Rolling Semicircle

Suppose that the semicircle begins upright ( $\theta = 0$ ), with the center of the diameter at $x = 0$ . The semicircle then proceeds to roll without slipping.

Coordinates of center of diameter (which would be the center of the circle if the circle was full):

$x_C = R \, \theta \\ y_C = R$

Coordinates of a point on the semicircle ( $\theta$ is effectively our state variable; each point has a constant $\phi$ ):

$x = R \, \theta + R \, sin (\theta + \phi) \\ y = R + R \, cos (\theta + \phi) \\ \dot{x} = R \,\dot{\theta} + R \, \dot{\theta} \, cos (\theta + \phi) \\ \dot{y} = -R \, \dot{\theta} \, sin (\theta + \phi)$

Velocity of a point on the semicircle:

$v^2 = \dot{x}^2 + \dot{y}^2 = 2 \, R^2 \, \dot{\theta}^2 + 2 \, R^2 \, \dot{\theta}^2 cos (\theta + \phi)$

Infinitesimal mass:

$dm = M \frac{d \phi}{\pi} = \frac{M}{\pi} \, d \phi$

Infinitesimal kinetic energy:

$dE = \frac{1}{2} \, dm \, v^2 = \frac{1}{2} \, \frac{M}{\pi} \, d \phi \, \Big [2 \, R^2 \, \dot{\theta}^2 + 2 \, R^2 \, \dot{\theta}^2 cos (\theta + \phi) \Big] \\ = \frac{M \, R^2 \, \dot{\theta}^2}{\pi} [1 + cos(\theta + \phi) ] \, d \phi$

Total kinetic energy:

$E =\frac{M \, R^2 \, \dot{\theta}^2}{\pi} \int_0^{\pi} [1 + cos(\theta + \phi) ] \, d \phi \\ = \frac{M \, R^2 \, \dot{\theta}^2}{\pi} [\pi -2 \, sin \theta ]$

Infinitesimal gravitational potential energy:

$dU = dm \, g \, y = \frac{M g}{\pi} \, d \phi \, [R + R \, cos (\theta + \phi)] = \frac{M g R}{\pi} \, [1 + cos (\theta + \phi)] \, d \phi$

The gravitational potential energy integral is the same as the kinetic energy integral, except for the constant:

$U = \frac{M g R}{\pi} [\pi -2 \, sin \theta ]$

System Lagrangian:

$L = E - U = \frac{M \, R^2 \, \dot{\theta}^2}{\pi} [\pi -2 \, sin \theta ] - \frac{M g R}{\pi} [\pi -2 \, sin \theta ]$

Equation of motion:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{\partial{L}}{\partial{\theta}}$

Evaluation step one:

$\frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{ 2 M \, R^2 \, \dot{\theta}}{\pi} [\pi -2 \, sin \theta ]$

Evaluation step two:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{ 2 M \, R^2 }{\pi} \Big [\dot{\theta} (-2 \, cos \theta \, \dot{\theta} ) + \ddot{\theta} (\pi -2 \, sin \theta ) \Big ] \\ = -\frac{ 4 M \, R^2 }{\pi} \, cos \theta \, \dot{\theta}^2 + 2 M \, R^2 \, \ddot{\theta} -\frac{ 4 M \, R^2 }{\pi} \, sin \theta \, \ddot{\theta}$

Evaluation step three:

$\frac{\partial{L}}{\partial{\theta}} = \frac{M \, R^2 \, \dot{\theta}^2}{\pi} [-2 \, cos \theta ] - \frac{M g R}{\pi} [ -2 \, cos \theta ] \\ = -\frac{2 M \, R^2 \, \dot{\theta}^2}{\pi} cos \theta + 2 \frac{M g R}{\pi} cos \theta$

Equating:

$-\frac{ 4 M \, R^2 }{\pi} \, cos \theta \, \dot{\theta}^2 + 2 M \, R^2 \, \ddot{\theta} -\frac{ 4 M \, R^2 }{\pi} \, sin \theta \, \ddot{\theta} = -\frac{2 M \, R^2 \, \dot{\theta}^2}{\pi} cos \theta + 2 \frac{M g R}{\pi} cos \theta \\ 2 M \, R^2 \, \ddot{\theta} -\frac{ 4 M \, R^2 }{\pi} \, sin \theta \, \ddot{\theta} = \frac{2 M \, R^2 \, \dot{\theta}^2}{\pi} cos \theta + 2 \frac{M g R}{\pi} cos \theta \\ R \pi \, \ddot{\theta} - 2 R \, sin \theta \, \ddot{\theta} = R \dot{\theta}^2 cos \theta + g cos \theta$

Final result:

$\ddot{\theta} = \frac{R \dot{\theta}^2 cos \theta + g \, cos \theta}{R \pi - 2 R \, sin \theta }$

Here is a plot of theta vs time (generated from numerical integration) for a semicircle diameter of 2 meters. The period is about 3.453 seconds, which seems reasonable for an object of that size. You can see that it rocks back and forth between $\theta = 0$ (standing on one end of the diameter) and $\theta = \pi$ (standing on the other end).

This is what I did bcuz i was lazy:

It is likely that the period would vary depending on the initial angle the wire is stood on (initial $\theta$ ). But since I was lazy, I assumed that the initial angle is small. $\theta \sim 0$ . The c.g. of the wire lies $c = \frac{2}{\pi}$ away from the center of the semicircle. Say the angle the wire is standing at angle $\theta$ . Then the torque generated by both the weight and normal force would be $\tau = mg\theta c = mg\theta \frac{2}{\pi}$ , assuming small $theta$ and where $m$ is the mass of the wire. The moment of inertia about the contact point between the wire and ground is $I = 2m$ . Hence the angular acceleration $\theta'' = \frac{tau}{I} = \frac{g}{\pi} \theta = k \theta$ . Hence the period of oscillation is $\frac{2\pi}{\sqrt{k}} = \sqrt{\frac{2\pi^3}{5}} \sim 3.52$

This differs from Steven's answer due to the initial angle (mine being close to 0 and Steven's being $\frac{\pi}{2}$ .

In all the steps shown below, M = 1 kg, and R =1m.

In the plot, the unit of angular displacement is in radians. I have left out some steps in this solution. If necessary, I will add them later. I do not usually post solutions here so please do point out any mistakes/typos.