Centrifugal force is an inertial or pseudo-force experienced by bodies in a non-inertial rotating reference frame, directed radially outward from the axis of rotation. It arises purely as a consequence of the acceleration of the coordinate system rather than any physical interaction between masses.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of centrifugal force, distinction between inertial and rotating frames, magnitude mω<sup>2</sup>r, simple numericals on revolving bodies, and effect of Earth's rotation on apparent weight.
PGT extension
Detailed transformation of coordinates from an inertial to a rotating frame, derivation of fictitious forces including centrifugal and Coriolis forces, and latitudinal variation of acceleration due to gravity.
Prerequisites
Newton's Laws of Motion in inertial frames
Kinematics of circular motion and angular velocity vector ω
Centripetal acceleration and non-inertial reference frames
Nature of Centrifugal Force
It is a fictitious or pseudo-force that appears only when observations are made from a non-inertial rotating frame of reference.
It has no physical origin in terms of fundamental interactions (like gravitational or electromagnetic forces); no reaction pair exists in an inertial frame.
Its magnitude is given by Fcf = mω2r, where r is the perpendicular distance from the axis of rotation.
Direction and Vector Form
The centrifugal force is directed radially outward, away from the instantaneous center of rotation or axis of rotation.
In vector notation, it is expressed as Fcf = -m(ω × (ω × r)), which simplifies to mω2r⊥ directed outward.
It acts along the line extending from the rotation axis through the position of the particle.
Centrifugal vs Centripetal Force
Centripetal force is a real inward force (tension, gravity, normal force) required to keep a body moving in a curved path in an inertial frame.
Centrifugal force is directed outward and is utilized in rotating frames to apply D'Alembert's principle, treating equilibrium problems dynamically.
In a rotating frame, the real centripetal force is balanced by the centrifugal force, giving a net zero acceleration in that frame.
Effect of Earth's Rotation on Gravity
Earth is a non-inertial rotating frame with angular speed ω ≈ 7.29 × 10-5 rad/s.
Objects on Earth's surface experience a centrifugal force outward perpendicular to the polar axis, reducing the effective gravitational force.
The effective acceleration due to gravity g' varies with latitude λ, being minimum at the equator and maximum at the poles.
Derivations
Derivation of Centrifugal Force from Rotating Reference Frame Transformation
Conditions: A coordinate system S' rotating with uniform angular velocity ω relative to an inertial frame S.
Express the position vector in terms of unit vectors of the rotating frame: $r = x'\hat{i}' + y'\hat{j}' + z'\hat{k}'$.
Take the time derivative in the inertial frame noting that unit vectors change with time: (dî'/dt) = ω × î'.
Apply the operator relation (d/dt)inertial = (d/dt)rotating + ω × to velocity and then to acceleration.
Equate m ainertial = Freal and isolate the apparent acceleration in the rotating frame: arot = (Freal/m) - 2(ω × vrot) - ω × (ω × r).
Identify the term -m(ω × (ω × r)) as the centrifugal force.
Result: Fcf = -m(ω × (ω × r))
Variation of Effective Acceleration Due to Gravity with Latitude
Conditions: Earth modeled as a sphere of radius R rotating with angular velocity ω at latitude λ.
Consider a body of mass m at latitude λ. The distance from the axis of rotation is $r = R \cos\lambda$.
The centrifugal force acting on the body is outward along the radius of the circular path, with magnitude Fc = mω2(R cosλ).
The true gravitational force mg acts toward the center of the Earth.
Resolve the centrifugal force into components: mgccosλ radially inward and mgcsinλ tangentially.
Write the net effective weight mg' vector sum: g' = g - ω2R cos2λ (using scalar approximation for small corrections).
