Troubleshooting a Discrepancy in Atwood Cylinder Solution

[Abhishek11235]

Homework Statement

The problem is in attached screenshot. Now,I solved this using force/torque method. However ,I got different solution as given in solution manual. Where I have gone wrong?

Homework Equations

The Attempt at a Solution

Applying F=ma to cylinder:
$$mg-T=ma$$
Applying ##\tau = I\alpha## to cylinder about CoM,
$$TR= I\alpha$$
Since the string doesn't slip we have
$$\alpha= a/R $$

Solving the above 3 with ##I=MR^2/2## we have,
$$Mg= 3Ma/2 \implies a=2g/3$$
However,the solution is ##a=g/2##. Where I have gone wrong?

 

[haruspex]

Since the string doesn't slip we have

That equation describes the cylinder's motion relative to the string, i.e. where a is the acceleration relative to the string.

 

[Abhishek11235]

Can you describe it in detail?

That equation describes the cylinder's motion relative to the string, i.e. where a is the acceleration relative to the string.

 

[Abhishek11235]

That equation describes the cylinder's motion relative to the string, i.e. where a is the acceleration relative to the string.

Please check my reasoning. The velocity of string is due to
1) The block moves down which pulls out string from cylinder
2) The cylinder moves down which unrolls thread

Hence ,total velocity of thread = 2×velocity of cylinder=##\omega ×R## since there is no slipping

 

[haruspex]

The velocity of string is due to
1) The block moves down which pulls out string from cylinder

Yes.

The cylinder moves down which unrolls thread

No. Even if the string were stationary the cylinder would move down unrolling the thread.