# Acceleration of a mass connected to 2 different disks

1. Dec 9, 2008

### cashmoney805

1. The problem statement, all variables and given/known data
The attachment gives the setup, but I can do the first part (a). The second question (not in the attachment) is "Now consider the same system as part (a) except M = .250 kg. After the system is released from rest, what will the acceleration of the .250kg mass be?

2. Relevant equations
sum of torques = T1*(r1) - T2*(r2) = I*angular acceleration
sum of forces on M= Mg-T1=Ma
sum of forces on m = T2-mg=ma

3. The attempt at a solution
I multiplied the two force equation by the r of the disk that each mass is attached to, then added all three equations together. The tensions cancel, giving:
RMg-rmg = RMa + rma + I*ang acc

I don't know what to do from here though. Maybe I = I big disk + I small disk. I don't know what ang acc is though. I would usually set it to a/r, but there are two different r's in this case. Any help is greatly appreciated!

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2. Dec 9, 2008

### Staff: Mentor

Good.
Careful! The masses have different accelerations. How do the accelerations relate?

Yep!

Good--there are two different accelerations, also.

3. Dec 9, 2008

### cashmoney805

Let's see.

Is it Ma = MR(ang acc)
ma = mr(ang acc)?

If so, does the final equation, after adding all three, look like:

g(RM + rm) = ang acc (R(^2)M + r(^2)m + .5MR(^2) + .5mr(^2) ?

If so, you can calculate ang acceleration. Then do you just multiply that by R to get a of block 1?

edit: actually the final equation should be (I think)
g(RM - rm) = ang acc (R(^2)M + r(^2)m + .5MR(^2) + .5mr(^2)

oh boy, edit again. the M and m next to .5 should be masses of the disks

Last edited: Dec 9, 2008
4. Dec 9, 2008

### Staff: Mentor

That looks OK. (Except for using the same letters to represent different masses!)
Yep.