Rotational Inertia and Net Torque with Friction

[Isabel1747]

Homework Statement

A motor drives a disk initially at rest through 23.4 rotations in 5.0 s. Assume the vector sum of the torques caused by the force exerted by the motor and the force of friction is constant. The rotational inertia of the disk is 4.0 kg⋅m2. When the motor is switched off, the disk comes to rest in 12 s.
1) What is the magnitude of torque created by the force of friction?
2) What is the magnitude of torque caused by the force exerted by the motor?

Relevant Equations

Wf = Wi + at
T = Ia

I converted the amount of rotations completed in 5 seconds into radians.
23.4 rot * 2pi = 147 rad
I found the angular acceleration of the object in the first 5 seconds it was speeding up.
Wf = Wi + at
a = 5.881 rad/s^2
I then used the moment of inertia given in the problem to solve for torque.
T = Ia
T = 23.5 Nm
I also found the angular acceleration of the object slowing down in 12 seconds.
Wf = Wi + at
a = -2.45 rad/s^2
I then used the moment of inertia given in the problem to solve for torque.
T = Ia
T = -9.8 Nm

The answer to the problem is
1) 20 Nm
2) 67 Nm

I am confused because to get these correct answers I can multiply my frictional torque by 2 for the first problem. For the second problem, I can add the magnitude of the frictional torque to the speeding up torque, and then multiply by 2. I feel like I'm missing a factor of 2 somewhere, but I cannot figure it out.
I just would like to know where the numbers come from/how to finish solving this problem!

 

[gneill]

I found the angular acceleration of the object in the first 5 seconds it was speeding up.
Wf = Wi + at
a = 5.881 rad/s^2

Hmm. I don't see any values stated for Wf or Wi. So exactly how did you accomplish this deduction. Hint: I think that this may be where your factor of 2 discrepancy arises.

 

[gneill]

Oh! By the way, Welcome to Physics Forums Isabel!

 

[Isabel1747]

Hmm. I don't see any values stated for Wf or Wi. So exactly how did you accomplish this deduction. Hint: I think that this may be where your factor of 2 discrepancy arises.

I found Wf after the first 5 seconds to be 29.4 rad/s...I found this by dividing the total radians traveled by the time it took. I used this value as the Wf in the speeding up equation, and the Wi in the slowing down equation. Is that incorrect? and thanks for the welcome!

 

[gneill]

Think back to your linear motion kinematics. If a body is undergoing constant acceleration from rest, what's the formula that you would use to predict the distance attained after a given time? The angular motion equations are analogous.

 

[haruspex]

dividing the total radians traveled by the time it took

That will give you the average rotation rate over the 5 seconds. Wf is the maximum rate.

 

[Isabel1747]

Think back to your linear motion kinematics. If a body is undergoing constant acceleration from rest, what's the formula that you would use to predict the distance attained after a given time? The angular motion equations are analogous.

I would use xf = xi + vi*t + .5at^2 to find the distance after a given time...or I could use vf^2 = vi^2 + 2ad... I do not know how this relates...do I need to find tangential acceleration of the disk?? If so, how do I find the radius??

 

[haruspex]

I would use xf = xi + vi*t + .5at^2 to find the distance after a given time...or I could use vf^2 = vi^2 + 2ad... I do not know how this relates...do I need to find tangential acceleration of the disk?? If so, how do I find the radius??

AS @gneill posted, the equation is analogous. In the linear acceleration equation, replace distance by angular distance, velocity by angular velocity, etc.

In fact, you could just plug in the radius as 'r' and it will cancel out.

 

[gneill]

I would use xf = xi + vi*t + .5at^2 to find the distance after a given time...or I could use vf^2 = vi^2 + 2ad... I do not know how this relates...do I need to find tangential acceleration of the disk??

The first equation is appropriate. Since the initial angular distance and velocity are zero, the angular equivalent becomes:
$\theta_f = \frac{1}{2}\alpha t^2$

You're effectively given the final angular "distance" and the time, you can then determine the angular acceleration that occurred. You do not need the tangential acceleration of the disk. Besides, you are not even given the radius of the disk...

That $\frac{1}{2}$ explains your factor of two discrepancy.

 

[Isabel1747]

The first equation is appropriate. Since the initial angular distance and velocity are zero, the angular equivalent becomes:
$\theta_f = \frac{1}{2}\alpha t^2$

You're effectively given the final angular "distance" and the time, you can then determine the angular acceleration that occurred. You do not need the tangential acceleration of the disk. Besides, you are not even given the radius of the disk...

That $\frac{1}{2}$ explains your factor of two discrepancy.

Thank you both so much!

 

[gneill]

You're most welcome. Glad we could help you.