# Moment of Inertia and torque?

1. Dec 6, 2009

### miamirulz29

1. The problem statement, all variables and given/known data
The combination of an applied force and a frictional force produces a constant total torque of 39.9Nm on a wheel rotating about a fixed axis. The applied force acts for 4.5s, during which time the angular speed of the wheel increases from 3 rad/s to 12 rad/s. The applied force is then removed. The wheel comes to rest in 72s.
A. What is the moment of inertia of the wheel? Answer in units of kgm^2
B. What is the magnitude of the frictional torque? Answer in Nm
C. What is the total number of revolutions of the wheel?

2. Relevant equations
$$\sum$$$$\tau$$ = I(moment of inertia) * $$\alpha$$

3. The attempt at a solution
A. 39.9 = 2I
I = 19.95
That is correct.
B. T = (19.95)(9/72) = 2.49375
That is incorrect. What am I doing wrong?
C. I need some help, do not know where to start.
Sorry if the equations look bad. This the first time I am using Latex and I still don't know exactly how to use it correctly.

2. Dec 6, 2009

### miamirulz29

Well I just I figured out Part b. Instead of doing 12-3/72, I had to do 12/72. Can somebody explain that to me please.

3. Dec 6, 2009

### nrqed

You need to consider the part of the motion when only the frictional torque is acting. The wheel then slows down from 12 rad/s to 0 rad/s in 72 sec. This is why you need to divide 12 rad/s by 72 s to get alpha

4. Dec 6, 2009

### miamirulz29

Oh right, thank you. Any help for part C? Just a way for me to get started please.

5. Dec 6, 2009

### nrqed

You're welcome.

First you need to find the total angle of rotation during the acceleration part and during the deceleration part (you must have seen the formula $\theta = \omega_i t + 1/2 \alpha t^2$). Then add the two angles for the total angle and fivide by 2 Pi to get the number of revolutions

6. Dec 6, 2009

### miamirulz29

So could I do this: theta = (1/2)(12/72)(72^2) for the decelerating part and for the accelerating part could I use the other formula: 12^2 - 9^2 / 2(12-3/4.5). Then add those together and divide by 2pi?

7. Dec 6, 2009

### nrqed

For the decelerating part, you are missing one term since omega_i is not zero.

For the accelerating part, it sounds good except that you mant 3^2 instead of 9^2.

8. Dec 6, 2009

### miamirulz29

Yes I meant 3^2 instead of 9^2. But isn't omega_i zero because is it come to rest when it decelerates.

9. Dec 6, 2009

### nrqed

It comes to rest at the end of the decelarating part so omega final is zero. But when it started decelarating, its omega was 12 rad/s, so that's the value of omega_i for the decelerating pat