# Tough Rotational Motion Problem

1. Oct 21, 2007

### Tlocc

1. The problem statement, all variables and given/known data
A uniform, solid cylinder of mass M and radius R rotates on a frictionless horizontal axle. Two equal masses hang from light/weightless cords wrapped around the cylinder. If the system is released from rest, find:
A. The tension in each cord.
B. The acceleration of each mass after the masses have descended a distance of H.

2. Relevant equations
Torque=T(tension)R
Torque=I x Alpha
T=mg-ma
T=Torque/R
Alpha=A(tangential)\R
I=((1/2)mr^2)

3. The attempt at a solution
Based on the relevant equations, I deduced that the tangential acceleration is the downwards acceleration since it is perpendicular to the radius. Combining equations as follows I retrieved my findings for a.
T=(I x Alpha)\R
TR=(I x Alpha)
TR=(Ia)\R
TR^2=Ia
(TR^2)\I=a
(TR^2)\((1\2)mr^2)=a
(2T)\m=a

I don't know what the answer is as it is not in the back of the book nor was it given during class so I don't know how far or close I am to the answer. If I'm right, let me know. If I'm wrong, it would be greatly appreciated if you could show me where I went wrong or if I was forgetting something.

2. Oct 21, 2007

### Tlocc

Hmm I'm wondering why nobody answered me and also a quick comment:
I'm not sure if this acceleration would be constant or not..

3. Oct 22, 2007

### Vijay Bhatnagar

If two masses are equal why will they move at all?

4. Oct 22, 2007

### Tlocc

the two equal masses are hanging from the big mass

5. Oct 22, 2007

### Vijay Bhatnagar

The cylinder (Mass M) is rotating on a fixed axis and the equal masses are connected at the ends of a rope wrapped around the cylinder. So why should the masses move. If the masses were unequal, the heavier one would move down.

6. Oct 22, 2007

### Tlocc

they aren't attached to the same cord

7. Oct 22, 2007

### Tlocc

there are two seperate cords

8. Oct 22, 2007

### Vijay Bhatnagar

In that case what you have done is basically correct except that there are two cords carrying equal weights, hence equal tensions T. Net torque on the cylinder will be 2TR and the relation will be 4T/M = a. Solve this one and mg - T = Ma to get T and a. a will be constant. Is B) finding acceleration or is it velocity?

9. Oct 22, 2007

### Tlocc

B was finding acceleration which was what my attempt was because I found T via the relevant questions.