Question

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) 

 

Hint:

For problems where an object loses contact with a circular path, the key condition is that the normal force becomes zero ($N=0$). Combine this with the conservation of energy to find the required height or speed.

Answer 1

Correct Answer: 1

Let the block of mass 'm' lose contact at a point A, which is at a vertical height 'h' from the top.
Let $\theta$ be the angle that the radius to point A makes with the vertical. The vertical distance dropped is $h = R(1 - \cos\theta)$.
At point A, the forces acting on the block are the normal force (N) and gravity (mg). The component of gravity towards the center of the hemisphere is $mg\cos\theta$.
The net force towards the center provides the centripetal force:
$mg\cos\theta - N = \frac{mv^2}{R}$.
The block loses contact when the normal force N becomes zero.
At $N=0$, we have $mg\cos\theta = \frac{mv^2}{R}$, which simplifies to $v^2 = gR\cos\theta$. (Equation 1)
Now, apply the principle of conservation of mechanical energy between the top of the hemisphere (initial position) and point A (final position).
Initial Energy (at top): $E_i = K.E._i + P.E._i = 0 + mgR$ (taking the base of the hemisphere as the reference level).
Final Energy (at A): $E_f = K.E._f + P.E._f = \frac{1}{2}mv^2 + mg(R-h) = \frac{1}{2}mv^2 + mgR\cos\theta$.
Equating initial and final energies:
$mgR = \frac{1}{2}mv^2 + mgR\cos\theta$.
$mgR(1 - \cos\theta) = \frac{1}{2}mv^2$.
$v^2 = 2gR(1 - \cos\theta)$. (Equation 2)
Now, equate the two expressions for $v^2$ from Equation 1 and Equation 2.
$gR\cos\theta = 2gR(1 - \cos\theta)$.
$\cos\theta = 2 - 2\cos\theta$.
$3\cos\theta = 2 \implies \cos\theta = \frac{2}{3}$.
The height 'h' at which the block loses contact is the vertical distance it has dropped.
$h = R(1 - \cos\theta) = R\left(1 - \frac{2}{3}\right) = \frac{R}{3}$.
Given the radius R = 3 m:
$h = \frac{3}{3} = 1$ m.