A small oscillations of a rod on the cylinder

Question

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?

Answer 1

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 $\vec{M}$.

Answer 2

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: $\frac{d}{dt} (2 B \dot{\phi}) = - 2 A \phi$