Question

A particle $P$ is moving in a circle of radius $r$ with uniform speed $v$. $C$ is the center of the circle and $AB$ is diameter. The angular velocity of $P$ about $A$ and $C$ is in the ratio

• A. $4:1$
• B. $1:1$
• C. $2:1$
• D. $1:2$

Answer 1

The correct option is D

$1:2$

Step 1. Given data:

Radius of the circle around which particle $P$ is moving = $r$

Speed of $P=v$

Diameter of the circle = $AB$

Step 2. Formula used:

The angular velocity of a particle about any point is given by:

$\omega =\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$r$}\right.$, where, $v=$ speed of the particle, $r=$distance of the particle from the point

Here, given the speed of the particle is constant.

Step 3. Calculations:

Angular speed of $P$ about point $A$, ${\omega }_A=\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$AB$}\right.=\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$2r$}\right.$, ( As $AB=2r$)

Angular speed of $P$ about point $C$, ${\omega }_C=\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$BC$}\right.=\raisebox{1ex}{$v$}\!\left/ \!\raisebox{-1ex}{$r$}\right.$, ( As $BC=r$)

Hence, $$\begin{gathered} \frac{{\omega }_A}{{\omega }_C}=\frac{1}{2} \\ {\omega }_A:{\omega }_C=1:2 \end{gathered}$$

Thus, the correct option is option D.