For an Atwood's with a cylinder and a block, why do the objects have the same accel?

  • Thread starter wu7
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  • #1
wu7
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Homework Statement



One pulley, on one side we have a block with mass m.
On the other side we have a cylinder with mass m.
Cylinder (radius R) has unlimited string (massless, negligible thickness, no slippage)

So you can imagine two blocks falling as more string unravels from the cylinder.

I am told that these two objects have the same acceleration downwards. Why?

Homework Equations



F=ma
ma=mg-T

torque? T*R=I*[itex]\alpha[/itex]

The Attempt at a Solution



The problem made it sound like this was a quick, obvious argument.
I proceeded to a longer argument, finding that the accelerations of both masses were (2/3)g downward:

starting with the cylinder-
TR=I*[itex]\alpha[/itex]
T*R2 = (1/2)M*R2 * a
Finding T= (1/2)ma, then plugging T back into the standard F=ma equations.
Then I find the acceleration of the block to be (2/3)g as well.

Is this even right? If it is, was there a way to show that the objects' accelerations were the same without finding the actual accelerations?

The reason I don't think this is right is because I did the problem using conservation of energy using the assumption that the objects fell down at the same acceleration.
With conservation of energy I got acceleration=(1/2)g (conserving KE, PE, angular KE)
 

Answers and Replies

  • #2
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I think you don't need to use torque.
By F= m * a(center of mass) you get that their accel will only depend on forces applied to them. In both cases block and cylinder you have F = T - mg.
Then => m * aBlock = m * aCylinder => aBlock = aCylinder
 

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