Circular Constant Acceleration Formula

[CyclicCircle]

A constant tangential force of magnitude 12N is applied to the rim of a stationary, uniform circular flywheel of mass 100kg and radius 0.5m. Find the speed at which the flywheel is rotating after it has completed 25 revolutions?

I know that this can be done using work-energy. But since a constant tangential force is applied, I tried using kinematic equations.

Initial angular velocity $\omega = 0$, angular displacement $\theta = 25 \times 2\pi = 50\pi$.

If $\alpha$ is the angular acceleration, $mr\alpha = 12$, $(100)(0.5)\alpha = 12$, $\alpha = 0.24$.

Final velocity $\Omega^2 = \omega^2 + 2\alpha \theta$, which gives $\Omega = 8.68$. But the correct answer is apparently 12.3.

 

[andrewkirk]

It is not clear what you are doing in the second last line.

First calculate the Torque $\tau$ applied by the force to the flywheel.
Then find the Moment of Inertia $I$ of the flywheel.
Then the angular acceleration is $\alpha=\tau/I$. It is not 0.24.

If you do this and then substitute into the last line of your post, you should get 12.3.

 

[CyclicCircle]

Thank you so much.

I used the fact that tangential acceleration = radius*(angular acceleration) and F = ma to obtain force = mass*radius*(tangential acceleration). What is the error in this?

 

[haruspex]

Thank you so much.

I used the fact that tangential acceleration = radius*(angular acceleration) and F = ma to obtain force = mass*radius*(tangential acceleration). What is the error in this?

It is a disc, not a ring. You need to use the right formula for its moment of inertia.

 

[CyclicCircle]

Thanks. I completely forgot about mass distribution!