# Two masses on frictionless table with one string connecting both masses on ground and the other

Tags:
1. Oct 12, 2014

### Latao Manh

1. The problem statement, all variables and given/known data

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?

2. Relevant 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 $v_{cm}-r\omega$
3. 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 $-T/4m$.

2. Oct 12, 2014

### Satvik Pandey

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 $a$?

Last edited: Oct 12, 2014
3. Oct 12, 2014

### 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.

4. Oct 12, 2014

### Satvik Pandey

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.

Last edited: Oct 12, 2014
5. Oct 12, 2014

### 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...?

6. Oct 12, 2014

### 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?

7. Oct 13, 2014

### 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...