An interesting COM problem

I was solving a problem

Here is the solution
The velocity of that small disc $v$ is with respect to hemisphere frame .
And the velocity of that $v_{0}$ of that hemisphere is in ground frame.

So, $F_{x} =0$
So let's conserve momentum in $x$ direction $0=-mv_{0}+m(v \cos \theta -v_{0})$ After this conserving energy as always $mgR=\frac{1}{2}mv_{0}^{2}+\frac{1}{2}m(v_{0}^{2}+v^{2}+2vv_{0} \cos (π- \theta))+ mg(R-R \cos \theta)$
And the last equation normal to the inclined plane when it losses contact normal reaction will be zero $N=0$
Which gives the equation $\frac{mv^{2}}{R}=mg \cos \theta$
After solving these 3 equations patiently gives $\cos \theta= \sqrt{3}-1, 2,-1-\sqrt{3}$
The only reasonable value of $\cos \theta$ seems to be $\large \boxed{cos \theta=\sqrt{3}-1}$

#Mechanics

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

@Karan Chatrath @Steven Chase @Krishna Karthik in case you guys are interested.

Talulah Riley - 8 months, 4 weeks ago

Hmmm nice one

Krishna Karthik - 8 months, 4 weeks ago

@Krishna Karthik,mention[6497226:Krishna Karthik]

SRIJAN Singh - 8 months, 4 weeks ago

@SRIJAN Singh Yes; what do you need?

Krishna Karthik - 8 months, 4 weeks ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$