The Centripetal Force Formula, expressed as F = mv²/r, gives the inward force required to keep an object moving in a circular path at constant speed. This formula is a core concept in Class 11 Physics (NCERT Chapter 5 — Laws of Motion and Chapter 6 — Work, Energy and Power context, with circular motion detailed in Chapter 4). It is equally vital for JEE Main, JEE Advanced, and NEET, where circular motion problems appear regularly. This article covers the formula expression, variable definitions, derivation, a complete physics formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Centripetal Force Formulas at a Glance
Quick reference for the most important centripetal force and circular motion formulas.
- Centripetal Force: \( F_c = \dfrac{mv^2}{r} \)
- In terms of angular velocity: \( F_c = m\omega^2 r \)
- In terms of time period: \( F_c = \dfrac{4\pi^2 m r}{T^2} \)
- Centripetal acceleration: \( a_c = \dfrac{v^2}{r} = \omega^2 r \)
- Relation between linear and angular velocity: \( v = \omega r \)
- Time period of circular motion: \( T = \dfrac{2\pi r}{v} = \dfrac{2\pi}{\omega} \)
- Frequency: \( f = \dfrac{1}{T} = \dfrac{v}{2\pi r} \)
What is Centripetal Force Formula?
The Centripetal Force Formula defines the net inward force that acts on any object travelling along a circular path. The word “centripetal” comes from Latin, meaning “centre-seeking.” Whenever an object moves in a circle, its velocity direction changes continuously. This change in direction means the object is accelerating, even if its speed stays constant. Newton’s second law tells us that acceleration requires a net force. That net force always points toward the centre of the circular path. We call it the centripetal force.
In NCERT Class 11 Physics, circular motion is introduced in Chapter 4 (Motion in a Plane). The centripetal force formula is formally stated as part of uniform circular motion. It applies to a wide range of real-world situations. These include a car turning on a curved road, a satellite orbiting Earth, a stone tied to a string swung in a circle, and electrons orbiting a nucleus in the Bohr model. Understanding this formula is essential for scoring well in CBSE board exams and for cracking JEE and NEET.
The Centripetal Force Formula is \( F_c = \dfrac{mv^2}{r} \). Here, \( m \) is mass, \( v \) is linear speed, and \( r \) is the radius of the circular path. It can also be written as \( F_c = m\omega^2 r \), where \( \omega \) is angular velocity.
Centripetal Force Formula — Expression and Variables
The standard expression for the centripetal force is:
\[ F_c = \frac{mv^2}{r} \]
An equivalent form using angular velocity \( \omega \) is:
\[ F_c = m\omega^2 r \]
A third form using the time period \( T \) is:
\[ F_c = \frac{4\pi^2 m r}{T^2} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( F_c \) | Centripetal Force | Newton (N) |
| \( m \) | Mass of the object | Kilogram (kg) |
| \( v \) | Linear (tangential) speed | Metre per second (m/s) |
| \( r \) | Radius of circular path | Metre (m) |
| \( \omega \) | Angular velocity | Radian per second (rad/s) |
| \( T \) | Time period of revolution | Second (s) |
| \( a_c \) | Centripetal acceleration | Metre per second squared (m/s²) |
Derivation of the Centripetal Force Formula
Consider an object of mass \( m \) moving with constant speed \( v \) along a circle of radius \( r \). At two nearby points on the circle, the velocity vectors have the same magnitude but different directions. The change in velocity \( \Delta v \) over a small time \( \Delta t \) points toward the centre.
Step 1: For a small angle \( \Delta\theta \), the magnitude of the change in velocity is \( \Delta v = v\,\Delta\theta \).
Step 2: The arc length covered is \( \Delta s = r\,\Delta\theta \), so \( \Delta\theta = \dfrac{\Delta s}{r} = \dfrac{v\,\Delta t}{r} \).
Step 3: Centripetal acceleration is \( a_c = \dfrac{\Delta v}{\Delta t} = \dfrac{v\,\Delta\theta}{\Delta t} = \dfrac{v^2}{r} \).
Step 4: Applying Newton’s second law: \( F_c = m\,a_c = \dfrac{mv^2}{r} \).
Since \( v = \omega r \), substituting gives \( F_c = m\omega^2 r \). This completes the derivation.
