Rolling without slipping over a plank

[(Ron)^2=-1]

Homework Statement

This is just a general case I'm having trouble trying to imagine:
https://lh3.googleusercontent.com/-mTyOwzfLy0E/VluaNlxEddI/AAAAAAAAAEk/2Creguw3xzY/w530-h174-p-rw/Screenshot%2Bfrom%2B2015-11-29%2B21%253A34%253A50.png

Suppose there is a cylinder, kind of like a yo-yo, that is being pulled by a massless string. The string does not slip and the yo-yo is rolling without slipping on top of a plank, there is no friction between the plank and the horizontal surface.

Homework Equations

Torque, Pure rotating condition

The Attempt at a Solution

My question is: since the plank does slips over the horizontal surface, then Is the velocity of the cylinder's centre of mass = 0 ? (Assuming Tension < Static Friction between the cylinder and the plank)

It's worth mentioning that the problem doesn't say anything about friction between the plank and the yo-yo, but I thought there must be friction because of the rolling without slipping condition

Thank you!

 

[haruspex]

Looks like you included a diagram or link to a diagram, but it's broken. So please clarify the set-up. Is it:
Smooth horizontal surface.
Plank lies on horizontal surface.
Cylinder on plank, axis horizontal and at right angles to plank, rolling contact.
String wrapped around cylinder, being drawn off horizontally, parallel to the plank, from the top of the cylinder.
?

 

[(Ron)^2=-1]

https://lh3.googleusercontent.com/-FQFbA0n6chNSwQwUE9jUwaEgA_EDDnwEXjPbTQBRg=w1366-h768-rw
Here is the diagram. If you can't see it here is the set-up
The cylinder is on top of the plank, and the plank lies in a smooth horizontal frictionless surface. The cylinder has a radius R and also a inner radius a, the string is wrapped around the inner radius (a) and its been pulled horizontally, parallel to the surface and the plank.

Thank you for your help!

 

[(Ron)^2=-1]

Looks like you included a diagram or link to a diagram, but it's broken. So please clarify the set-up

Here is the diagram, sorry I was having trouble with the link.

 

[Herman Trivilino]

(Assuming Tension < Static Friction between the cylinder and the plank)

It looks like the force $\vec{F}$ is the tension force you mentioned. Aren't that, and the static friction exerted on the cylinder by the plank, the only forces acting on the cylinder in the horizontal direction?

 

[haruspex]

Ah, that's rather different.

Is the velocity of the cylinder's centre of mass = 0 ?

I don't see how the cylinder's centre of mass could fail to move. The only external horizontal force on the cylinder+plank is to the right, so the mass centre of that system must go to the right. If the cylinder's centre stays put then the plank goes to the left.
Also, I think you mean acceleration, not velocity.

 

[(Ron)^2=-1]

It looks like the force $\vec{\displaystyle F}$ is the tension force you mentioned. Aren't that, and the static friction exerted on the cylinder by the plank, the only forces acting on the cylinder in the horizontal direction?

Yes, you're right Mister. Although the problem says nothing about friction between the plank and the cylinder, I thought because of the condition of pure rotation there must be static friction, otherwise the cylinder would just slip.

 

[(Ron)^2=-1]

Ah, that's rather different.

I don't see how the cylinder's centre of mass could fail to move. The only external horizontal force on the cylinder+plank is to the right, so the mass centre of that system must go to the right. If the cylinder's centre stays put then the plank goes to the left.
Also, I think you mean acceleration, not velocity.

What I thought was that because the plank can slide over the horizontal smooth surface and assuming Force < Max static friction, then the cylinder will start rotating but the centre of mass will not undergo translational motion and then the velocity will be = 0 (all with respect to a reference frame fixed on the cylinder's centre of mass)

 

[haruspex]

Although the problem says nothing about friction between the plank and the cylinder, I thought because of the condition of pure rotation there must be static friction

That's right, there is. But the static frictional force never exceeds the force necessary to prevent slipping. Thus, it is surely less than F.

 

[(Ron)^2=-1]

(all with respect to a reference frame fixed on the cylinder's centre of mass)

Sorry here I meant a reference frame fixed to the initial position

 

[haruspex]

assuming Force < Max static friction

actual static friction <= max static friction
actual static friction < F (since the cylinder must accelerate to the right)
How F compares with max static friction we don't know and don't care.

 

[(Ron)^2=-1]

actual static friction <= max static friction
actual static friction < F (since the cylinder must accelerate to the right)
How F compares with max static friction we don't know and don't care.

Thank you so much!, I finally got the idea. So the centre of mass must, indeed, have acceleration in order for the pure rotation condition to hold.

 

[haruspex]

Thank you so much!, I finally got the idea. So the centre of mass must, indeed, have acceleration in order for the pure rotation condition to hold.

Yes, although I think it would also accelerate to the right if there were no friction between cylinder and plank. F would be the only horizontal force acting.