A particle P is moving in a circle of radius | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The angular velocity $\omega$ of a particle moving with velocity $v$ at a distance $r_{\perp}$ from a point is given by $\omega = \frac{v_{\perp}}

Vedclass · Accepted Answer

$1:2$

Vedclass · Answer

(B) The angular velocity $\omega$ of a particle moving with velocity $v$ at a distance $r_{\perp}$ from a point is given by $\omega = \frac{v_{\perp}}{r_{\perp}}$,where $v_{\perp}$ is the component of velocity perpendicular to the position vector.At point $B$,the velocity $v$ is perpendicular to the diameter $AB$.For point $A$,the distance is $AB = 2a$. Thus,the angular velocity about $A$ is $\omega_A = \frac{v}{2a}$.For point $C$,the distance is $CB = a$. Thus,the angular velocity about $C$ is $\omega_C = \frac{v}{a}$.The ratio of angular velocities is $\frac{\omega_A}{\omega_C} = \frac{v/2a}{v/a} = \frac{1}{2}$.

$A$ particle $P$ is moving in a circle of radius $a$ with a uniform speed $v$. $C$ is the centre of the circle and $AB$ is a diameter. When passing through $B$,the angular velocity of $P$ about $A$ and $C$ are in the ratio:

Similar Questions

If a particle moves in a circle describing equal angles in equal times,its velocity vector

$A$ particle is moving with constant speed in a circular path. When the particle turns by an angle $90^{\circ}$,the ratio of instantaneous velocity to its average velocity is $\pi : x \sqrt{2}$. The value of $x$ will be $.........$

The circular motion of a particle with constant speed is

Two cars of masses $m_1$ and $m_2$ are moving in circles of radii $r_1$ and $r_2$,respectively. Their speeds are such that they make complete circles in the same time $t$. The ratio of their centripetal acceleration is

$A$ particle is moving in a circle with uniform speed. Its motion is

Vedclass Test Series

Exam Paper Generator

Online Exam Module