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(D) The angular speed $\omega$ of a particle in uniform circular motion is related to its time period $T$ by the formula $\omega = \frac{2\pi}{T}$.Giv

Vedclass · Accepted Answer

$1: 1$

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(D) The angular speed $\omega$ of a particle in uniform circular motion is related to its time period $T$ by the formula $\omega = \frac{2\pi}{T}$.Given that the time periods of particles $A$ and $B$ are the same,i.e.,$T_A = T_B = T$.Therefore,the angular speed of particle $A$ is $\omega_A = \frac{2\pi}{T_A} = \frac{2\pi}{T}$.Similarly,the angular speed of particle $B$ is $\omega_B = \frac{2\pi}{T_B} = \frac{2\pi}{T}$.Taking the ratio of the angular speeds,we get $\frac{\omega_A}{\omega_B} = \frac{2\pi/T}{2\pi/T} = 1$.Thus,the ratio of the angular speed of $A$ to that of $B$ is $1: 1$.

Two particles $A$ and $B$ are moving in uniform circular motion in concentric circles of radii $r_{A}$ and $r_{B}$ with speeds $v_A$ and $v_B$ respectively. The time period of rotation is the same. The ratio of angular speed of $A$ to that of $B$ will be

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