Question

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 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 210 complete revolutionsAt what rate is the flywheel spinning when the power comes back on(in rpm)

Answer 1

Complete Question

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 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 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? 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:

$\theta=274rev$

Explanation:

From the question we are told that:

Angular velocity $\omega=510rpm$

Mass $m=40.kg$

Diameter d $75=>0.75m$

Off Time $t=40.0s$

Oscillation at Power off $N=210$

Generally the equation for Angular displacement is mathematically given by

 $\theta_{\infty}=\frac{w+w_0}{t}t$

 $w=\frac{2*\theta_{\infty}}{t}-w_0$

 $w=\frac{28210}{40*(\frac{1}{60})}-510$

 $w=120rpm$

Generally the equation for Time to come to rest is mathematically given by

 $t=(\frac{\omega_0}{\omega_0-\omega})t$

 $t=(\frac{510}{510-120rpm})(40.0)(\frac{1}{60})$

 $t=0.87min$

Therefore Angular displacement is

 $\theta =(\frac{120+510}{2})0.87$

 $\theta=274rev$