Question

A particle mass $m$ is attached to a thin uniform rod of length ${}^{\prime }{a}^{\prime }$ at a distance of $\frac{a}{4}$ from the mid point C as shown in the figure. The mass of the rod is ${}^{\prime }4m^{\prime }$. The moment of inertia of the combined system about an axis passing through 'O' and perpendicular to the rod is:

• A. $\frac{27}{48}{ma}^2$
• B. $\frac{51}{48}{ma}^2$
• C. $\frac{64}{48}{ma}^2$
• D. $\frac{91}{48}{ma}^2$

Answer 1

The correct option is D $\frac{91}{48}{ma}^2$

Moment of inertia of the rod alone about the given axis=$M\frac{l^2}{3}$

$=\left(4m\right)\frac{a^2}{3}=\frac{4}{3}m{a}^2$

Moment of inertia of the particle about the given axis=$m{r}^2=m(\frac{3a}{4}{)}^2=\frac{9}{16}m{a}^2$

Thus total moment of inertia of the system=$I_1+I_2$

$=\frac{91}{48}m{a}^2$