Question 574: Centre of Mass and Rotational Motion

Question

Two identical particles each of mass '$m$' are separated by a distance '$d$'. The axis of rotation passes through the midpoint of '$\mathrm{d}$' and is perpendicular to the length $\mathrm{d}$. If '$\mathrm{K}$' is the average rotational kinetic energy of the system, then the angular frequency is

Accepted Answer

Moment of inertia $\mathrm{I}=2 \mathrm{~m}\left(\frac{\mathrm{d}}{2}ight)^2=\frac{\mathrm{md}^2}{2}$
Kinetic energy $\mathrm{k}=\frac{1}{2} \mathrm{I} \omega^2$
$\begin{aligned}
& 	herefore \omega^2=\frac{2 k}{I}=2 \mathrm{k} \cdot \frac{2}{\mathrm{md}^2}=\frac{4 \mathrm{k}}{\mathrm{md}^2} \
& 	herefore \omega=\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}
\end{aligned}$

Answer

$2 \mathrm{~d} \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}$

Answer

$\frac{\mathrm{d}}{2} \sqrt{\frac{\mathrm{K}}{\mathrm{m}}}$

Answer

$\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{K}}{\mathrm{m}}}$

Answer

$\frac{d}{4} \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}$

Centre of Mass and Rotational Motion

Two identical particles each of mass 'm' are separated by a distance 'd'. The axis of rotation passes through the midpoint of '\mathrm{d}' and is perpendicular to the length \mathrm{d}. If '\mathrm{K}' is the average rotational kinetic energy of the system, then the angular frequency is

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