A high-speed flywheel in a motor is spinning at 450 rpm when a power failure sud

Question

A high-speed flywheel in a motor is spinning at 450 rpm when a power failure suddenly occurs. The flywheel has mass 45.0 kg and diameter 70.0 cm . The power is off for 27.0 s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 150 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?

How many revolutions would the wheel have made during this time?

Explanation / Answer

Initial rate of revolution = ri = 450 rpm

Mass of flywheel =M = 45 kg

Radius of flywheel = R = diameter/2 = 70/2 = 35 cm = 0.35 m

Time for which power off = t = 27 s = 27/60 = 0.45 min

No. of revolution in 0.45 min = No = 150 revolutions

Let rf ( final rate) be the rate of revolution after time 0.45 min

using relation

No = rit + at2/2

we can find a ( acceleration in revolution/min2 )

putting available values

150 = 450 * 0.45 +a(0.45)2/2

150-202.5 = a(0.45)2/2

-52.5 * 2/(0.45*0.45) =

a = -518.519 rev/min2

now using

rf = ri +at

we have

rf = 450- 518.519 * 0.45 = 216.67 rpm

To calculate time to stop , again we can use

rf = ri +at

if flywheel stopped then rf = 0 so

0 = 450 - 518.519t

t = 450/518.519 = 0.868 min = 52.08 sec

So flywheel stops in 52.08 sec after power off

To find No. of revolutions we can use

rf2 - ri2 = 2aN

( N = no. of revolutions )

putting the values

0 - (450)2 = 2 *-518.519 * N

N = (450)2/1037.038 = 195.27 revolutions

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