Question

A body consisting of a cone and hemisphere on the same base then greatest height of the one the equilibrium may be stable is ?
1.. √2
2.. √5
3.. √7
4... √3

Answer 1

Answer:

Height will be √3 times radius

Step-by-step explanation:

As the body is symmetrical about the vertical axis, its centre of gravity will lie on this axis. Considering two parts of the body, viz. hemisphere and cone. Let bottom of the hemisphere be the axis reference.

Hemisphere:

y-coordinate of center of graving of hemisphere.

$y_{1}=\frac{5}{8} r \text {. }$

Mass of hemisphere :

$m_{1}=\rho \frac{2}{3} \pi r^{3}$

For cone :

Y-coordinate of centre of gravity of cone

$y_{2}=r+\frac{h}{4}$

$Mass of cone \quad m_{2}=\rho \frac{1}{3} \pi r^{2} h$

Distance of centre of gravity of the combined body from O is,

$\bar{y}=\frac{m_{1} y_{1}+m_{2} y_{2}}{m_{1}+m_{2}}=\frac{\rho\left(\frac{2}{3}\right) \pi r^{3} \times\left(\frac{5}{8}\right) r+\rho\left(\frac{1}{3}\right) \pi r^{2} h \times\left[r+\left(\frac{h}{4}\right)\right]}{\rho\left(\frac{2}{3}\right) \pi r^{3}+\rho\left(\frac{1}{3}\right) \pi r^{2} h}$

$\bar{y}=\frac{\left(\frac{5}{12}\right) r^{4}+\left(\frac{1}{3}\right) r^{3} h+\left(\frac{1}{12}\right) r^{2} h^{2}}{\left(\frac{2}{3}\right) r^{3}+\left(\frac{1}{3}\right) r^{2} h}$

Condition for stable equilibrium :

Center of gravity of a body should be below the common face BC or maximum it coined with it. Therefore,

$\bar{y}=\frac{\left(\frac{5}{12}\right) r^{4}+\left(\frac{1}{3}\right) r^{3} h+\left(\frac{1}{12}\right) r^{2} h^{2}}{\left(\frac{2}{3}\right) r^{3}+\left(\frac{1}{3}\right) r^{2} h}$

On solving,

$r^{2}=h^{2} \\&h=\sqrt{3} \cdot r \\&h=1.739 r$

So height  is √3 times greater than radius of cone.