Question

A solid hemisphere of radius R is mounted on a solid cylinder of the same radius and density as shown. The height of the cylinder is such that the resulting centre of the mass is O. If the radius of the sphere is $2\sqrt{2}m$ and the height of the cylinder is n metre. Find n.

Answer 1

Hint: Make the point O, which is the centre of mass of the system, as the origin (0,0). Then use the formula for centre of mass of a system of two bodies i.e $M{{x}_{com}}={{M}_{1}}{{x}_{1}}+{{M}_{2}}{{x}_{2}}$, where M is total mass of the system, xcom is the position of centre of mass of the system, $x_1$ and $x_2$ are the positions of the centre of masses of $M_1$ and $M_2$.

Complete step by step answer:
The centre of mass for a solid hemisphere is at a distance of $\dfrac{3R}{8}$ units from the centre of the circular base along the axis passing through this centre and perpendicular to the circular base. The centre of mass of a cylinder is at its centre i.e. at a distance of half of the height from the centre of the circular and perpendicular to the surface of the top.
We know the centre of mass of the hemisphere and the cylinder. Let the mass of the hemisphere be M1 and the mass of the cylinder be M2. Then by using the centre of mass formula $M{{x}_{com}}={{M}_{1}}{{x}_{1}}+{{M}_{2}}{{x}_{2}}$ ……(1) , where M is total mass of the system, xcom is the position of centre of mass of the system, $x_1$ and $x_2$ are the positions of the centre of masses of $M_1$ and $M_2$. In this case, $M={{M}_{1}}+{{M}_{2}}$…….(2). Let the origin (0,0) be point O and also it is given to the centre of mass of the system. $\Rightarrow {{x}_{com}}=0$.
${{x}_{1}}=\dfrac{3R}{8}$ and we have to find $x_2$. Before that let us find $M_1$ and $M_2$.
The hemisphere and the cylinder have the same radius (R) and same density ($\rho $). The formula for density is $\text{ }\!\!\rho\!\!\text{ =}\dfrac{\text{mass}}{\text{volume}}$ . Therefore, $\text{ }\!\!\rho\!\!\text{ =}\dfrac{\text{mass}}{\text{volume}}=\dfrac{{{M}_{1}}}{{{V}_{1}}}=\dfrac{{{M}_{2}}}{{{V}_{2}}}$, where $V_1$ and $V_2$ are the volumes of the hemisphere and the cylinder respectively.
$\Rightarrow {{V}_{2}}=\dfrac{2}{3}\pi {{R}^{3}}$ and $\Rightarrow {{V}_{2}}=\pi {{R}^{2}}h$ (h is the height of the cylinder)
$\Rightarrow {{M}_{1}}=\rho {{V}_{1}}=\dfrac{2}{3}\rho \pi {{R}^{3}}$ and ${{M}_{2}}=\rho {{V}_{2}}=\rho \pi {{R}^{2}}h$
Substitute the values of M1 and M2 in equation (2).
$\Rightarrow M=\dfrac{2}{3}\rho \pi {{R}^{3}}+\rho \pi {{R}^{2}}h$
Substitute the values of M, M1 and M2 in equation (1).
Therefore, $\left( \dfrac{2}{3}\rho \pi {{R}^{3}}+\rho \pi {{R}^{2}}h \right){{x}_{com}}=\dfrac{2}{3}\rho \pi {{R}^{3}}{{x}_{1}}+\rho \pi {{R}^{2}}h{{x}_{2}}$
We had assumed xcom to be 0. ${{x}_{1}}=\dfrac{3R}{8}$. Substitute these values in the above equation.
$\Rightarrow \left( \dfrac{2}{3}\rho \pi {{R}^{3}}+\rho \pi {{R}^{2}}h \right).(0)=\dfrac{2}{3}\rho \pi {{R}^{3}}.\left( \dfrac{3R}{8} \right)+\rho \pi {{R}^{2}}h{{x}_{2}}$
$\Rightarrow (0)=\dfrac{{{R}^{2}}}{4}+h{{x}_{2}}$
Therefore, ${{x}_{2}}=-\dfrac{{{R}^{2}}}{4h}$. But we know that x2 will be half of the height of the cylinder. But since it is down the origin, the value of x2 will be negative. This implies, ${{x}_{2}}=-\dfrac{h}{2}$
$\Rightarrow -\dfrac{h}{2}=-\dfrac{{{R}^{2}}}{4h}\Rightarrow {{h}^{2}}=\dfrac{{{R}^{2}}}{2}$
Therefore, $h=\dfrac{R}{\sqrt{2}}=\dfrac{2\sqrt{2}}{\sqrt{2}}=2m$. Hence, the height of the cylinder i.e. n is 2m.

Note: Here, there were only two bodies so the centre of mass of the system will be on the joining the centre of masses of both the bodies. If there are more bodies then we have to solve the question using vectors i.e. $M{{\overrightarrow{r}}_{com}}={{M}_{1}}{{\overrightarrow{r}}_{1}}+{{M}_{2}}{{\overrightarrow{r}}_{2}}+{{M}_{3}}{{\overrightarrow{r}}_{3}}+.......$.