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A small cylinder in a pipe

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Homework Statement


A small solid cylinder of radius r lies inside a large tube of internal radius R. Show that the angular velocity of the small tube $$\omega$$ is related to $$ \psi $$, the angle between the centre of mass of the solid cylinder, the centre of the large tube and the vertical by;

$$\omega = \dfrac{R-r}{r} \dfrac{d\psi}{dt}$$

The find the period of oscillation of the small cylinder about its equilibrium position.

Homework Equations



See below.

The Attempt at a Solution



I cannot quite imagine the whole setup I assumed that the objects should touch otherwise the angular velocities of the cylinder and the pipe should not have any relation. I tried to apply the conservation of angular momentum, by assuming that we start turning one cylinder and this one then exert torque on the other cylinder.... I did not work.

For the second part I assume I would want to use the energy method, but being able to visualise the problem clarifying the set up I am stuck.

Could someone please explain the setup to me and maybe give some hints.
 

Answers and Replies

  • #2
CWatters
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I cannot quite imagine the whole setup..
When I read it I imagined the large tube is stationary with the small cylinder rolling inside it (bit like a skateboarder in a half pipe).

Edit: Particularly this bit...
The find the period of oscillation of the small cylinder about its equilibrium position.
 
  • #3
haruspex
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When I read it I imagined the large tube is stationary with the small cylinder rolling inside it (bit like a skateboarder in a half pipe).
That certainly fits with the equation to be proved.
 
  • #4
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Ok so I went thought through this again, and I came to the following conclusions;

I) To get the relation between $$\omega$$ and derivative of $$\psi$$, you assume that there is no slipping. Hence the center of mass of the small cylinder should move with the same velocity as the edge of the cylinder. Now the CM of the cylinder moves with $$\psi * (R-r)$$ and the edge of the cylinder moves with $$\omega *r$$ equating this and differentiating gives the required expression.

II) For the second part use the energy method. There are Pe, Ke, Kr. Potential energy is $$mgh=mg(R-r)(1-cos(\psi))$$, the velocity of CM, $$v=( \dot{\psi})(R-r)$$, therefore the Ke due to the motion of CM, $$Ke = 0.5*v^2*m$$. The last component of the energy is rotational kinetic energy Kr, $$Kr=0.5*I*\omega^2$$. This yields the equation of motion;
$$\dots{\psi}=\dfrac{2*g*sin(\psi)}{3(R-r)}$$ and then for small angles this simplifies to; $$\omega = \sqrt{\dfrac{2*g}{3*(R-r)}}$$

I don't have the solutions to the problem, so I would appreciate if someone could tell me if this is correct.
 
  • #5
haruspex
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Ok so I went thought through this again, and I came to the following conclusions;

I) To get the relation between $$\omega$$ and derivative of $$\psi$$, you assume that there is no slipping. Hence the center of mass of the small cylinder should move with the same velocity as the edge of the cylinder. Now the CM of the cylinder moves with $$\psi * (R-r)$$ and the edge of the cylinder moves with $$\omega *r$$ equating this and differentiating gives the required expression.

II) For the second part use the energy method. There are Pe, Ke, Kr. Potential energy is $$mgh=mg(R-r)(1-cos(\psi))$$, the velocity of CM, $$v=( \dot{\psi})(R-r)$$, therefore the Ke due to the motion of CM, $$Ke = 0.5*v^2*m$$. The last component of the energy is rotational kinetic energy Kr, $$Kr=0.5*I*\omega^2$$. This yields the equation of motion;
$$\dots{\psi}=\dfrac{2*g*sin(\psi)}{3(R-r)}$$ and then for small angles this simplifies to; $$\omega = \sqrt{\dfrac{2*g}{3*(R-r)}}$$

I don't have the solutions to the problem, so I would appreciate if someone could tell me if this is correct.
I think there is a typo or two in there, and a confusing reuse of omega, but the end result looks right.
 

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