Question

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. The angular velocity of particle when it is at point B about (A) and (C) are in the ratio:
A. 1:1
B. 1:2
C. 2:1
D. 4:1

Answer 1

Hint:As per question a particle (at point B) is moving in a circular path with a uniform speed v thus we need to find angular velocity of particle with the help of linear velocity with respect to points two points one is C and other is A. Now as given the diameter of circle is AB (2a length) with center at point at C So, the distance of point B from C is a (BC) and from A is 2a (AB). As per question we need to find the ratio between both the angular velocity one with distance BC and other is AB.

Formula used:
Angular velocity of any body is given as,
$\omega =\dfrac{\Delta \theta }{\Delta t}$
where \[\omega \] is angular velocity, and $\Delta \theta $ is is change in displacement, $\Delta t$ is change in time or can say rate change in displacement $\dfrac{\Delta \theta }{\Delta t}$.
Linear velocity is given as,
$v=\dfrac{\Delta s}{\Delta t}$
where v is linear velocity and $\Delta s$ is change in linear displacement, $\Delta t$ is change in time or can say rate change in linear displacement such as $\dfrac{\Delta s}{\Delta t}$ .
Angular displacement = arc/radius=$\theta =\dfrac{s}{r}$
where s is an arc covered with particles and r is the radius of the circle in which the body is moving.

Complete step by step solution:
A particle p present at point B on circle and moving on a circumference of circle having radius a (CB), where point C is the centre of the circle and AB is a diameter of circle is given. The particle is moving in this circle with uniform linear velocity v (rate change of linear displacement is constant).

By using the formula we get
Linear velocity $v=\dfrac{\Delta s}{\Delta t}$
Similarly
Angular velocity $\omega =\dfrac{\Delta \theta }{\Delta t}$
By establishing a relation in between linear velocity v and in angular velocity $\omega $
$\omega =\dfrac{v}{r}$
Here, angular speed about point A.
Here the radius of circle AB = 2a
\[{{\omega }_{A}}=\dfrac{v}{AB} \\
\Rightarrow {{\omega }_{A}}=\dfrac{v}{2a} \\ \]
Angular velocity at point c
\[{{\omega }_{C}}=\dfrac{v}{Bc} \\
\Rightarrow {{\omega }_{C}}=\dfrac{v}{a} \\ \]
By taking a ration of angular velocity at point And angular velocity at point c
Ratio \[\dfrac{{{\omega }_{A}}}{{{\omega }_{C}}}\]
\[\dfrac{{{\omega }_{A}}}{{{\omega }_{C}}}=\dfrac{\dfrac{v}{2a}}{\dfrac{v}{a}} \\
\Rightarrow \dfrac{{{\omega }_{A}}}{{{\omega }_{C}}}=\dfrac{a}{2a} \\
\therefore \dfrac{{{\omega }_{A}}}{{{\omega }_{C}}}=\dfrac{1}{2} \]

Here the correct option is B.

Note: From result we can conclude that the angular velocity of a particle at B has great angular velocity as compared to angular velocity of particle B with respect to C point (angular velocity of C is equal to two times of angular velocity of particle when particle is about point A). In the given question velocity is uniform thus, angular velocity is totally dependent on distance r and relation between angular velocity and distance is inversely proportional so for small distance angular velocity will be more and vice versa.