Question

.A uniform sphere of mass m radius r starts rolling down without slipping from the top of another larger sphere of radius R. Find the angular velocity of the sphere after it leaves the surface of the larger sphere.

Answer 1

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.

So,

mgh = ½ mv² + ½ I?²

Or,

mg(R+r) (1-cos?) = ½ mv²+(1/5 )mv²

= (7/10) mv²

Or,

(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)}] .