Lagrangian mechanics: a cylinder rolling inside another
I consider a large hollow cylinder with an inner radius R, which remains stationary. Inside it, a smaller solid cylinder of radius r, mass m, and moment of inertia I (about its own center, G) rolls.
My primary generalized coordinate will be \theta, the angle that the line connecting the center of the large cylinder (O) to the center of the small cylinder (G) makes with the vertical. Initially, I also consider \alpha, the angle of rotation of the small cylinder about its own axis G. The total angular velocity of the small cylinder in an inertial frame is \Omega_s = \dot\theta + \dot\alpha, where \dot\theta accounts for the rotation of the line OG.
The kinetic energy T has two components.
- The translational kinetic energy of the center of mass G. Since G moves in a circle of radius (R-r) around O, its speed is (R-r)\dot\theta. This gives \frac{1}{2}m(R-r)^2\dot\theta^2.
- The rotational kinetic energy of the small cylinder about its center G. This is \frac{1}{2}I\Omega_s^2 = \frac{1}{2}I(\dot\theta + \dot\alpha)^2.
So, the total kinetic energy is:
T = \frac{1}{2}m(R-r)^2\dot\theta^2 + \frac{1}{2}I(\dot\theta + \dot\alpha)^2
For the potential energy V, I take the reference level (V=0) to be when G is at the same horizontal level as O. The center G is at a vertical distance (R-r)\cos(\theta) below O. Thus, the potential energy is:
V = -mg(R-r)\cos(\theta)
A key aspect of this problem is the constraint that the small cylinder rolls without slipping. This condition relates \dot\theta and \dot\alpha. The velocity of the contact point P on the small cylinder (relative to G) is r(\dot\theta + \dot\alpha) tangentially. The velocity of G itself is (R-r)\dot\theta. For no slipping, the point P must have zero velocity in the lab frame. This leads to the constraint:
(R-r)\dot\theta + r(\dot\theta + \dot\alpha) = 0
Which simplifies to:
R\dot\theta + r\dot\alpha = 0
From this, I can express \dot\alpha in terms of \dot\theta:
\dot\alpha = -\frac{R}{r}\dot\theta
The absolute angular velocity of the small cylinder becomes \Omega_s = \dot\theta + \dot\alpha = \dot\theta - \frac{R}{r}\dot\theta = \frac{r-R}{r}\dot\theta.
I substitute this constraint back into the kinetic energy expression. The term for rotational kinetic energy becomes \frac{1}{2}I\left(\frac{r - R}{r}\right)^2\dot\theta^2.
The total kinetic energy, now solely in terms of \dot\theta, is:
T = \frac{1}{2}m(R-r)^2\dot\theta^2 + \frac{1}{2}I\left(\frac{R - r}{r}\right)^2\dot\theta^2 = \frac{1}{2}\left[m(R - r)^2 + I \left(\frac{R - r}{r}\right)^2\right]\dot\theta^2
The Lagrangian \mathcal L = T - V is:
\mathcal L = \frac{1}{2}\left[m(R - r)^2 + I\left(\frac{R - r}{r}\right)^2\right]\dot\theta^2 + mg(R - r)\cos(\theta)
To find the equation of motion, I use the Euler-Lagrange equation for the coordinate \theta:
\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial \mathcal L}{\partial \dot\theta}\right) - \frac{\partial \mathcal L}{\partial \theta} = 0
Calculating the derivatives:
\begin{aligned} \frac{\partial \mathcal L}{\partial \dot\theta} & = \left[m(R - r)^2 + I\left(\frac{R - r}{r}\right)^2\right]\dot\theta\\ \frac{\partial \mathcal L}{\partial \theta} & = -mg(R - r)\sin(\theta) \end{aligned}
Substituting these into the Euler-Lagrange equation gives the equation of motion:
\left[m(R - r)^2 + I\left(\frac{R - r}{r}\right)^2\right]\ddot\theta + mg(R - r)\sin(\theta) = 0
This equation describes the oscillation of the small cylinder. For small angles \theta, where \sin(\theta) \approx \theta, the motion is simple harmonic. If I consider a solid cylinder where I = \frac{1}{2}mr^2, the term I\left(\frac{R - r}{r}\right)^2 simplifies to \frac{1}{2}m(R - r)^2.
The equation of motion then becomes:
\frac{3}{2}m(R - r)^2\ddot\theta + mg(R - r)\theta = 0
The angular frequency f for these small oscillations is:
f = \frac{1}{2\pi}\sqrt{\frac{2g}{3(R - r)}}
For more insights into this topic, you can find the details here.