
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
163.2k+ views
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.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Charging and Discharging of Capacitor

Other Pages
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
