# Determine the Angular Velocity

1. Dec 5, 2014

### Kirushanthy

1. The problem statement, all variables and given/known data

2. Relevant equations
How do I find the angular velocity because I got a value of of -99.1?

3. The attempt at a solution
I used the force equation for M1 and M2. So Fx for M1: Fx= T-um1g=m1a. Fx for M2: Fx= m2gsin(theta)-T-um2gcos(theta)=m2a. Since the 2 accelerations are the same, I used the 2 equations to solve for a and i got a value of 5.415m/s^2. Angular acceleration is equal to a/r so i did 5.415/0.20= 27.075. Since angular speed should be angular acceleration multiplied by time, I found time by using the xf=3=1/at^2, where I got t= 1.05s. When I multiply the values, I get 28.5. What am I doing wrong?I have my exam tomorrow so I would really appreciate it if someone can help me before thent!

2. Dec 5, 2014

### ehild

The tensions are not the same in the pieces of the string, as they must drive the rotation of the pulley.

3. Dec 5, 2014

### guitarphysics

Conceptually your solution seems to be fine, I agree with everything you did. Check again, maybe it's an arithmetic mistake.
EDIT: Oooh never mind, my bad. ehild is right, I think you need to consider that a difference in the two tensions will provoke a rotation in the pulley. Didn't read carefully enough, sorry!

Last edited: Dec 6, 2014
4. Dec 6, 2014

### sugz

Hi,
How do we know the tension is different
How do we know that the tension is different have been used to the tension being the same when the rope is over a massless pulley? Can you or anyone else elaborate on this concept.?

5. Dec 6, 2014

### ehild

But the pulley is not massless now. It has moment of inertia I=0.41 kgm2. The rim of the pulley moves together with the string, it has the same linear acceleration and angular acceleration α=a/r. The equation between torque and angular acceleration is τ=Iα. One string turns the pulley in one direction, the other rotates it to the opposite direction. The net torque is difference of tension multiplied by the radius of the pulley. For nonzero torque, the tensions must be different.

6. Dec 6, 2014

### Kirushanthy

How are they different? They are the same rope and so tension is uniform throughout the rope. In the textbook, they did a similar example, but it wasn't on an incline.

7. Dec 6, 2014

### ehild

The rope interacts with the pulley. In the piece in contact with the pulley, the tension changes, as the pulley is accelerated, and it has mass and angular momentum.

8. Dec 6, 2014

### Kirushanthy

Hi ehild,

Its very kind of you to help me! I understand this now that you have explained it. I also approached it another way in the meantime using torque and momentum as change in momentum is equal to net torque and got the acceleration as negative. The only way I was able to get the correct answer for the question was when I ignored the negative sign on acceleration. I will post my steps here, can you please tell me where I made a mistake?

http://imgur.com/4sYgrBu

9. Dec 6, 2014

### ehild

Check the signs. The sign of angular momentum in line 1 is inconsistent with the sign of the torque in line 5. The net torque must increase the angular momentum.