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

The correct answer is 1 : 2.

Let us understand the motion of the particle step-by-step:

Step 1: Angular velocity about the center C
The particle P is moving in a circle of radius a with a uniform speed v.
The center of the circle is C. The distance of P from the center C is always equal to the radius, which is a.
The velocity vector of P is always perpendicular to the radius vector pointing from C to P.
Therefore, the angular velocity of P about the center C, denoted as ${\omega }_C$, is given by the relation:
${\omega }_C=\frac{v}{a}$

Step 2: Distance from A to B
AB is a diameter of the circle. Since the radius is a, the length of the diameter AB is:
$r=2a$

Step 3: Angular velocity about point A when P is at B
When the particle P passes through point B, its distance from point A is equal to the diameter of the circle, which is 2a.
Since P is moving along the circular path, at point B, its velocity vector v is tangent to the circle at B. Since AB is a diameter (passing through the center C and point B), the tangent at B is perpendicular to the line AB.
Thus, the component of velocity of P perpendicular to the line joining A and P (which is AB at this instant) is the full speed v.
The angular velocity of P about point A, denoted as ${\omega }_A$, is therefore:
${\omega }_A=\frac{v}{2a}$

Step 4: Find the ratio
We need to find the ratio of the angular velocity of P about A to that about C when passing through B:
$\frac{{\omega }_A}{{\omega }_C}=\frac{\left(\frac{v}{2a}\right)}{\left(\frac{v}{a}\right)}=\frac{1}{2}$

Thus, the ratio of the angular velocity of P about A and C is 1 : 2.