Question

A solid hemisphere ($m,R$) has the moment of inertia about axis $1$ (diameter of hemisphere)

• A. $\frac{m{R}^2}{5}$
• B. $\frac{2}{5}m{R}^2$
• C. $\frac{2}{3}m{R}^2$
• D. $\frac{173}{320}m{R}^2$

Answer 1

The correct option is A $\frac{m{R}^2}{5}$

Here first we assume that solid hemisphere is constructed by small discs so now.

Mass 'dm' of disc $= \dfrac{\text{mass of hemisphere}}{Vol. of hemisphere} \times vol. of disc.

$dm=\left(\frac{m}{\frac{2}{3}\pi R^3}\right)$ $\left(\pi y^{2}dx\right)$

$dm=\rho \left(\pi y^{2}dx\right)$ $\left[where\rho =\frac{M}{\frac{2}{3}\pi R^3}\right]$

$d{I}_{y{y}^{\prime }}=\left(dm\right)\frac{y^2}{4}+\left(dm\right)x^2$ $[I_{y{y}^{\prime }}=I_{cm}+m{r}^2]$

$I_{y{y}^{\prime }}={\int }_0^R\left(dm\right)\frac{y^2}{4}+{\int }_0^{R}dm{x}^2$

$={\int }_0^R\left(\rho \pi y^{2}dx\right)\frac{y^2}{4}+{\int }_0^R\left(\rho \pi y^{2}dx\right)x^2$

$=\frac{\rho \pi }{4}{\int }_0^R(R^2-x^2)dx+\rho y{\int }_0^R(R^2{x}^2-x^4)dx$

$=\frac{\rho \pi }{4}[R^5+\frac{R^5}{5}-2R^2\frac{R^3}{3}]+$$[\rho \pi R^2\left(\frac{R^3}{3}\right)-\rho \pi \left(\frac{R^5}{5}\right)]$

$=\rho \pi R^5[\frac{1}{4}+\frac{1}{20}-\frac{1}{4}.\frac{2}{3}]$ $+\rho \pi R^5[\frac{1}{3}-\frac{1}{5}]$

$=\rho \pi R^5\left(\frac{4}{15}\right)$

But $\rho =\frac{3m}{2\pi R^3}$

$I_{y{y}^{\prime }}=\frac{3m}{2\pi R^3}.\pi R^5\frac{4}{15}=\frac{2}{5}m{R}^2$

$\therefore I_1=\frac{2}{5}m{R}^2$