The Uniform Circular Motion Formula describes the motion of an object travelling along a circular path at a constant speed, and it is a foundational concept in Class 11 Physics (NCERT Chapter 4 — Motion in a Plane). This set of formulas covers centripetal acceleration, centripetal force, angular velocity, time period, and frequency. These expressions appear regularly in CBSE board exams as well as JEE Main, JEE Advanced, and NEET. In this article, we cover every key formula, their derivations, a complete formula sheet, three progressive solved examples, common mistakes, and expert exam tips for 2025-26.
Key Uniform Circular Motion Formulas at a Glance
Quick reference for the most important formulas in uniform circular motion.
- Angular velocity: \( \omega = \dfrac{2\pi}{T} = 2\pi f \)
- Linear (tangential) speed: \( v = r\omega \)
- Centripetal acceleration: \( a_c = \dfrac{v^2}{r} = r\omega^2 \)
- Centripetal force: \( F_c = \dfrac{mv^2}{r} = mr\omega^2 \)
- Time period: \( T = \dfrac{2\pi r}{v} = \dfrac{2\pi}{\omega} \)
- Frequency: \( f = \dfrac{1}{T} \)
- Angular displacement: \( \theta = \omega t \)
What is the Uniform Circular Motion Formula?
Uniform circular motion (UCM) occurs when an object moves along a circular path with a constant speed. Although the speed remains constant, the velocity changes continuously because its direction changes at every point. This change in velocity means the object experiences a net acceleration directed towards the centre of the circle. This inward acceleration is called centripetal acceleration.
The Uniform Circular Motion Formula is the collection of mathematical expressions that relate the radius of the circular path, the speed of the object, the time period, the frequency, and the forces involved. These formulas are introduced in NCERT Class 11 Physics, Chapter 4 (Motion in a Plane), and they form the basis for understanding planetary motion, satellite orbits, banking of roads, and the motion of charged particles in magnetic fields.
In CBSE board exams, UCM questions typically carry 2–5 marks. In JEE and NEET, UCM concepts are embedded in problems on circular dynamics, vertical circular motion, and electromagnetism. A thorough understanding of the Uniform Circular Motion Formula is therefore essential for every science student.
Uniform Circular Motion Formula — Expression and Variables
1. Angular Velocity
\[ \omega = \frac{2\pi}{T} = 2\pi f \]
2. Linear Speed
\[ v = r\omega = \frac{2\pi r}{T} \]
3. Centripetal Acceleration
\[ a_c = \frac{v^2}{r} = r\omega^2 \]
4. Centripetal Force
\[ F_c = \frac{mv^2}{r} = mr\omega^2 \]
5. Time Period
\[ T = \frac{2\pi r}{v} = \frac{2\pi}{\omega} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( \omega \) | Angular velocity | rad/s |
| \( v \) | Linear (tangential) speed | m/s |
| \( r \) | Radius of circular path | m |
| \( T \) | Time period | s |
| \( f \) | Frequency | Hz (s−1) |
| \( a_c \) | Centripetal acceleration | m/s² |
| \( F_c \) | Centripetal force | N (kg·m/s²) |
| \( m \) | Mass of the object | kg |
| \( \theta \) | Angular displacement | rad |
Derivation of Centripetal Acceleration
Consider an object moving with constant speed \( v \) along a circle of radius \( r \). At two nearby points P and Q separated by a small time \( \Delta t \), the velocity vectors have the same magnitude but different directions. The change in velocity \( \Delta v \) is directed towards the centre. Using the geometry of similar triangles:
\[ \frac{\Delta v}{v} = \frac{\Delta s}{r} \]
Dividing both sides by \( \Delta t \) and taking the limit as \( \Delta t \to 0 \):
\[ a_c = \frac{v^2}{r} \]
Since \( v = r\omega \), we can substitute to obtain the alternate form \( a_c = r\omega^2 \). Multiplying by mass \( m \) gives the centripetal force \( F_c = mv^2/r \). This derivation is directly referenced in NCERT Class 11, Chapter 4.
