Question

A clown balances a small spherical grape at the top of his bald head, which also has the shape of a sphere. After drawing sufficient applause, the grape starts from rest and rolls down without slipping. It will leave contact with the clown's scalp when the radial line joining it to the center of curvature makes what angle with the vertical?

Answer 1

Answer:

48.189°

Explanation:

Let us say,

Radius of curvature of the clown's bald head is = R

Angle where the grape leaves the contact with the head is (with vertical) = θ

Height from the top of the head at which the contact is lost = y

Mass of the grape = m

Velocity of the grape at the point where it loses contact = v

So,

Using the Conservation of Work and Energy, we can say that there is 0 Work done on the system,

W = ΔK + ΔU

So,

$0=(\frac{1}{2}mv^{2}-0)+(mgy-0)\\v^{2}=2gy\\Now,\\y=R-Rcos\theta\\y=R(1-cos\theta)\\So,\\v^{2}=2[R(1-cos\theta)]g$

Now, using this at the point where contact is lost,

$N=-m(\frac{v^{2}}{R})+mg.cos\theta\\N=-m[2g(1-cos\theta)]+mg.cos\theta$

At that point the Normal force will be zero, because the contact is lost.

So,

On putting, N = 0 we get,

$N=-m[2g(1-cos\theta)]+mg.cos\theta\\0=-m[2g(1-cos\theta)]+mg.cos\theta\\2g-2g.cos\theta=g.cos\theta\\3g.cos\theta=2g\\cos\theta=\frac{2}{3}\\\theta=48.189\,degrees$

Therefore, the angle at which the grape lose contact with the bald head is at 48.189° from vertical.