Result: g' = g - Rω2cos2λ
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Magnitude of Centrifugal Force
$$F_{cf} = m \omega^2 r$$
m = mass, $\omega = angular velocity$, r = distance from rotation axis Newton (N)
Uniform circular motion or rotating frame
Vector Expression for Centrifugal Force
$$F_{cf} = -m \omega \times (\omega \times r)$$
m = mass, $\omega = angular velocity vector$, r = position vector Newton (N)
General rotation axis
Effective Acceleration due to Gravity at Latitude λ
$$g' = g - R \omega^2 \cos^2(\lambda)$$
g = true gravity, R = Earth radius, $\omega = Earth angular speed$, $\lambda = latitude$ m/s2
Spherical Earth approximation
Centrifugal Acceleration
$$a_{cf} = \omega^2 r = \frac{v^2}{r}$$
v = tangential velocity, r = radius m/s2
Rigid body rotation
Equatorial Reduction in Gravity
$$\Delta g = R \omega^2$$
Δ$g = maximum reduction at equator (\lambda = 0)$, R = radius, $\omega = angular speed$ m/s2
At the equator
Centrifugal Potential Energy
$$V_{cf} = -\frac{1}{2} m \omega^2 r^2$$
Vcf = effective potential due to centrifugal force Joule (J)
Conservative pseudo-force field in rotating frame
Condition for Object to Float at Equator
$$\omega = \sqrt{\frac{g}{R}}$$
$\omega = critical angular velocity$, g = acceleration due to gravity, R = radius rad/s
Apparent weight becomes zero
Centrifugal Force in Cylindrical Coordinates
$$F_{cf} = m \omega^2 r \hat{r}$$
r = radial distance, &hat;r = unit radial vector Newton (N)
Cylindrical coordinate system aligned with rotation axis
Graphs and experiments
Variation of Effective Gravity with Latitude
Draw: A plot of effective g' (y-axis) versus Latitude λ (x-axis) ranging from 0° to 90°.
Interpret: Shows minimum value g - Rω2 at the equator ($\lambda = 0°)$ and maximum value equal to true g at the poles ($\lambda = 90°)$.
Exam use: Frequently asked conceptual questions regarding where a body weighs the least.
Rotating Bucket of Water
Draw: A parabolic free surface of liquid inside a cylindrical container rotating about a vertical axis.
Interpret: The parabolic profile $z = (\omega^{2}r^{2})/(2g) results from the vector balance of gravitational$ and centrifugal forces acting on the fluid elements.
Exam use: Tested in PGT examinations for fluid statics in rotating frames.
Centrifuge Working Principle Diagram
Draw: A schematic of test tubes in a centrifuge rotor showing denser particles moving outward.
Interpret: Demonstrates how enhanced centrifugal force (much greater than standard gravity) accelerates phase separation based on density differences.
Exam use: Application-based multiple-choice questions in TGT/PGT.
Solved numericals
TGT core
A mass of 2 kg is tied to a string of length 0.5 m and rotated in a horizontal circle at 120 revolutions per minute. Calculate the centrifugal force acting on the mass in the rotating frame.
Calculate the angular speed with which the Earth must rotate on its axis so that a person at the equator weighs three-fifths of their normal weight. Take $g = 10 m/s^{2}$ and Earth's radius R = 6400 km.
At the equator, effective weight is given by mg' = mg - mω2R.
Given condition: mg' = (3/5)mg, which implies mω2$R = mg - (3/5)mg = (2/5)mg$.
Cancel mass m and solve for ω: ω2$R = (2/5)g \Rightarrow \omega = \sqrt(2g / (5R))$.
A smooth horizontal disc rotates with a constant angular velocity ω about a vertical axis passing through its center. A small particle of mass m is placed on the disc at a distance r from the axis. What is the time taken by the particle to slide off the edge if the radius of the disc is R and the coefficient of static friction is μ? Assume initial velocity is zero relative to the disc.
In the rotating frame of the disc, the forces acting on the particle are centrifugal force (mω2r) radially outward and maximum static friction (μmg) radially inward.
The particle begins to slide when centrifugal force exceeds maximum static friction: mω2r > μmg, or at any position r, the radial equation of motion in the rotating frame is m(d2r/dt2) = mω2r - μmg (assuming sliding has commenced with kinetic friction μk, or pure outward acceleration if friction is overcome).
For threshold of sliding starting from $r = r_{0}$ where μm$g = m\omega^{2}r_{0}$, r0 = μg / ω2.
If placed at r = 0, friction holds it until mω2$r = \mumg$. For purely outward radial motion once sliding starts with negligible friction: d2r/dt2 - ω2r = 0.
Solving the differential equation r(t) = A eωt + B e-\omega t with initial conditions r(0) = r0 and dr/dt(0) = 0 yields the transit time to reach R.
Centrifugal force applies strictly within non-inertial rotating frames of reference.