Complete Physics Formula Sheet — Circular Motion
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Angular Velocity | \( \omega = 2\pi f = 2\pi/T \) | f = frequency, T = time period | rad/s | Class 11, Ch 4 |
| Linear Speed | \( v = r\omega \) | r = radius, ω = angular velocity | m/s | Class 11, Ch 4 |
| Centripetal Acceleration | \( a_c = v^2/r = r\omega^2 \) | v = speed, r = radius | m/s² | Class 11, Ch 4 |
| Centripetal Force | \( F_c = mv^2/r \) | m = mass, v = speed, r = radius | N | Class 11, Ch 5 |
| Time Period | \( T = 2\pi r/v \) | r = radius, v = speed | s | Class 11, Ch 4 |
| Frequency | \( f = 1/T \) | T = time period | Hz | Class 11, Ch 4 |
| Angular Displacement | \( \theta = \omega t \) | ω = angular velocity, t = time | rad | Class 11, Ch 4 |
| Banking of Roads (ideal) | \( \tan\theta = v^2/(rg) \) | θ = banking angle, g = 9.8 m/s² | dimensionless (angle) | Class 11, Ch 5 |
| Conical Pendulum Period | \( T = 2\pi\sqrt{L\cos\theta/g} \) | L = string length, θ = half-angle | s | Class 11, Ch 5 |
| Minimum Speed at Top (vertical circle) | \( v_{min} = \sqrt{rg} \) | r = radius, g = gravitational acceleration | m/s | Class 11, Ch 5 |
| Orbital Speed (satellite) | \( v_o = \sqrt{GM/r} \) | G = gravitational constant, M = Earth mass | m/s | Class 11, Ch 8 |
Uniform Circular Motion Formula — Solved Examples
Example 1 (Class 9–10 Level): Finding Centripetal Force
Problem: A stone of mass 0.5 kg is tied to a string and whirled in a horizontal circle of radius 1.2 m with a constant speed of 4 m/s. Calculate the centripetal force acting on the stone.
Given: m = 0.5 kg, r = 1.2 m, v = 4 m/s
Step 1: Write the centripetal force formula: \( F_c = \dfrac{mv^2}{r} \)
Step 2: Substitute the values: \( F_c = \dfrac{0.5 \times (4)^2}{1.2} \)
Step 3: Calculate: \( F_c = \dfrac{0.5 \times 16}{1.2} = \dfrac{8}{1.2} \approx 6.67 \) N
Answer
Centripetal Force = 6.67 N (directed towards the centre of the circle).
Example 2 (Class 11–12 Level): Angular Velocity and Time Period
Problem: A particle moves in a circle of radius 0.8 m. It completes 5 revolutions per second. Find (a) the angular velocity, (b) the linear speed, and (c) the centripetal acceleration.
Given: r = 0.8 m, f = 5 Hz
Step 1: Find angular velocity using \( \omega = 2\pi f \):
\( \omega = 2\pi \times 5 = 10\pi \approx 31.42 \) rad/s
Step 2: Find linear speed using \( v = r\omega \):
\( v = 0.8 \times 10\pi = 8\pi \approx 25.13 \) m/s
Step 3: Find centripetal acceleration using \( a_c = r\omega^2 \):
\( a_c = 0.8 \times (10\pi)^2 = 0.8 \times 100\pi^2 = 80\pi^2 \approx 789.6 \) m/s²
Answer
(a) ω = 10π ≈ 31.42 rad/s (b) v = 8π ≈ 25.13 m/s (c) a⊂c⊂ = 80π² ≈ 789.6 m/s²
Example 3 (JEE/NEET Level): Banking of Roads
Problem: A car of mass 1200 kg travels along a banked circular road of radius 200 m. The banking angle is 15°. Find the ideal speed at which the car should travel so that no friction is needed. Take g = 10 m/s².
Given: r = 200 m, θ = 15°, g = 10 m/s²
Step 1: Use the ideal banking formula: \( \tan\theta = \dfrac{v^2}{rg} \)
Step 2: Rearrange for v: \( v = \sqrt{rg\tan\theta} \)
Step 3: Substitute values:
\( v = \sqrt{200 \times 10 \times \tan 15^\circ} \)
\( \tan 15^\circ \approx 0.2679 \)
\( v = \sqrt{200 \times 10 \times 0.2679} = \sqrt{535.8} \approx 23.15 \) m/s
Step 4: Convert to km/h for context: \( 23.15 \times 3.6 \approx 83.3 \) km/h
Answer
The ideal speed for no friction = 23.15 m/s (approximately 83.3 km/h). The mass of the car is not needed for this calculation because it cancels out.
