Solving for Acceleration of Mass on Cylinder w/ Frictionless String

[Latao Manh]

Homework Statement

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A solid cylinder of mass m and radius r lies flat on frictionless horizontal table, with a massless string running halfway around it, as shown in Fig. 8.50. A mass also of mass m is attached to one end of the string, and you pull on the other end with a force T. The circumference of the cylinder is sufficiently rough so that the string does not slip with respect to it. What is the acceleration of the mass m attached to the end of the string?

Homework Equations

Torque formula:
$rT-rF = I\alpha = mr^2\alpha/2$
$T-F = mr\alpha/2$
where $\alpha$ is angular acceleration.

acceleration of center-of-mass formula for cylinder:
$a_{cm-of-cylinder} = (T+F)/m$

Formula that describes tangential velocity for upper string and lower string (that I am not really sure of):
$v_{cm}+r\omega$ or $v_{cm}-r\omega$

The Attempt at a Solution

What I tried is differentiating the last formula above and using that as tangential acceleration formula, and get the solution using all the three formula. But I reached a wrong answer. The answer is $-T/4m$.

 

[Satvik Pandey]

Formula that describes tangential velocity for upper string and lower string (that I am not really sure of):
vcm+rωv_{cm}+r\omega or vcm−rωv_{cm}-r\omega

Hi Latao.Welcome to PF.:)
I think you need to find relation between $a_{cm}$,$\alpha$ and $a$(acceleration of bead).
As the string is not slipping,so at the point of contact of cylinder and string their accelerations are equal.
Can you find relation between $a_{cm}$,$\alpha$ and $a$?

 

[Latao Manh]

I really do not know how to progress further. So I know that acceleration of the length of the upper string is equal in magnitude to the acceleration of the length of the lower string, because string's lengths are conserved.

But I am not sure how equations should be written.

 

[Satvik Pandey]

I really do not know how to progress further. So I know that acceleration of the length of the upper string is equal in magnitude to the acceleration of the length of the lower string, because string's lengths are conserved.

Let 1 be the point on string and 2 be the point on the cylinder.Let 1 and 2 be in contact with each other.
As the string is not slipping so the accelerations of point 1 and 2 are equal.Are't they?
Also the acceleration of point 1 is equal to the acceleration of mass 'm'.
Can you find the acceleration of point 2?
It has two acceleration one due to translation and second due to rotation.
If you have any problem,feel free to ask me.

 

[NTW]

I find this problem very interesting, but beyond my abilities... However, I believe that a way of solving it could be by a = T/m, the 'm' in the denominator being the sum of the small mass, plus one-half of the cylinder mass (halved to compensate for the 'pulley geometry') plus the angular inertia associated with the increasing angular velocity of the cylinder.

But, even if that approach were right, I don't know how to proceed...

Any comment...?

 

[Satvik Pandey]

Hi NTW. Latao has made two correct equations in #post1.

From Newton's second can you find acceleration of mass 'm'?
The acceleration of mass 'm' is equal to acceleration of point 1. Can you tell me why?
As the string is not slipping so the accelerations of point 1 and 2 are equal.
Point 2 has two acceleration one due to translation and second due to rotation.
Acceleration due to translation is $a_{cm}$ (towards right)
and due to rotation it has tangential acceleration (towards left)
Can you find what is the tangential acceleration of point 2?

 

[NTW]

Thanks, Satvik. I don't doubt that Latao is in the right way to solve the problem. I was just proposing an alternative approach, somewhat based on the Atwood machine, as acceleration = T/(sum of masses involved) but my difficulty is to express the inertia of the cylinder's rotation in terms of inertial mass... I believe there is a way for that 'conversion', but I can't find it...