Atwood's machine with a cylinder

[dnp33]

Homework Statement

a massless string of negligible thickness is wrapped around a uniform cylinder of mass m and raidus r. the string passes up over a massless pulley and is tied to a block of mass m at its other end.
the system is released from rest. what are the accelerations of the block and the cylinder? assume that the string does not slip with respect to the cylinder.
Use conservation of energy (after applying a quick F=ma argument to show that the two objects move downward with the same acceleration)

Homework Equations

F=ma
K=[tex]\frac{1}{2}[/tex]mv²+[tex]\frac{1}{2}[/tex]Iw²
P=mgd
where w=angular frequency
and I=[tex]\frac{1}{2}[/tex]mr² is the moment of inertia.

The Attempt at a Solution

I wrote an F=ma equation for each mass, and because they are the same mass should undergo the same acceleration
I wasn't sure if that was as in depth as the question required.

Then i wrote a conservation of energy equation for the system
mgd + mgd = [tex]\frac{1}{2}[/tex]mv² + [tex]\frac{1}{2}[/tex]mv² + [tex]\frac{1}{2}[/tex]Iw²

and solved for velocity, where i used the kinematics equation
v²=2ad
to solve for the acceleration, which i found to be
[tex]\frac{4}{5}[/tex]g

the problem is i really have no idea if I'm right, or if there was an error in my reasoning.
any feed back would be appreciated.
Thanks.

 

[anathema]

Your solution is correct and well-reasoned. Using F=ma to show that the block and cylinder have the same acceleration is a good first step, and using conservation of energy to solve for the acceleration is the correct approach. Your use of the kinematics equation to solve for the final velocity is also correct.

One thing you could consider adding to your solution is a brief explanation of why the block and cylinder have the same acceleration. This is because the string is assumed to not slip with respect to the cylinder, meaning that the string and the cylinder move together as one object. Since the string is also attached to the block, the block will have the same acceleration as the cylinder.

Overall, your solution is well thought out and correct. Keep up the good work!