Question

A thin rod of mass $0.3kg$ and length $0.2m$ is suspended horizontally by a metal wire which passes through its centre of mass and is perpendicular to its length. The rod is set to torsional oscillation with a period of $2s$. The thin rod is replaced by an equilateral triangular lamina which is suspended horizontally from its centre of mass. If its period of oscillations is found to be $10s$, the moment of inertia of the triangular lamina about the axis of suspension is approximately.

• A. $5.0\times {10}^{-3}kg{m}^2$
• B. $0.5\times {10}^{-3}kg{m}^2$
• C. $0.5\times {10}^{-4}kg{m}^2$
• D. $2.5\times {10}^{-2}kg{m}^2$

Answer 1

The correct option is D $2.5\times {10}^{-2}kg{m}^2$

$\omega =\sqrt{\frac{C}{I}}$(torsional oscilation equation)

let ${\omega }_r$, ${\omega }_l$ be frequencies of rod and lamina respectively and

${\omega }_r=\pi$ and ${\omega }_l=\frac{\pi }{5}$

$\sqrt{\frac{C}{I_r}}=\pi$ and $\sqrt{\frac{C}{I_l}}=\frac{\pi }{5}$

$I_r=(\frac{{\pi }^2}{C}{)}^2$ and $I_l=(\frac{{\pi }^2}{25C}{)}^{-1}$

$I_l=25I_r$

$I_r=\frac{1}{12}m{l}^2=\frac{1}{12}0.3\times (0.2{)}^2={10}^{-3}{I}_l=25\times {10}^{-3}=2.5\times {10}^{-2}kg-m^2$