Question

9.16) A flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . 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 240 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?

Answer 1

a)

using 2nd equation of angular motion

x = wt - 0.5 at ^2

240 * 2* 3.14 = 500* 2* 3.14 / 60 * 39 - 0.5* a* 39^2

a = 0.7 rad/s^2

using 1st equation of angular motion

w' = w - at = 500* 2* 3.14 / 60 - 0.7* 39 =24.96 rad/s

w' = 238.46 rev/min

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b)

using 1st equation of angular motion

t = w / a = 500 * 2* 3.14 / 60 /0.7

t = 74.76 s

total revolution

x = w^2 / (2 a) = 1956.27 rad

x = 311.5 rev

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