
A body moves with constant angular velocity on a circle. Magnitude of angular acceleration.
A. \[r{\omega ^2}\]
B. Constant
C. Zero
D. None of the above
Answer
233.1k+ views
Hint:The rate of change of angular velocity is called angular acceleration. When a body is moving in a circular path with constant speed than at every point the direction of linear velocity changes.
Formula used:
\[v = \omega r\]
where v is the linear velocity, \[\omega \] is the angular velocity and r is the distance of the point from the axis of rotation.
\[\alpha = \dfrac{{d\omega }}{{dt}}\]
Here \[\alpha \] is the angular acceleration of the particle and \[\dfrac{{d\omega }}{{dt}}\] is the rate of change of angular velocity.
Complete step by step solution:
When a body is moving in a circular path then the position of the body along the circular path changes with time. As we know, the rate of change of the linear position of the body with respect to time is called linear velocity. The rate of change of the angular position with respect to time is called the angular velocity.
It is given that the particle is moving in a circular path with constant angular velocity. Let at any time \[{t_1}\]the angular velocity is \[{\omega _1}\]. Then at any other time \[{t_2}\]the angular velocity is \[{\omega _2}\]. We need to find the angular acceleration of a particle on a circular path.

Image: Particle’s uniform circular motion
Let the angular acceleration of the body is \[\alpha \]. The distance of the body from the axis of rotation, i.e. from the center of the circular path is equal to the radius of the circular path. Using the formula for angular acceleration,
\[\alpha = \dfrac{{{\omega _2} - {\omega _1}}}{{{t_2} - {t_1}}}\]
As the angular velocity is constant \[{\omega _1} = {\omega _2}\]
\[\alpha = \dfrac{{\omega - \omega }}{{{t_2} - {t_1}}}\]
\[\Rightarrow \alpha = \dfrac{0}{{{t_2} - {t_1}}}\]
\[\therefore \alpha = 0\,rad/{s^2}\]
Hence, the angular acceleration of the body is equal to zero.
Therefore, the correct option is C.
Note: We should be careful while using the relation between the linear velocity and the angular velocity as it is for the instantaneous velocity. So, if we find the linear velocity of the accelerated body in a circular path then the obtained linear velocity will be corresponding to the angular velocity at that instant in time.
Formula used:
\[v = \omega r\]
where v is the linear velocity, \[\omega \] is the angular velocity and r is the distance of the point from the axis of rotation.
\[\alpha = \dfrac{{d\omega }}{{dt}}\]
Here \[\alpha \] is the angular acceleration of the particle and \[\dfrac{{d\omega }}{{dt}}\] is the rate of change of angular velocity.
Complete step by step solution:
When a body is moving in a circular path then the position of the body along the circular path changes with time. As we know, the rate of change of the linear position of the body with respect to time is called linear velocity. The rate of change of the angular position with respect to time is called the angular velocity.
It is given that the particle is moving in a circular path with constant angular velocity. Let at any time \[{t_1}\]the angular velocity is \[{\omega _1}\]. Then at any other time \[{t_2}\]the angular velocity is \[{\omega _2}\]. We need to find the angular acceleration of a particle on a circular path.

Image: Particle’s uniform circular motion
Let the angular acceleration of the body is \[\alpha \]. The distance of the body from the axis of rotation, i.e. from the center of the circular path is equal to the radius of the circular path. Using the formula for angular acceleration,
\[\alpha = \dfrac{{{\omega _2} - {\omega _1}}}{{{t_2} - {t_1}}}\]
As the angular velocity is constant \[{\omega _1} = {\omega _2}\]
\[\alpha = \dfrac{{\omega - \omega }}{{{t_2} - {t_1}}}\]
\[\Rightarrow \alpha = \dfrac{0}{{{t_2} - {t_1}}}\]
\[\therefore \alpha = 0\,rad/{s^2}\]
Hence, the angular acceleration of the body is equal to zero.
Therefore, the correct option is C.
Note: We should be careful while using the relation between the linear velocity and the angular velocity as it is for the instantaneous velocity. So, if we find the linear velocity of the accelerated body in a circular path then the obtained linear velocity will be corresponding to the angular velocity at that instant in time.
Recently Updated Pages
Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

