JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Laws of Motion

A small block slides down from the top of hemisphere of radius

R = 3 m

as shown in the figure. The height '

h

' at which the block will lose contact with the surface of the sphere is ____ m.
(Assume there is no friction between the block and the hemisphere)

Answer: 2

Solution

The block leaves the surface when the normal force falls to zero, so at that point the radial component of gravity alone supplies the centripetal acceleration:

m g cos θ = \frac{m v ^{2}}{R}, cos θ = \frac{h}{R}

From the top of the hemisphere to that point the block descends through

R - h

, so energy conservation gives

v^{2} = 2 g (R - h)

. Substituting,

g \cdot \frac{h}{R} = \frac{2 g ( R - h )}{R} \Rightarrow h = \frac{2 R}{3}

With

R = 3

m, the block loses contact at a height of

2

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer

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