# 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 $\large \boxed{cos \theta=\sqrt{3}-1}$ Note by Talulah Riley
9 months ago

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## Comments

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

- 9 months ago

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Hmmm nice one

- 9 months ago

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@Krishna Karthik,mention[6497226:Krishna Karthik]

- 9 months ago

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@SRIJAN Singh Yes; what do you need?

- 9 months ago

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