At what height (in metres) from the ground would a body sliding over a curved surface of a semi-circular cylinder of radius 6m, kept horizontally on the ground, lose its contact with the surface ?

Answer

4

Explanation

We have, `F = mg costheta - N`
When the ball loses contact, the normal force (N) = 0
`therefore F = mg cos theta`
since the ball is moving in a circular path,
`F = (mv^(2))/R`
`mg cos theta = (mv^(2))/R`
`therefore v^(2) = Rg cos theta`
From conservation of energy,
`mgR = 1/2 mv^(2) + mgh`
`Rg = 1/2 Rg cos theta + gh`
`h = R - R/2 cos theta` From
figure we have,
`cos theta = h/R`
`therefore h = R- R/2 xx h/R`
`therefore (3h)/2 = R`
`therefore h = (2R)/3 = (2 xx 6)/2`
`therefore h = 4m`