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Imagine a homogeneous uniform cylinder whose cross section is not necessarily a circle (it can be an ellipse) but some other figure with area $A$. The cylinder oscillates like a see-saw around its midpoint so its angular velocity is $\dot{\theta}=\frac{d\theta}{dt}$, where $\theta$ is the angle between the cylinder and the horizontal.

We have the following relations between its volume $V$, its density $\rho$, its length $L$ and the area $A$ of its cross section

$$m=(m_{L}+m_{R})=\rho V=\rho LA$$

where $m_{R}$ is the mass of the right half of the cylinder and $m_{L}$ is the mass of the left half of the cylinder (left and right of the midpoint around which it is oscillating). In general $m_{R}\neq m_{L}$ since there is a slight asymmetry between the two sides of the cylinder.

What are expressions for the angular momentum and torque of the gravitaty force in terms of $m_{L},m_{R},\dot{\theta},\rho$ and $A$ assuming the center of mass of each half is located approximately a quarter of the length $L$ of the cylinder?

For the torque I get using the definition and assuming the center of mass at $L/4$,

$$\tau=m_{L}g\cos\theta\frac{L}{4}-m_{R}g\cos\theta\frac{L}{4}=\frac{g\cos\theta (m_{L}^{2}-m_{R}^{2})}{4\rho A}$$

How would I compute the angular momentum? I assume $v=\dot{\theta}L/2$?

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  • $\begingroup$ I am confused by your first equation. Should it be m=(mL+mR)=pV=pLA? and if the cylinder is oscillating about the midpoint wouldn't mL=mR? $\endgroup$ – J. Shupperd Sep 21 '16 at 20:22
  • $\begingroup$ I fixed the mass density formula. No $$m_{L}\neq m_{R}$$ because there is a slight asymmetry between the left and right hand sides. $\endgroup$ – JennyToy Sep 21 '16 at 20:45
  • $\begingroup$ So does the Area change as you go along the cylinder? Is this where the asymmetry comes from? $\endgroup$ – J. Shupperd Sep 21 '16 at 20:49
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    $\begingroup$ In the first sentence you stipulate "homogeneous uniform cylinder" then later you stipulate that there is an "slight asymmetry". So which is it to be? $\endgroup$ – MaxW Sep 21 '16 at 21:07
  • $\begingroup$ Perhaps I should explain that the cylinder models a candle that is burning (slowly) at both ends while oscillating. The asymmetry comes from the fact that the lower end of the candle is being burnt at a slightly faster rate then the upper end. $\endgroup$ – JennyToy Sep 21 '16 at 21:17
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Divide mentally the cylinder into two halves: the half to the left of the oscillation midpoint and the half to the right of the oscillation midpoint.

Lets call the length of the left part of the cylinder $L_{l}$ and the length of the right part $L_{r}$. The total length $L=L_{r}+L_{l}$.

We know that the mass of the left half is $$m_{l}=\rho V_{l}=\rho AL_{l}$$, hence $$L_{l}=\frac{m_{l}}{\rho A}\label{1}\tag{1}$$

Similarly for the right side

$$m_{r}=\rho V_{r}=\rho AL_{r}$$ and $$L_{r}=\frac{m_{r}}{\rho A}\label{2}\tag{2}$$

$A$ denotes the area of the cross section of the cylinder (same on left and right halves of the cylinder).

The angular momentum is given by the product of the momentum of inertia with the angular velocity of each cylinder half so

$$AM=I_{l}\dot{\theta}+I_{r}\dot{\theta}$$

Since $I_{r}=\frac{m_{r}L_{r}^{2}}{3}, I_{l}=\frac{m_{l}L_{l}^{2}}{3}$ the angular momentum becomes

$$AM=\frac{m_{r}^{3}+m_{l}^{3}}{3\rho^{2} A^{2}}\dot{\theta}$$

where I used (\ref{1}) and (\ref{2}) to replace $L_{l},L_{r}$.

To derive the total torque due to the gravity force add up the torque from the left to the torque from the right halves of the cylinder, taking into account that the gravity force acts at the midpoint. Since torque is $$\tau=m_{l}\vec{r_{l}}\times \vec{F}+m_{r}\vec{r_{r}}\times\vec{F}$$ we get

$$\tau=g\cos\theta(m_{l}L_{l}/2-m_{r}L_{r}/2)=\frac{m_{l}^{2}-m_{r}^{2}}{2\rho A}g\cos\theta$$

where I assume the center of mass of each side of the cylinder lies at the middle of it so at $L_{r}/2$ or $L_{l}/2$ and hence $\vec{r_{r}}=L_{r}/2(\cos\theta,\sin\theta,0),\vec{r_{l}}=L_{l}/2(-\cos\theta,-\sin\theta,0)$ and then I used again (\ref{1}) and (\ref{2}).

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