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(B) The angular velocity $\omega$ of a particle moving with velocity $v$ at a distance $r$ from a point,where the velocity vector is perpendicular to

Vedclass · Accepted Answer

$1 : 2$

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(B) The angular velocity $\omega$ of a particle moving with velocity $v$ at a distance $r$ from a point,where the velocity vector is perpendicular to the position vector,is given by $\omega = \frac{v}{r}$.When the particle is at point $B$,its velocity $v$ is perpendicular to the diameter $AB$.For point $C$ (the center),the distance is $r_C = a$. Thus,$\omega_C = \frac{v}{a}$.For point $A$,the distance is $r_A = AB = 2a$. Thus,$\omega_A = \frac{v}{2a}$.The ratio of angular velocity with respect to $A$ and $C$ is $\frac{\omega_A}{\omega_C} = \frac{v/2a}{v/a} = \frac{1}{2}$.Therefore,the ratio is $1 : 2$.

$A$ particle $P$ moves in a circle of radius $a$ with a constant speed $v$. $C$ is the center of the circle,and $AB$ is a diameter. When the particle passes through point $B$,what is the ratio of its angular velocity with respect to $A$ and $C$ respectively?

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