A flywheel in a motor is spinning at 530 rpm when a power failure suddenly occur

Question

A flywheel in a motor is spinning at 530 rpm when a power failure suddenly occurs. The flywheel has mass 40.0kg and diameter 75.0cm . The power is off for 39.0 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 250 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?

Explanation / Answer

A flywheel in a motor is spinning at 550rpm when a power failure suddenly occurs. The flywheel has mass 40.0kg and diameter 75.0cm . The power is off for 38.0s , 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 200 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? 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^2 200 revs = 9.167rps * 38s + 0.5*a*38^2 a = -0.205 revs/sec/sec (note -ve sign indicates slowing down) Qu a.) v = u + at v = 9.167 + (-0.205)*38 = 9.167 - 7.807 rps = 1.36 rps So when the power comes on the flywheel is doing 1.36 revs/sec = 81.6 rpm Qu b.) Time to stop = u / a = 9.167rps / 0.205 rpsps = 44.7 seconds Number of revs in 44.7 seconds s = 9.167*44.7 + 0.5*(-0.205)*(44.7)^2 s = 409.75 - 204.8 revs = 205 revolutions (3 s.f.)

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