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

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

 

[kerimek]

I would first like to commend you on your clear and organized approach to solving this problem. Your calculations for the moment of inertia and torque appear to be correct.

For part d, we can use the equation for angular acceleration, α = τ/I, where τ is the torque and I is the moment of inertia. We know the values for τ and I, so we can solve for α. Then, we can use the equation ωf = ωi + αt, where ωf is the final angular velocity (which is 0 since the wheel comes to a stop), ωi is the initial angular velocity (which is 75 rad/s), α is the angular acceleration we just calculated, and t is the time we are trying to find. Solving for t, we get t = (ωf - ωi)/α. Plugging in the values, we get t = (0 - 75 rad/s)/0.69 m/s^2 = 108.7 s. This is the time it takes for the wheel to come to a stop.

For part e, we can use the equation θ = ωit + 1/2 αt^2, where θ is the angular displacement (in radians), ωi is the initial angular velocity, α is the angular acceleration we calculated, and t is the time we just found. Since we want to find the number of revolutions, we can convert the angular displacement to revolutions by dividing by 2π. So, θ = 75 rad/s * 108.7 s + 1/2 * 0.69 m/s^2 * (108.7 s)^2 = 4,088.7 rad. Converting to revolutions, we get 4,088.7 rad / 2π rad/rev = 650.6 revolutions. Therefore, the wheel goes through approximately 651 revolutions as it comes to a stop.

I hope this helps you further in solving this problem. It is important to remember to always check your units and equations to make sure they are consistent and to use the appropriate equations for the given scenario. Keep up the good work!