A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occur

Question

A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 33.0 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 (rpm) ?

(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 (seconds)?

(c) N = (rev)?

Explanation / Answer

initial angular velocity, wi = 590 rpm = 590 x 2pi rad / 60 s

wi = 61.78 rad/s

angle revolved, theta = 240 rev = 240 x 2pi rad = 1507.96 rad

t = 33 s

a) using theta = wi*t + alpha*t^2 /2

1507.96 = (61.78 x 33) + alpha(33^2 / 2)

544.5 alpha = - 530.78

alpha = - 0.975 rad/s^2

now using wf = wi + alpha*t

wf = 61.78 + (-0.975 x 33)

wf = 29.61 rad/s = 29.61 x (60 / 2pi ) rpm = 282.77 rpm

b) now using wf = wi + alpha*t

0 = 61.78 + (-0.975 x t)

t = 63.36 s .....Ans

c)
Using wf^2 - wi^2 = 2(alpha)(theta)

0^2 - 61.78^2 = 2(-0.975)(theta)

theta = 1957.32 rad =1957.32 /2pi rev = 311.52 rev

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