Question

A small block of mass m starts sliding down from rest along the smooth surface of a fixed hollow hemisphere of same mass m. Find the distance of centre of mass of block and hemisphere from centre of hemisphere C when block m separates from the surface of hemisphere.

• A. The distance of the centre of mass of the block and hemisphere from the centre of hemisphere C is $\frac{R\sqrt{69}}{15}$.
• B. The distance of the centre of mass of the block and hemisphere from the centre of hemisphere C is $\frac{R\sqrt{5}}{15}$.
• C. The distance of the centre of mass of the block and hemisphere from the centre of hemisphere C is $\frac{8R}{15}$.
• D. The distance of the centre of mass of the block and hemisphere from the centre of hemisphere C is $\frac{R\sqrt{69}}{5}$.

Answer 1

Answer: (A)

The distance of the centre of mass of the block and hemisphere from the centre of hemisphere C is $\frac{R\sqrt{69}}{15}$.

Answer 2

1. Condition for Separation: The block separates from the hemisphere when the normal reaction force $N$ becomes zero. At this point, the radial component of gravity provides the centripetal force: $mg \cos \theta = m \frac{v^2}{R}$.

2. Energy Conservation: The block starts from rest at height $R$. Using conservation of mechanical energy, the kinetic energy $v^2$ at an angle $\theta$ is related to the initial state: $mgR = \frac{1}{2}mv^2 + mg(R \cos \theta)$, which simplifies to $v^2 = 2gR(1 - \cos \theta)$.

3. Solving for $\theta$: Substituting the expression for $v^2$ into the separation condition: $mg \cos \theta = m \frac{2gR(1 - \cos \theta)}{R}$. This yields $\cos \theta = 2/3$.

4. Block's Position: At the point of separation, the coordinates of the block relative to the center of the hemisphere C are: $y_{block} = R \cos \theta = R \left(\frac{2}{3}\right) = \frac{2R}{3}$ $x_{block} = R \sin \theta = R \sqrt{1 - \cos^2 \theta} = R \sqrt{1 - \left(\frac{2}{3}\right)^2} = \frac{R\sqrt{5}}{3}$

5. Hemisphere's Center of Mass: The center of mass of a hollow hemisphere of radius $R$ is located at a distance $R/2$ from its center along the axis of symmetry. Assuming the axis of symmetry is vertical, its coordinates are $(0, R/2)$. The mass of the hemisphere is given as $M=4m$ from the figure.

6. System's Center of Mass: The total mass of the system (block + hemisphere) is $m_{total} = m + M = m + 4m = 5m$. The coordinates of the center of mass $(X_{CM}, Y_{CM})$ are: $X_{CM} = \frac{m \cdot x_{block} + M \cdot x_{hemisphere}}{m_{total}} = \frac{m \left(\frac{R\sqrt{5}}{3}\right) + 4m (0)}{5m} = \frac{R\sqrt{5}}{15}$ $Y_{CM} = \frac{m \cdot y_{block} + M \cdot y_{hemisphere}}{m_{total}} = \frac{m \left(\frac{2R}{3}\right) + 4m \left(\frac{R}{2}\right)}{5m} = \frac{\frac{2mR}{3} + 2mR}{5m} = \frac{8R}{15}$

7. Distance from C: The distance of the system's center of mass from C is: $d = \sqrt{X_{CM}^2 + Y_{CM}^2} = \sqrt{\left(\frac{R\sqrt{5}}{15}\right)^2 + \left(\frac{8R}{15}\right)^2} = \sqrt{\frac{5R^2}{225} + \frac{64R^2}{225}} = \sqrt{\frac{69R^2}{225}} = \frac{R\sqrt{69}}{15}$.