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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