Acceleration of a block on a pulley with mass

Mar 2013
5
0
A 1.4 kg block and a 2.7 kg block are attached to opposite ends of a light rope. The rope hangs over a solid, frictionless pulley that is 25 cm in diameter and has a mass of 0.72 kg. When the blocks are released, what is the acceleration of the lighter block?

I think I've figured out that the force equation for the 1.4 kg block is
T1-m1g=m1a,

and for the 2.7 kg block:
T2-m2g=m2a.

I believe the torque equation for the pulley is torque=rF, where F would be (T2+T1), I think. And that is set equal to I*alpha, or (mr^2)*(a/r).

But, I really have no idea what to do beyond that. I tried solving for a by rearranging the force equations to equal T1 and T2, and then plugged in those values into T1 and T2 in the torque equation, but my answer was wrong. Any help or clarification would be greatly appreciated.
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
Couple of points:

1. Be careful of the sign of a - it's positive for one block and negative for the other.

2.The net force that creates torque on the pulley is the difference of the two tensions.

3. The value for I for the pulley depends on its mass distribution - are you sure it's mr^2?
 
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Mar 2013
5
0
Oh, I had assumed I=mr^2, but that was my mistake. Also, point number 2 is very helpful. I ended up figuring it out because we had a question on the test like that today, and somehow the logic came to me in the middle of the test, oddly enough. One point is to make use of the fact that angular acceleration is equal to linear acceleration over the radius, right? And then solve accordingly with the equations you get when you solve the force equations of the blocks for T1 and T2?
 
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