JEE Mains Model Paper 12

Let V be the velocity and q is the angle made by the radius vector with the vertical at the
instant when the ball break off the sphere.
Let it happens at a vertical height h below the top. Therefore
We have mgh =

1

2

mv2 +

1

2

Iω2
⇒ mg (R + r) (1 - cosθ) =

1

2

m (rω)2 +

1

2

Iω2 .... (i)
At the break off, mg cosθ = N +

mv2

(R+r)

Substituting N = 0 and v = rω we get mg cosθ =

m(rω)2

(R+r)

... (ii)
Solving equations (i) and (ii), we get ω = √

10g(R+r)

17r2