Net Torque / Rotational Dynamics Pulley System help

[elsternj]

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

Homework Equations

$$\sum$$F=ma
$$\sum$$$$\tau$$ = I$$\alpha$$
I = md²

The Attempt at a Solution

Let's take downward to be positive

For the block:
$$\sum$$F=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.

 

[Doc Al]

For the block:
$$\sum$$F=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.

 

[elsternj]

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.

 

[Doc Al]

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.