A particle of mass $m$ is attached to a thin uniform rod of length $\mathrm{a}$ and mass $4 m$. The distance of the particle from the centre of mass of the rod is $a / 4$. The moment of inertia of the combination about an axis passing through $\mathrm{O}$ normal to the rod is

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Aicharya · Accepted Answer

Moment of inertia
$=m\left(\frac{3 a}{4}\right)^{2}+m_{1} \frac{a^{2}}{3}$
For the centre of rod
$\left(\frac{m_{1} a^{2}}{12}+\frac{m_{1} a^{2}}{4}\right)=\frac{m_{1} a^{2}}{3}$
$\therefore \quad m_{1}=4 m$
Total $I=m\left(\frac{3 a}{4}\right)^{2}+\frac{4 m a^{2}}{3}$
$=\frac{9 m a^{2}}{16}+\frac{4 m a^{2}}{3}$
$=\frac{(27+64)}{48} m a^{2}=\frac{91}{48} m a^{2}$

A particle of mass $m$ is attached to a thin uniform rod of length $\mathrm{a}$ and mass $4 m$. The distance…

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