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In the equilibrium, the centripetal force must be equal to the down word force.
So,
mv2/(R+r) = mg cos?
When the body is rolling without slipping, in accordance to principle of conservation of energy, potential energy must be in the form of translational energy plus rotational kinetic energy.
mgh = ½ mv2 + ½ I?2
Or,
mg(R+r) (1-cos?) = ½ mv2+(1/5 )mv2
= (7/10) mv2
(10/7) mg (1-cos?) = mg cos?
Again,
mv2 = (10/7) mg (R+r) (1-cos?)
10/7 = 17/7 cos?
Or, cos? = 10/17
The velocity of sphere radius r at the instant when it leaves contact with surface of fixed sphere will be,
v = v[g(R+r) cos?] =v[(10/17){g(R+r)}]
So the angular velocity will be,
? = v/r
=v[(10/17r2){g(R+r)}]
From the above observation, we conclude that, the angular velocity of sphere radius r at the instant when it leaves contact with surface of fixed sphere would bev[(10/17r2){g(R+r)}] .
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