Question

A small body of mass m slides down from the top of a hemisphere of radius r the surface of the block and the hemisphere a frictionless the height at which the body lose contact with the surface of the hemisphere is

Answer 1

Thus, the body will lose contact 2/3rd of the way up.

Mass of the body = m (Given)

Radius of the body = r (Given)

F = mgsinθ - N

When the ball loses contact, the normal force (N) = 0

= F = mgsinθ

As the ball is moving in a circular path, thus

F = mv² / R

=  mgsinθ = mv² / R

= v² = Rgsinθ

According to principle of conservation of energy -

mgR = 1/2mv² + mgh

Rg = 1/2Rgsinθ + gh

h = R - 1/2Rsinθ

h = R - ½Rh / R

3h/2 = R

h = 2R/3

Thus, the body will lose contact 2/3 of the way up.

Answer 2

Explanation:

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