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)

1. Conservation of Energy

As the block slides from the top of the hemisphere (radius $$R = 3\text{ m}$$) down to a height $$h$$, it drops a vertical distance of $$R - h$$. Equating its loss in potential energy to its gain in kinetic energy gives:

$$mg(R - h) = \frac{1}{2}mv^2 \implies v^2 = 2g(R - h)$$

2. Condition for Losing Contact

The centripetal force equation along the radial direction is:

$$mg\cos\theta - N = \frac{mv^2}{R}$$

The block leaves the surface when the normal force drops to zero ($$N = 0$$). Substituting $$\cos\theta = \frac{h}{R}$$ yields:

$$mg\left(\frac{h}{R}\right) = \frac{mv^2}{R} \implies v^2 = gh$$

3. Solve for Height ($$h$$)

Equating the two expressions for $$v^2$$:

$$2g(R - h) = gh$$

$$2R - 2h = h \implies 3h = 2R \implies h = \frac{2}{3}R$$

Plugging in $$R = 3\text{ m}$$:

$$h = \frac{2}{3} \times 3 = 2\text{ m}$$

Conclusion

The block will lose contact with the hemisphere at a height of 2 m.