CBSE Exam Tips 2025-26
- Memorise both forms of each formula. CBSE often gives problems where you must choose between \( v^2/r \) and \( r\omega^2 \). Knowing both saves time.
- State the direction of centripetal force. In 3-mark answers, always mention that centripetal force (and acceleration) points towards the centre. Examiners deduct marks for missing this.
- Use consistent units. Convert rpm to rad/s before substituting into formulas. Multiply rpm by \( 2\pi/60 \) to get rad/s.
- Distinguish centripetal from centrifugal. Centripetal force is a real force directed inward. Centrifugal force is a pseudo-force in a rotating frame. CBSE questions frequently test this distinction in 2025-26 exams.
- We recommend drawing a free-body diagram for every circular motion problem. It helps identify which real force (tension, normal reaction, gravity, friction) provides the centripetal force in each scenario.
- For banking problems, remember the ideal banking angle formula \( \tan\theta = v^2/(rg) \) does not involve mass. If mass is given, it is extra data meant to confuse you.
Common Mistakes to Avoid
- Confusing speed with velocity. In UCM, speed is constant but velocity is not. Never say “velocity is constant” in circular motion — it changes direction at every instant.
- Forgetting to convert rpm to rad/s. A very common error is substituting the rpm value directly into \( \omega \). Always convert: \( \omega = 2\pi n/60 \), where n is in rpm.
- Using diameter instead of radius. The formula uses the radius r, not the diameter d. If the problem gives diameter, halve it before substituting.
- Treating centripetal force as a separate force. Centripetal force is not a new type of force. It is the net resultant of existing forces (tension, gravity, friction, normal reaction) directed towards the centre.
- Incorrect direction of centripetal acceleration. Students sometimes draw the acceleration tangentially. It always points radially inward, perpendicular to the velocity vector in UCM.
JEE/NEET Application of the Uniform Circular Motion Formula
In our experience, JEE aspirants encounter the Uniform Circular Motion Formula in at least 2–3 questions per paper, often disguised within larger problems. Here are the three most important application patterns to master.
Pattern 1: Vertical Circular Motion
JEE problems frequently combine UCM with energy conservation. At the top of a vertical circle, the minimum speed condition requires that gravity alone provides centripetal force:
\[ mg = \frac{mv_{min}^2}{r} \implies v_{min} = \sqrt{rg} \]
At the bottom, the normal reaction exceeds gravity by the centripetal term: \( N = mg + mv^2/r \). These two conditions appear in nearly every JEE vertical circle question.
Pattern 2: Charged Particle in a Magnetic Field (NEET/JEE)
When a charged particle of charge q and mass m moves perpendicular to a magnetic field B, the magnetic force provides centripetal force:
\[ qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB} \]
This is a direct application of the Uniform Circular Motion Formula. NEET 2024 and JEE Main 2024 both featured questions on this concept. The radius of the circular path is called the cyclotron radius or Larmor radius.
Pattern 3: Conical Pendulum and Satellite Orbits
For a conical pendulum, the horizontal component of tension provides centripetal force while the vertical component balances gravity. For satellites, gravitational force provides centripetal force:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \implies v_o = \sqrt{\frac{GM}{r}} \]
Our experts suggest practising all three patterns with timed conditions. Recognising which force acts as the centripetal force is the key skill tested in JEE Advanced.
For a deeper understanding of energy in wave motion, visit our article on the Wave Power Formula. You can also explore the Heat Transfer Formula and the Photon Energy Formula for related Class 11–12 Physics concepts. The official NCERT syllabus and textbooks are available at ncert.nic.in.
FAQs on Uniform Circular Motion Formula
Explore our complete collection of physics formulas at the Physics Formulas hub. For related topics, read our detailed guides on the Shear Modulus Formula and the Wave Power Formula. These resources together cover the full scope of Class 11 Mechanics for CBSE and competitive exams.