Determine the acceleration of the cylinder axis if there is no slip

[Davidllerenav]

Homework Statement

A ratchet runs through the surface of a large thin-walled cylinder so that if radius vector always forms, with respect to point $O$, an angle $\varphi$ with the vertical. The mass of the ratchet is $\eta$ times smaller than that of the cylinder. Determine the acceleration of the cylinder axis if there is no slip.

Relevant Equations

$I=\eta mR^2$

So I first wrote the moment of inertia of the cylinder, since it says that it is thin-walled, I think that its moment of inertia is $I=\eta mR^2$. After that I wrote the sum of torques, I think that there are three forces that cause torque, the two forces of friction, the one caused by the surface and the other between the ratchet and the cylinder and the projection of the weight of the ratchet. So I have $mg\sin\theta R+Fr_1R-Fr_2R=I\alpha$ The free body diagram of the cylinder gives me that $N_1=m\eta g$, the free body diagram of the ratchet gives me that $N_2=mg\cos\theta$ thus I have $mg(\sin\theta R+\mu R-k\cos\theta)=\eta m R^2\alpha$. Since $a=\alpha R$ I get $mg(\sin\theta R+\mu R-k\cos\theta)=\eta m Ra$. After that I don't know what to do. I need to eliminate the two friction coefficients $\mu$ and $k$. I also don't know how exactly the sum of forces on x for both bodies are behaving, can you help me please?

 

[jbriggs444]

After that I wrote the sum of torques, I think that there are three forces that cause torque, the two forces of friction, the one caused by the surface and the other between the ratchet and the cylinder and the projection of the weight of the ratchet. So I have $mg\sin\theta R+Fr_1R-Fr_2R=I\alpha$

If you are looking at torques on the cylinder alone then why should the weight of the ratchet count?

 

[haruspex]

I am unclear what this ratchet looks like and what it is doing. It doesn't seem to be what I would call a ratchet. How is it staying at the same angle? Is something else exerting a force on it?

Is your I the moment of inertia of the cylinder about its centre? If so, why do you have a factor η in there? Isn't your m the mass of the cylinder?

 

[jbriggs444]

If so, why do you have a factor η in there? Isn't your m the mass of the cylinder?

m seems to be the mass of the ratchet. One can see it used in that way in the $mg\sin(\theta)$ term.

I have assumed some sort of motor or hamster wheel causing the ratchet to continually climb the sloping side of the cylinder to stay at its assigned angle.

 

[Davidllerenav]

I am unclear what this ratchet looks like and what it is doing. It doesn't seem to be what I would call a ratchet. How is it staying at the same angle? Is something else exerting a force on it?

Is your I the moment of inertia of the cylinder about its centre? If so, why do you have a factor η in there? Isn't your m the mass of the cylinder?

I'm confused by the problem statement too.
I guess that it is moving up, so the cylinder is rotating with oposite direction.
No, m is the mass of the ratchet.

 

[Davidllerenav]

If you are looking at torques on the cylinder alone then why should the weight of the ratchet count?

Wouldn't its projection cause torque?

 

[Davidllerenav]

m seems to be the mass of the ratchet. One can see it used in that way in the $mg\sin(\theta)$ term.

I have assumed some sort of motor or hamster wheel causing the ratchet to continually climb the sloping side of the cylinder to stay at its assigned angle.

Yes, I think that's what its happening.

 

[jbriggs444]

Wouldn't its projection cause torque?

The weight of the ratchet is a force on the ratchet. (I am regarding it as an opaque mechanism climbing the wheel). What you want is the force of the ratchet on the cylinder. But you have already captured the relevant portion of that as the tangential friction between ratchet and cylinder.

 

[Davidllerenav]

The weight of the ratchet is a force on the ratchet. (I am regarding it as an opaque mechanism climbing the wheel). What you want is the force of the ratchet on the cylinder. But you have already captured the relevant portion of that as the tangential friction between ratchet and cylinder.

So the only forces creating torque on the cylinder are the two frictions?

 

[jbriggs444]

So the only forces creating torque on the cylinder are the two frictions?

Yes. [Assuming you've selected the reference axis down the middle of the cylinder]

 

[Davidllerenav]

Yes. [Assuming you've selected the reference axis down the middle of the cylinder]

Ok. Do I need to do the sum of forces on x?

 

[jbriggs444]

Ok. Do I need to do the sum of forces on x?

By "x", I assume you mean the ratchet. Yes, I would expect you to need a force balance there in order to help determine the frictional force between ratchet and cylinder.

 

[Davidllerenav]

By "x", I assume you mean the ratchet. Yes, I would expect you to need a force balance there in order to help determine the frictional force between ratchet and cylinder.

Ok, thanks. Also, my teacher gave me the answer, it was somethin like $a=\frac {g\sin\theta}{\eta\cos\theta}$, I'm not sure how the denominator actually was, but it had $\eta$ and $\cos\theta$ involved. Maybe this can help.

 

[jbriggs444]

Ok, thanks. Also, my teacher gave me the answer, it was somethin like $a=\frac {g\sin\theta}{\eta\cos\theta}$, I'm not sure how the denominator actually was, but it had $\eta$ and $\cos\theta$ involved. Maybe this can help.

[repaired LaTeX in quote]
A back of the envelope calculation gives me something along those lines... $a=\frac{g \sin \theta}{\ mumble\ \eta \ mumble\ \cos \theta}$

I eliminated most of the algebra by choosing a particular reference axis to eliminate the torque from the floor and defining the system boundaries to eliminate friction between ratchet and cylinder. [I have gotten lazy in my old age -- solving simultaneous equations is too much work].

Then there is only one torque. Two places where a rate of change of angular momentum can manifest. And one equation to solve for a.