Complete Circular Motion Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Centripetal Force | \( F_c = \dfrac{mv^2}{r} \) | m = mass, v = speed, r = radius | N | Class 11, Ch 4 |
| Centripetal Force (angular form) | \( F_c = m\omega^2 r \) | m = mass, ω = angular velocity, r = radius | N | Class 11, Ch 4 |
| Centripetal Force (period form) | \( F_c = \dfrac{4\pi^2 m r}{T^2} \) | m = mass, r = radius, T = time period | N | Class 11, Ch 4 |
| Centripetal Acceleration | \( a_c = \dfrac{v^2}{r} \) | v = speed, r = radius | m/s² | Class 11, Ch 4 |
| Angular Velocity | \( \omega = \dfrac{v}{r} = \dfrac{2\pi}{T} \) | v = speed, r = radius, T = period | rad/s | Class 11, Ch 4 |
| Time Period | \( T = \dfrac{2\pi r}{v} \) | r = radius, v = speed | s | Class 11, Ch 4 |
| Frequency of Revolution | \( f = \dfrac{1}{T} \) | T = time period | Hz | Class 11, Ch 4 |
| Banking of Roads (angle) | \( \tan\theta = \dfrac{v^2}{rg} \) | v = speed, r = radius, g = gravity | dimensionless | Class 11, Ch 5 |
| Maximum speed on banked road | \( v_{max} = \sqrt{rg\tan\theta} \) | r = radius, g = gravity, θ = bank angle | m/s | Class 11, Ch 5 |
| Vertical Circle — Min speed at top | \( v_{top} = \sqrt{gr} \) | g = gravity, r = radius | m/s | Class 11, Ch 6 |
| Vertical Circle — Min speed at bottom | \( v_{bottom} = \sqrt{5gr} \) | g = gravity, r = radius | m/s | Class 11, Ch 6 |
Centripetal Force Formula — Solved Examples
Example 1 (Class 9-10 / Basic Level)
Problem: A stone of mass 0.5 kg is tied to a string and whirled in a horizontal circle of radius 2 m. The stone moves with a constant speed of 4 m/s. Find the centripetal force acting on the stone.
Given:
- Mass, \( m = 0.5 \) kg
- Speed, \( v = 4 \) m/s
- Radius, \( r = 2 \) m
Step 1: Write the centripetal force formula: \( F_c = \dfrac{mv^2}{r} \)
Step 2: Substitute the given values: \( F_c = \dfrac{0.5 \times (4)^2}{2} \)
Step 3: Calculate the numerator: \( 0.5 \times 16 = 8 \)
Step 4: Divide by radius: \( F_c = \dfrac{8}{2} = 4 \) N
Answer
The centripetal force acting on the stone is 4 N, directed toward the centre of the circular path.
Example 2 (Class 11-12 Level)
Problem: A car of mass 1200 kg travels along a circular road of radius 50 m. The car completes one full revolution in 31.4 seconds. Calculate (a) the angular velocity, (b) the linear speed, and (c) the centripetal force required.
Given:
- Mass, \( m = 1200 \) kg
- Radius, \( r = 50 \) m
- Time period, \( T = 31.4 \) s
Step 1: Find angular velocity using \( \omega = \dfrac{2\pi}{T} \):
\( \omega = \dfrac{2 \times 3.14}{31.4} = \dfrac{6.28}{31.4} = 0.2 \) rad/s
Step 2: Find linear speed using \( v = \omega r \):
\( v = 0.2 \times 50 = 10 \) m/s
Step 3: Apply the centripetal force formula \( F_c = \dfrac{mv^2}{r} \):
\( F_c = \dfrac{1200 \times (10)^2}{50} = \dfrac{1200 \times 100}{50} = \dfrac{120000}{50} = 2400 \) N
Verification using angular form: \( F_c = m\omega^2 r = 1200 \times (0.2)^2 \times 50 = 1200 \times 0.04 \times 50 = 2400 \) N ✓
Answer
(a) Angular velocity = 0.2 rad/s | (b) Linear speed = 10 m/s | (c) Centripetal force = 2400 N
Example 3 (JEE/NEET Level)
Problem: A particle of mass 200 g moves in a vertical circle of radius 1.5 m. At the topmost point of the circle, the speed of the particle is 6 m/s. Find (a) the centripetal acceleration at the top, (b) the net centripetal force at the top, and (c) the tension in the string at the top. (Take g = 10 m/s²)
Given:
- Mass, \( m = 200 \text{ g} = 0.2 \) kg
- Radius, \( r = 1.5 \) m
- Speed at top, \( v = 6 \) m/s
- \( g = 10 \) m/s²
Step 1: Find centripetal acceleration at the top:
\( a_c = \dfrac{v^2}{r} = \dfrac{(6)^2}{1.5} = \dfrac{36}{1.5} = 24 \) m/s²
Step 2: Find net centripetal force required:
\( F_c = m\,a_c = 0.2 \times 24 = 4.8 \) N
Step 3: At the topmost point, both tension \( T \) and weight \( mg \) act downward (toward the centre). So the net inward force equals \( T + mg \):
\( T + mg = F_c \)
\( T = F_c – mg = 4.8 – (0.2 \times 10) = 4.8 – 2.0 = 2.8 \) N
Step 4: Since \( T = 2.8 \) N > 0, the string remains taut at the top. The particle successfully completes the vertical circle.
Answer
(a) Centripetal acceleration = 24 m/s² | (b) Net centripetal force = 4.8 N | (c) Tension in string = 2.8 N
CBSE Exam Tips 2025-26
- State the direction explicitly. In CBSE board answers, always mention that centripetal force acts toward the centre. Examiners award a dedicated mark for direction in 3-mark problems.
