A uniform ball of radius r is placed on the top of a sphere of radius R = 10 r . It is given a slight push due to which it starts rolling down the sphere without slipping. The spin angular velocity of the ball when it breaks off from the sphere is ω = p q g r , where g is the acceleration due to gravity and p and q are the smallest integers. What is the value of p + q ?

Step-by-step solution

mv 2 R + r = mg cos θ mgh = 1 2 mv 2 + 1 2 I ω 2 mg R + r 1 - cos θ = 1 2 mv 2 + 1 5 mv 2 = 7 1 0 mv 2 1 0 7 mg 1 - cos θ = mg cos θ mv 2 = 1 0 7 mg R + r 1 - cos θ 1 0 7 = 1 7 7 cos θ or cos θ = 1 0 1 7 v = g R + r cos θ = 1 0 1 7 g R + r and ω = v r = 1 0 g R + r 1 7 r 2 = 110 g 17 r ⇒ p + q = 127