Question

A uniform solid hemisphere of density $\rho_1$ and radius $R$ and a solid cone of density $\rho_2$ and radius, and height each $R$ are glued as shown. Center of mass of system lies at common centre of circular interface then $\frac{\rho_2}{\rho_1}$ is ________.

Answer 1

3

Answer 2

Let the common center of the circular interface be the origin. The center of mass of a solid hemisphere of radius $R$ from its base is at $z_1 = \frac{3R}{8}$. The mass of the hemisphere is $m_1 = \rho_1 \frac{2}{3}\pi R^3$. The center of mass of a solid cone of height $R$ from its base is at $z_2 = -\frac{R}{4}$. The mass of the cone is $m_2 = \rho_2 \frac{1}{3}\pi R^3$. For the system's center of mass to be at the origin, $m_1 z_1 + m_2 z_2 = 0$. Substituting these values gives: $\left(\rho_1 \frac{2}{3}\pi R^3\right) \left(\frac{3R}{8}\right) + \left(\rho_2 \frac{1}{3}\pi R^3\right) \left(-\frac{R}{4}\right) = 0$ Simplifying this equation leads to $\frac{\rho_2}{\rho_1} = 3$.