- Convert units before substituting. We recommend converting grams to kilograms and centimetres to metres before plugging values into \( F_c = mv^2/r \). Unit errors are the most common source of lost marks.
- Know all three formula forms. CBSE 2025-26 papers may give time period \( T \) instead of speed \( v \). Keep \( F_c = 4\pi^2 mr/T^2 \) memorised for such cases.
- Distinguish centripetal from centrifugal force. Centripetal force is real and acts inward. Centrifugal force is a pseudo force in a rotating frame. CBSE frequently asks students to explain this distinction.
- Draw free body diagrams. For banking-of-roads and vertical circle questions, always draw the free body diagram first. It helps identify which forces contribute to the centripetal force equation.
- Revise Chapter 4 numerical patterns. Our experts suggest practising at least 10 NCERT exemplar problems on circular motion before the board exam. Questions on satellites and banking appear almost every year.
Common Mistakes to Avoid
| Mistake | Why It Is Wrong | Correct Approach |
|---|---|---|
| Treating centripetal force as a separate, additional force | Centripetal force is not a new type of force. It is the net resultant of existing forces (tension, gravity, normal force, friction) directed inward. | Identify which physical force(s) provide the centripetal force in each situation. |
| Squaring only the number, not the unit of velocity | Writing \( v^2 = 4 \text{ m/s}^2 \) when \( v = 4 \text{ m/s} \) gives wrong dimensional analysis. | Write \( v^2 = 16 \text{ m}^2/\text{s}^2 \) to keep units consistent. |
| Confusing radius with diameter | Using the diameter of the circular path instead of the radius doubles the denominator and halves the answer. | Always use \( r = \text{diameter}/2 \) when the diameter is given. |
| Adding centrifugal force in an inertial frame | Centrifugal force only appears in a rotating (non-inertial) reference frame. Adding it in ground-frame calculations gives wrong results. | In ground-frame problems, use only real forces and set their net inward component equal to \( mv^2/r \). |
| Forgetting that centripetal force does no work | Since centripetal force is always perpendicular to velocity, it does zero work. Many students incorrectly use it in energy equations. | Use energy conservation separately. Do not include centripetal force in work-energy theorem calculations. |
JEE/NEET Application of Centripetal Force Formula
In our experience, JEE aspirants encounter centripetal force in at least two to three questions per paper. NEET typically includes one circular motion question, often linked to satellite motion or the Bohr model. Below are the three most important application patterns.
Pattern 1 — Banking of Roads and Conical Pendulum
JEE frequently combines the centripetal force formula with trigonometry. For a banked road with angle \( \theta \) and no friction, the normal force component provides the centripetal force:
\[ N\sin\theta = \frac{mv^2}{r} \quad \text{and} \quad N\cos\theta = mg \]
Dividing these two equations gives \( \tan\theta = v^2/(rg) \). JEE questions often ask for the optimal banking angle or the maximum safe speed. Mastering this derivation is essential for JEE Mains scoring.
Pattern 2 — Vertical Circle and Energy Conservation
A classic JEE/NEET problem combines centripetal force with energy conservation. At the top of a vertical circle of radius \( r \), the minimum condition for maintaining contact is \( T = 0 \), which gives \( mg = mv^2/r \), so \( v_{min} = \sqrt{gr} \). Using energy conservation from the bottom:
\[ \frac{1}{2}mv_{bottom}^2 = \frac{1}{2}mv_{top}^2 + mg(2r) \]
Substituting \( v_{top} = \sqrt{gr} \) gives \( v_{bottom,min} = \sqrt{5gr} \). This result appears directly in NEET and JEE Main multiple-choice questions.
Pattern 3 — Satellite Orbital Motion
For NEET, the centripetal force formula connects directly to gravitational force for satellites. Setting gravitational force equal to centripetal force:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac{GM}{r}} \]
This orbital velocity formula is a direct NEET Biology-Physics crossover topic. It also leads to Kepler’s third law. In our experience, students who understand this derivation can answer satellite questions in under 60 seconds during the exam.
FAQs on Centripetal Force Formula
Explore More Physics Formulas
We hope this comprehensive guide on the Centripetal Force Formula has helped you understand the concept thoroughly. To strengthen your Physics preparation further, explore these related formula articles on ncertbooks.net:
- Angular Velocity Formula — learn how \( \omega \) connects to centripetal force and circular motion period.
- Normal Force Formula — understand how normal force acts as the centripetal force in banking and loop problems.
- Instantaneous Speed Formula — revise how linear speed \( v \) is defined and measured at any point on a curved path.
For the complete list of Class 11 and Class 12 Physics formulas, visit our Physics Formulas Hub. You can also refer to the official NCERT textbook available at ncert.nic.in for the standard derivations and problems on circular motion.