Calculating Moment of Inertia of Grinding Wheel - 45 N Brake Applied

[rent981]

Here is the problem I am working on. Here is the work I have so far, does this look right.
A grinding wheel that has a mass of 65.0 kg and a radius of 0.500 m is rotating at 75.0 rad/s. (The carousel can be modeled as a disk and assume it rotates without friction on its axis)

a) What is the moment of inertia of the grinding wheel?

I=MR^2. So I=(65kg)(.5m^2)=16.25 kgm^2



b) A brake is applied to the outer edge with a force of 45 N. What is the resulting torque?

Torque=r*f. So (.5m)(45N)=22.5 Nm



c) What is the resulting angular acceleration of the grinding wheel?
Fr=mra. So 22.5=(65kg)(.5)(a). a=.69 m/s^2.





d) How much time passes until the wheel comes to a stop?

not sure what to do here. I know that its rotating at 75r/s. And a 45N brake is being applied. I don't know what relates time to radians and force.




e) How many revolutions does the wheel go through as it comes to a stop?


This can be determined by dividing the answer from d by its velocity.

any help will be greatly appreciated!

 

[rock.freak667]

d) How much time passes until the wheel comes to a stop?

not sure what to do here. I know that its rotating at 75r/s. And a 45N brake is being applied. I don't know what relates time to radians and force.




e) How many revolutions does the wheel go through as it comes to a stop?


This can be determined by dividing the answer from d by its velocity.

any help will be greatly appreciated!

You can assume the angular acceleration is constant, so you can use the equations of rotational motions

\omega= \omega_0 + \alpha t
\omega^2=\omega_0^2+2 \alpha \theta
\theta=\omega_0 t +\frac{1}{2}\alpha t^2