A light meter rod has two-point masses each of $2 \mathrm{~kg}$ fixed at its ends. If the system rotates about its centre of mass with an angular speed of $0.5 \mathrm{rad} \mathrm{s}^{-1}$, its rotational $\mathrm{KE}$ is

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Aicharya · Accepted Answer

Total mass of light meter rod including two point masses, $m=(2+2) \mathrm{kg}=4 \mathrm{~kg}$ Angular speed, $I=0.5 \mathrm{rad} \mathrm{s}^{-1}$ Length of rod $=1 \mathrm{~m}$

Moment of inertia about centre of mass, $I=2\left(\frac{1}{2}\right)^2+2\left(\frac{1}{2}\right)^2=\frac{2}{4}+\frac{2}{4}=1 \mathrm{~kg}-\mathrm{m}^2$ $\therefore$ Rotational kinetic energy $K_{\text {rot }}=\frac{1}{2} I \omega^2=\frac{1}{2} \times 1 \times(0.5)^2=\frac{0.25}{2}=0.125 \mathrm{~J}$

A light meter rod has two-point masses each of \(2 \mathrm{~kg}\) fixed at its ends. If the system rotates…

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