Pulley, center of mass

Mar 2016
5
0
Ohio
I'm working on the following problem:
A massless pulley of diameter 50mm has a massless cord around it connected to two hanging containers of sugar. For the part I'm working on, one container has mass 520g, and the other has mass 480g. They start out at rest at the same level. I need to find the acceleration of the center of mass when they are released. This is what I've done:
The center of mass has zero acceleration in the horizontal direction. For the vertical direction, I use F(net)=Ma(com), or a(com)=F(net)/M. M is just the total mass, or .52kg+.48kg= 1.00kg. Then I have F(net)= F1-F2=m1g - m2g=(m1-m2)g= (.52kg-.48kg)(9.8m/s^2) = .04kg*9.8m/s^2= .392N. So a(com)=.392N/1.00kg = .392m/s^2. However, the answer in the back of the book is -1.6x10^-2 m/s^2. What am I doing wrong (besides the sign convention)??? I thought maybe I needed the tension in the string for F(net), but I thought they would cancel. If it is the tension, how does that fit in? I never liked pulley problems...
 
Jun 2010
422
33
NC
Aippo... Hi..

Your setup is that of the classical Atwood Machine. But Atwood wanted "g."

I haven't looked at this in some time I'm useless for your question.
However, here is the "Atwood Solution." Might be of interest while
you are looking at your problem.

Atwood's Machine | THERMO Spoken Here!

Good Luck, JP
 
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ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
So a(com)=.392N/1.00kg = .392m/s^2.
So far so this is all good, but you are not done yet!

However, the answer in the back of the book is -1.6x10^-2 m/s^2. What am I doing wrong (besides the sign convention)???
The "trick" is that they asked for the acceleration of the center of mass, whereas what you found is the acceleration of each mass. The acceleration of the c.m. is the weighted average of the acceleration of the two masses:

a(c.m.) = [(-.392 m/s^2 x 0.52 Kg) + (.392 m/s^2 x 0.48 Kg)] /(0.52 Kg + 0.48 Kg)
 
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