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

Correct answer: (C) 2/(~d) √(K/m). Moment of inertia I=2 ~m (d/2 )^2=(md^2)/2 Kinetic energy K=1/2 I ω^2 ∴ ω^2=(2 ~K)/I=2 ~K · 2/(md^2)=(4 ~K)/(md^2) ∴ ω=2/(~d) √(k/m)

Q127 / 150 · MHT CET 2021 (23 Sep Shift 1) · Physics

Two identical particles each of mass ' $\mathrm{m}$ ' 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

A
$2 \mathrm{~d} \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}$
B
$\frac{\mathrm{d}}{2} \sqrt{\frac{\mathrm{K}}{\mathrm{m}}}$
Correct answer
C
$\frac{2}{\mathrm{~d}} \sqrt{\frac{\mathrm{K}}{\mathrm{m}}}$
D
$\frac{\mathrm{d}}{4} \sqrt{\frac{\mathrm{m}}{\mathrm{K}}}$

Solution

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

Hence, the correct answer is option: 'C'.

Solution

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