Uneven Disk Toppling Over

Centre of mass of a semicircular disk is 4R/3π.

Centre of mass of given system is 4R/9π.

Now using conservation of energy, ΔPE=ΔRKE. ( RKE stands for rotational kinetic energy)

8MgR/9π= 1/2 MR^2/2 w^2 (w is angular velocity)

w =√(32/9 g/πR)

A=32 B=9

A+B=41

*Here, I would like to use Lagrangian Mechanic which is different method from @A E to solve, well this is a way much longer then @A E 's one. His solution really better than mine~ *

We can use Lagrangian equation $\frac{d}{dt}(\frac{\partial L}{\partial \omega})=\frac{\partial L}{\partial \theta}$ only if the system moving by conservative force. In this problem, the system shall be effective within gravity force which is conservative.

Step 1:Find KE

First, consider KE of Lagrangian, we know every particle of the whole disk is turning at the same angular speed at the same time, ω.

So, KE seems like

$KE= \frac{1}{2} ω^2 \int r^2 dM + \frac{1}{2} ω^2 \int r^2 dm$

dM stands for the upper part and dm stands for the bottom part of the disk, consider it two have density 2 $\rho$ and $\rho$

Which leads to

$\rho = \frac{\frac{1}{3} M}{\frac {1}{2} \pi R^2}=\frac{2M}{3 \pi R^2}$

And so

$dM=2 \rho rdrd \theta,dm = \rho rdrd \theta$

So

$KE = \frac {1}{2} \omega ^2 \int r^2 (3 \rho rdrd \theta)=\frac{3}{2} \omega^2 \rho \int_0^\pi \int_0^R r^3 drd\theta$

Solve this, we get

$KE = \frac{3}{8} \pi \rho \omega^2 R^4 = \frac{1}{8}MR^2\omega^2$

Step 2: Find PE

Consider $\theta=0$ at initial position and $\theta=\pi$ at target position, and also the zero potential plane is horizontal and pass through the centre of the disk.

Use integration as above, we can get

$PE = \frac{2MgR}{9\pi}cos\theta$

Step 3: Write Lagrangian in form of L( $\theta,\omega$ )

As we know, by using Lagrangian definition $L=KE-PE$ , we get

$L = \frac{1}{8}MR^2\omega^2 - \frac{2MgR}{9\pi}cos\theta$

Step 4: Compute Lagrangian Equation

By using angular form of Lagrangian Equation $\frac{d}{dt} (\frac{\partial L}{\partial \omega})=\frac{\partial L}{\partial \theta}$ , we finally get

$\alpha = \frac{8g}{9\pi R}sin\theta$

Which $\alpha$ represent angular acceleration.

Last Step: Solve Differentiate Equation

As usual kinematic form, $vdv=adx$ , we can use angular form instead of this.

$\omega d\omega= \alpha d\theta$

Which lead us to

$\int \omega d\omega = \int \alpha d\theta$

$\frac{\omega^2}{2}= \int_0^\pi \frac{8g}{9\pi R}sin\theta d\theta$

$\omega = \sqrt( \frac{32g}{9\pi R})$

$A=32, B=9 ,A+B= \boxed {41}$

Apply conservation of energy as work done by non conservative and external force is 0 . K.Ei +P.Ei = K.Ef + P.Ef THE CENTER OF MASS OF EACH SEMI CIRCULAR DISC IS AT A DISTANCE OF 4R/3π . P.Ei = (2M/3)(4R/3π)g + (M/3)(-4R/3π)g = (M/3)(4R/3π)g K.Ei=0 P.Ef = -P.Ei . K.Ef=( (1/2) (2M/3)R.R/2+ (1/2)(M/3)R.R/2 )ω.ω ON PUTTING THESE VALUES IN THE EQUATION WE GET ω = √ 32g/9πR .