.167rps / 0.145rpsps = 63.2 seconds Number of revs in 63.2 seconds s = 9.167*63.2 + 0.5*(-0.145)*(63.2)^2 s = 289.7 revolutions (3 s.f.)

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