A body slides down an inclined surface which ends into a vertical loop of radius R = 40cm.

What must be the height H of the inclined surface for the body so that it does not leave contact with the surface even at the uppermost point of the loop? Assume friction to be absent.

Explanation

Let v be the velocity of the particle of the particle at the highest point. At the highest position the net force on the body is `(N + Mg)` towards the centre of the path, and which provides the required centripetal force.
Therefore, `N +mg = (mv)^(2)/R`
The body is not detached from the loop if `N ge 0`. In the limiting case, N = 0
That is ` mg = (mv)^(2)/R or v^(2) = gR`
Applying energy conservation at the initial and highest point of the loop, we get.
`mgH = mg(2R) + 1/2 (mv)^(2)`
Using `v^(2) = gR,` we obtain,
`mgH = mg (2R) + 1/2 m(gR) or`
`H = 5/2R = 2.5 R`
Putting `R = 40 cm = 0.4 m,`we get
`H = (2.5)(0.4) = 1` m.