An interesting COM problem

I was solving a problem

Here is the solution
The velocity of that small disc vv is with respect to hemisphere frame .
And the velocity of that v0v_{0} of that hemisphere is in ground frame.

So, Fx=0F_{x} =0
So let's conserve momentum in xx direction 0=mv0+m(vcosθv0)0=-mv_{0}+m(v \cos \theta -v_{0}) After this conserving energy as always mgR=12mv02+12m(v02+v2+2vv0cos(πθ))+mg(RRcosθ)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=0N=0
Which gives the equation mv2R=mgcosθ\frac{mv^{2}}{R}=mg \cos \theta
After solving these 3 equations patiently gives cosθ=31,2,13 \cos \theta= \sqrt{3}-1, 2,-1-\sqrt{3}
The only reasonable value of cosθ\cos \theta seems to be cosθ=31\large \boxed{cos \theta=\sqrt{3}-1}

Note by Talulah Riley
9 months ago

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

Talulah Riley - 9 months ago

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Hmmm nice one

Krishna Karthik - 9 months ago

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@Krishna Karthik,mention[6497226:Krishna Karthik]

SRIJAN Singh - 9 months ago

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@SRIJAN Singh Yes; what do you need?

Krishna Karthik - 9 months ago

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