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A small body of mass 'm' slides without friction from the top of a hemisphere of radius r as shown in the figure. At what height will the body be detached from the surface of the hemisphere?

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The small body of mass 'm' is placed at the top point A of a hemisphere of radius 'r' and centre O. When the body is sliding down the surface of the hemisphere: let us suppose that at an instant, it is at point B and possesses velocity v (see figure). In the position 'B', the following forces act the body:

(i) Weight mg acting vertically downwards, and

(ii) Normal reaction R acting along, the normal to the surface of the hemisphere at the point b.

While sliding at the point B, the body moves along a circular path of radius r(surface of the hemisphere) with velocity v. The required centripetal force is provided by the net component of the weight of the body along BO, i.e.,

mv2/r = mg cosθ - R

As the body slides, v increases and R decreases. The body gets detached from the surface of the hemisphere, when R becomes zero. Therefore when the body detaches from the hemisphere,

mv2/r = mg cosθ    ....(i)

Suppose that the body detaches at the point B, so that it is at a vertical distance AN = x below the top of the hemisphere. Then the velocity of the body at the point B is given by, v2 - 02 = 2gx or, v = √{2 gx}

When the body is at the point B, if it has a verticle height ON = h, above the horizontal, then

cosθ = ON/OB = h/R

In the equation (i) substituting for v and cosθ, we have

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