
If the body is moving in a circle of radius $r$ with a constant speed $v$. Its angular velocity is :
A. $\dfrac{{{v}^{2}}}{r} \\ $
B. $vr \\ $
C. $\dfrac{v}{r} \\ $
D. $\dfrac{r}{v}$
Answer
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Hint:As per question a body is moving with constant speed $v$ (rate change in displacement is constant or same) means body while moving is not changing its speed and the path in which both is moving in a circle whose radius is r as given. So, now we need to find the angular velocity $\omega$ (rate change of angular displacement) of the body because when the body moves in a circular path we consider angular velocity of body instead of linear velocity.. As per given question, we have constant velocity but need to find angular velocity so, we need to find a relation between both velocities.
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 arc covered with particles and r is radius of circle in which the body is moving.
Complete step by step solution:
A body moving in a circular motion with constant velocity v
Here we need to find the relation in between linear velocity v and in angular velocity
By using the formula we get
Linear velocity $v=\dfrac{\Delta s}{\Delta t}$
Similarly
Angular velocity $\omega =\dfrac{\Delta \theta }{\Delta t}$
Where $\omega $is the angular displacement
And we also know that ,
Linear displacement $\theta =\dfrac{s}{r}$
Here by finding a relation in between $\omega $and v
$\omega =\dfrac{\Delta \theta }{\Delta t}$
By using the formula of linear displacement
$\theta =\dfrac{s}{r}$
$\begin{align} & \Rightarrow \omega =\dfrac{\Delta \theta }{\Delta t} \\ & \Rightarrow \omega =\dfrac{\Delta \left( \dfrac{s}{r} \right)}{\Delta t} \\ & \Rightarrow \omega =\dfrac{\Delta s}{\Delta tr} \\ \end{align}$
In circular motion r is radius ,so it always constant
$\omega =\dfrac{\Delta s}{r\Delta t}$
here $v=\dfrac{\Delta s}{\Delta t}$
$\omega =\dfrac{v}{r}$ such as

Hence, option C is the correct answer.
Note:It is important to note the body is moving in a circular path and the tangent to the point of the circle represents the linear velocity. When the body moves along the direction of tangent and displace from one point to the other in the direction of tangent given displacement and rate change of that displacement give us linear velocity of body moving in circular path. When this velocity is divided by radius of circle given angular velocity of body moving in circular path.
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 arc covered with particles and r is radius of circle in which the body is moving.
Complete step by step solution:
A body moving in a circular motion with constant velocity v
Here we need to find the relation in between linear velocity v and in angular velocity
By using the formula we get
Linear velocity $v=\dfrac{\Delta s}{\Delta t}$
Similarly
Angular velocity $\omega =\dfrac{\Delta \theta }{\Delta t}$
Where $\omega $is the angular displacement
And we also know that ,
Linear displacement $\theta =\dfrac{s}{r}$
Here by finding a relation in between $\omega $and v
$\omega =\dfrac{\Delta \theta }{\Delta t}$
By using the formula of linear displacement
$\theta =\dfrac{s}{r}$
$\begin{align} & \Rightarrow \omega =\dfrac{\Delta \theta }{\Delta t} \\ & \Rightarrow \omega =\dfrac{\Delta \left( \dfrac{s}{r} \right)}{\Delta t} \\ & \Rightarrow \omega =\dfrac{\Delta s}{\Delta tr} \\ \end{align}$
In circular motion r is radius ,so it always constant
$\omega =\dfrac{\Delta s}{r\Delta t}$
here $v=\dfrac{\Delta s}{\Delta t}$
$\omega =\dfrac{v}{r}$ such as

Hence, option C is the correct answer.
Note:It is important to note the body is moving in a circular path and the tangent to the point of the circle represents the linear velocity. When the body moves along the direction of tangent and displace from one point to the other in the direction of tangent given displacement and rate change of that displacement give us linear velocity of body moving in circular path. When this velocity is divided by radius of circle given angular velocity of body moving in circular path.
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