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Let's have the next case. A rod (with mass $m$, length $L$ and a momentum of inertia $I$) at the initial time is located on a cylinder (with radius $R$) surface so that it's (rod's) center of mass lies on top of the surface. The сylinder is fixed. The rod moves along the surface without slip. There's need to find a frequency of small oscillations.

If I understand the task correctly, the center of rod's mass moves along the involute $$ x_{c} = R\sin(\varphi ) - R \varphi \cos (\varphi ), $$ $$ y_{c} = R\cos(\varphi) + R \varphi \sin(\varphi ). $$ So the lagrangian

$$ L = \frac{m l^{2}}{2} \dot \varphi^{2} \varphi^{2} + \frac{I\dot \varphi^{2}}{2} + mg( R\cos(\varphi) + R \varphi \sin(\varphi ) ) $$ near the equilibrium is approximately equal to

$$ L \approx \frac{I\dot \varphi^{2}}{2} + mgR \frac{\varphi^{2}}{2}. $$ But there isn't a solution describing small oscillations. Where did I make the mistake?

An illustration

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A picture would help to clarify the system. Your formula for the Lagrangian (v1) would have oscillatory solutions if the potential term has the opposite sign. Are you sure that you put the correct sign in front of the potential term? – Qmechanic Mar 23 '13 at 21:44

2 Answers 2

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If $\varphi$ is the angle of the rod then I get

$$ \ddot{\varphi} = \mbox{-} \frac{m r \varphi(g \cos\varphi+ 3 r \dot{\varphi}^2)}{I+m\,r^2 \varphi^2} $$

using vector calculus. Since $\ddot{\varphi} \propto -\varphi$ then the result is small angle linear harmonic motion. If C is the center of the rod then:

$$ \sum \vec{F} = m \frac{{\rm d}^2 \vec{r}_C}{{\rm d} t^2} $$ $$ \sum \vec{M}_C = I \vec{\alpha} $$

with $\vec{r}_C = \hat{n} r + \hat{e} r \theta$ the location of the c.g. relative to the center of the cylinder and $\vec{\alpha} = (0,0,\ddot{\varphi})$ the angular acceleration vector. Here $\hat{n}=(-\sin\varphi,\cos\varphi,0)$ is the unit vector from the center of the cylinder to the contact point and $\hat{e}=(-\cos\varphi,-\sin\varphi,0)$ is the unit vector from the contact point to the center of gravity of the rod.

You can add up the forces $\vec{F}$ of gravity, contact normal and contact friction (no-slip condition). The you can add the torques of those forces into $\vec{M}$.

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Why did you consider, that the $\mathbf n$ has the "minus" sign for it's X-component, and $\mathbf e$ has the "minus" signs for all of it's components? – user8817 Mar 24 '13 at 9:57
Starting from vertical and moving counter clockwise for a positive angle $\varphi$ produces the vector $\hat{n}$. $\hat{e}$ is chosen to point towards the c.g. again for a positive angle. – ja72 Apr 23 '13 at 4:07

Your energy conservation is of the form: $E= A\phi^2 + B\dot\phi^2 = const$

Take the time-derivative, then: $2A\phi\dot\phi + 2B\dot\phi\ddot\phi = 0$

Follows SHO equation: $\ddot\phi = - (A/B)\phi$,

and the angular frequency is $\omega=\sqrt{A/B}$

If you want to use the Lagrangian then $L = T -V = B\dot\phi^2 - A\phi^2$

The Euler-Lagrange equation is $\frac{d}{dt} (\partial{L}/\partial{\dot{q}}) = \partial{L}/\partial{q}$

so we arrive at the same result: $\frac{d}{dt} (2 B \dot{\phi}) = - 2 A \phi$

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Welcome @Maxim. Remember that you can use Latex formatting for better-looking equations. – fffred May 27 '13 at 17:04

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