A boy is moving with a constant speed v on a | vedclass

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Question

(B) Let the boy's velocity be $\vec{v}_B = v \hat{i}$.Let the position of the point mass on the circle of radius $R$ be given by $	heta = \omega t$,w

Vedclass · Accepted Answer

$\frac{v}{{2\pi R}}$

Vedclass · Answer

(B) Let the boy's velocity be $\vec{v}_B = v \hat{i}$.Let the position of the point mass on the circle of radius $R$ be given by $	heta = \omega t$,where $\omega = \frac{v}{R}$.The velocity of the point mass is $\vec{v}_P = -v \sin(\omega t) \hat{i} + v \cos(\omega t) \hat{j}$.The relative velocity is $\vec{v}_{rel} = \vec{v}_P - \vec{v}_B = (-v \sin(\omega t) - v) \hat{i} + v \cos(\omega t) \hat{j}$.The magnitude of relative velocity is $|\vec{v}_{rel}| = \sqrt{(-v \sin(\omega t) - v)^2 + (v \cos(\omega t))^2}$.$|\vec{v}_{rel}| = \sqrt{v^2 \sin^2(\omega t) + v^2 + 2v^2 \sin(\omega t) + v^2 \cos^2(\omega t)} = \sqrt{2v^2 + 2v^2 \sin(\omega t)} = v \sqrt{2(1 + \sin(\omega t))}$.The magnitude varies periodically with the angular frequency $\omega = \frac{v}{R}$.The time period of this variation is $T = \frac{2\pi}{\omega} = \frac{2\pi R}{v}$.The frequency of change is $f = \frac{1}{T} = \frac{v}{2\pi R}$.Thus,the correct option is $B$.

$A$ boy is moving with a constant speed $v$ on a small trolley towards a distant circle as shown in the figure. $A$ point mass is moving on the circle with a constant speed $v$. What is the frequency of change in the magnitude of the relative velocity of the point mass,as observed by the boy?

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