A high-speed flywheel in a motor is spinning at 500rpm when a power failure sudd

Question

A high-speed flywheel in a motor is spinning at 500rpm when a power failure suddenly occurs. The flywheel has mass 35.0kg and diameter 73.0cm . The power is off for 35.0s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 180 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?
C)How many revolutions would the wheel have made during this time?

Explanation / Answer

Use rotational kinematic eqn to do this.

Theta = theta0 +omega0*t + 1/2*alpha*t^2 theta0 = 0 t =35.0s

theta = 180*2*pi = 1130.97 rad.

omega0 = 500*2*pi/60 = 52.4 rad /s
So alpha = (theta - omega0*t)/(1/2*t^2) = (1130.97-52.4*35)/(1/2*35^2) = -1.14 rad/s^2

Now use omega = omega0 + alpha*t to find omega after 35s = 52.4 + (-1.14)*35 = 12.5 rad/s
b) Use omega = omega0 + alpha*t to find t for when omega = 0 So 0 = omega) + alpha*t

t = -omega0/alpha = -52.4/(-1.14) = 45.9s

c) Use theta = theta0 + omega0*t +1/2*alpha*t^2 = 0 + 52.4 * 45.9 + 1/2*(-1.14) * 45.9^2 = 1204.27 rad

1204.27 rad* 1 rev/2pi rad = 191.66 rev

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