Net Torque / Rotational Dynamics Pulley System help

AI Thread Summary
The discussion focuses on solving a physics problem involving a pulley system with a solid cylinder, a pulley, and a suspended block. The equations of motion for the block, cylinder, and pulley are analyzed, with particular attention to the correct application of torque and forces. There is clarification that the tension T2 does not exert a torque on the cylinder because it is attached to the point of rotation, leading to the consideration of friction in the equations. The participant expresses confusion about obtaining a numerical answer (3.27) instead of a formula in terms of M and R, indicating a need for further exploration of the solution. The conversation emphasizes the importance of correctly identifying forces and torques in rotational dynamics.
elsternj
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Homework Statement


A uniform, solid cylinder with mass M and radius 2R rests on a horizontal tabletop. A string is attached by a yoke to a frictionless axle through the center of the cylinder so that the cylinder can rotate about the axle. The string runs over a disk-shaped pulley with mass M and radius R that is mounted on a frictionless axle through its center. A block of mass M is suspended from the free end of the string (the figure ). The string doesn't slip over the pulley surface, and the cylinder rolls without slipping on the tabletop.

Find the magnitude of the acceleration of the block after the system is released from rest.
Express your answer in terms of the variables M, R, and appropriate constants

YF-10-62.jpg

Homework Equations


\sumF=ma
\sum\tau = I\alpha
I = md2

The Attempt at a Solution


Let's take downward to be positive

For the block:
\sumF=ma
mg - T1 = ma

For the cylinder:
Rotational:
\sum\tau = I\alpha
T22R = 1/2m(2R)2\alpha

Translational:
T2 - f = ma

For the Pulley:
T1R-T2R = I\alpha

Okay so I am fairly new to understanding torque. First I want to know if my equations are correct and if I am including everything. I'm having trouble moving on from here also. Any insight would be greatly appreciated.
 
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elsternj said:
For the block:
\sumF=ma
mg - T1 = ma
Looks good.

For the cylinder:
Rotational:
\sum\tau = I\alpha
T22R = 1/2m(2R)2\alpha
No. The tension T2 does not exert a torque about the cylinder's center of mass. What force does?

Translational:
T2 - f = ma
Good.

For the Pulley:
T1R-T2R = I\alpha
Good. You'll need to express I in terms of M and R, of course.
 
No. The tension T2 does not exert a torque about the cylinder's center of mass. What force does?

Ah, I see it doesn't exert a torque because it is actually attached to the point of rotation, right?
So... friction? -f(2R) = 1/2m(2R)2\alpha ?

So oddly enough the answer to this question is 3.27 which kind of confuses me because I was expecting it to have to be in terms of M and R like it asks and we don't have any numbers for our variables. But... i'll worry about that when I get to there I suppose.
 
elsternj said:
Ah, I see it doesn't exert a torque because it is actually attached to the point of rotation, right?
So... friction? -f(2R) = 1/2m(2R)2\alpha ?
Exactly. (Get rid of that minus sign, though.)

So oddly enough the answer to this question is 3.27 which kind of confuses me because I was expecting it to have to be in terms of M and R like it asks and we don't have any numbers for our variables. But... i'll worry about that when I get to there I suppose.
Solve it and see for yourself. It's quite possible that those M's and R's end up canceling. :wink:
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

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