Question

A high-speed flywheel in a motor is spinning at 450rpm when a power failure suddenly occurs. The flywheel has mass 36.0kg and diameter 73.0cm . The power is off for 30.0s 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.


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

Answer 1

an rotational analogue of uniformly De-ccelerated motion.
Torque ~ Force
Angular Velocity ~ Linear velocity.
Moment Inertia ~ mass

Answer 2

lol

Answer 3

\[E_K = \frac{1}{2}M \omega^2\] where M is the moment of inertia. But one does not need to explicitly compute it, since the rate is actually given

Answer 4

y so funny ?

Answer 5

Finally \[ -\frac{dE_K}{dt} = I\dot{\omega}\omega \] where I is the moment of inertia, and the analogue of second law of Newton is
\[ T = I\dot{\omega} \]

Answer 6

So compute from data \[\Large \dot{\omega}\] and you have the rate of decrease of the kinetic energy. Finally this rate will bring the wheel to rest - this gives you the number of rotations.

Answer 7

thank you :)