Classical mechanics problem

@Azimuddin Sheikh @Ram Mohith Here is my full solution

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 height of a tall man). The period is about 3.5 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).

Is this question same as that one in Mechanics

I got the value of angular velocity as, $\omega = 2 \sqrt{\dfrac{g(1 - \cos \theta)}{3R}}$ but I don't know whether it is correct or not. Here is the way I did it.

As the friction is sufficient to prevent slipping the semicircular disk can be assumed to be performing pure rotation about the bottom most point let's say "O". Now, to find $\omega$ I used the concept of the law of conservation of energy. The energy changes that take place here are only rotational kinetic energy and change in potential energy. To find the potential energy first we have to calculate by how much height did the center of mass displaced. By fixing the initial position of center of mass as datum line after rotating through $\theta$ the center of mass will move down by $h$ and I found it to be $h = R(\cos \theta - 1)$ which will be negative since the center of mass moved below the datum line. After rotating through $\theta$ the disk will acquire an angular velocity $\omega$ so it will have a rotational kinetic energy of $\dfrac12 I_o \omega^2$ where moment of inertia of the semicircular disc about point "O" is $I_o = \dfrac{mR^2}{2} + mR^2 = \dfrac32 mR^2$ by parrallel axis theorem. By law of conservation of energy, $\begin{aligned} \text{Total final energy} & = \text{Total initial energy} \\ \dfrac12 I_o \omega^2 + mgh & = 0 \\ \dfrac12 I_o \omega^2 + mgR(\cos \theta - 1) & = 0 \\ \dfrac12 I_o \omega^2 = mgR(1 - \cos \theta) & \\ \dfrac34 mR^2 \omega^2 = mgR(1 - \cos \theta) & \\ \omega^2 = \dfrac{4g(1 - \cos \theta)}{3R} & \\ \therefore \boxed{\omega = 2\sqrt{\dfrac{g(1 - \cos \theta)}{3R}}} & \\ \end{aligned}$

I think this may be wrong so please inform me if it is.

(@Steven Chase) sir can u check this out