The angular velocity ω is assumed uniform unless time-varying pseudo-forces (Euler force) are included.
In geophysical applications, Earth is assumed to be a rigid, spherically symmetric rotating body.
Common misconceptions
Common error: Centrifugal force is the reaction force to centripetal force according to Newton's Third Law.
Correct: Centripetal and centrifugal forces do not form a Newton's third law action-reaction pair because they act on the same body and belong to different reference frames.
Action-reaction pairs must act on two different bodies and be of the same physical nature.
Common error: Centrifugal force pulls objects outward in an inertial reference frame.
Correct: Centrifugal force does not exist in an inertial frame of reference; objects simply tend to move in a straight line due to inertia (Newton's first law).
Pseudo-forces only emerge when mathematical transformations are made to a non-inertial frame.
Quick revision
Centrifugal force is a fictitious outward radial force equal to mω2r present in rotating frames.
Magnitude of Centrifugal Force: $F_{cf} = m \omega^2 r$
Centrifugal force is a fictitious or pseudo-force experienced only in a non-inertial rotating frame of reference.
It has no physical origin or reaction pair in an inertial frame; it does not satisfy Newton's third law with centripetal force.
The magnitude of centrifugal force is given by $F_{cf} = m\omega^2 r$, directed radially outward from the axis of rotation.
Vector expression for centrifugal force is $F_{cf} = -m(\omega \times (\omega \times r)) = m\omega^2 r_{\perp}$.
Earth's rotation reduces the effective acceleration due to gravity, with the maximum reduction occurring at the equator ($\Delta g = R\omega^2$).
The effective acceleration due to gravity at latitude $\lambda$ is given by $g' = g - R\omega^2 \cos^2(\lambda)$.
At a critical angular velocity $\omega = \sqrt{g/R}$, the apparent weight of an object at the equator becomes zero (objects float).
Centrifugal potential energy is expressed as $V_{cf} = -\frac{1}{2}m\omega^2 r^2$ in a rotating frame.
Glossary
Centrifugal ForceA pseudo-force directed radially outward in a rotating (non-inertial) frame of reference with magnitude $m\omega^2 r$.
Centripetal ForceA real inward force directed toward the center of curvature required to keep a body moving in a circular path in an inertial frame.
Pseudo-ForceAn apparent force introduced to make Newton's laws applicable in non-inertial (accelerated or rotating) reference frames.
Angular Velocity ($\omega$)The rate of change of angular displacement, measured in radians per second (rad/s).
Effective Gravity ($g'$)The net acceleration experienced by an object on Earth's surface, accounting for both true gravitation and the centrifugal acceleration.
Latitude ($\lambda$)The angular distance of a point on Earth's surface north or south of the equator.
Centripetal Force vs Centrifugal Force
Parameter
Centripetal Force
Centrifugal Force
Nature
Real physical force (tension, gravity, normal)
Fictitious or pseudo-force
Reference Frame
Inertial frame of reference
Non-inertial rotating frame of reference
Direction
Radially inward toward the center
Radially outward away from the rotation axis
Origin
Fundamental interactions (gravitational, electromagnetic)
Kinematics of a rotating coordinate system
Newton's Third Law
Has a reaction pair acting on another body
No reaction pair exists in an inertial frame
Practice Quiz
Choose an option to reveal the explanation.
1. What is the nature of the centrifugal force in a rotating frame of reference?
Explanation: Centrifugal force is a fictitious or pseudo-force that appears only when observations are made from a non-inertial rotating frame of reference.
2. What is the magnitude of the centrifugal force acting on a body of mass $m$ at a perpendicular distance $r$ from the axis of rotation with angular velocity $\omega$?
Explanation: The magnitude of the centrifugal force is given by $F_{cf} = m\omega^2 r$, where $r$ is the perpendicular distance from the axis of rotation.
3. What is the direction of the centrifugal force experienced by an observer in a rotating frame?
Explanation: The centrifugal force is directed radially outward, away from the instantaneous center or axis of rotation.
4. Why do centripetal and centrifugal forces not form a Newton's third law action-reaction pair?
Explanation: Action-reaction pairs must act on two different bodies and be of the same physical nature. Centripetal and centrifugal forces act on the same body and across different frames.
5. What is the vector expression for centrifugal force in terms of angular velocity vector $\omega$ and position vector $r$?
Explanation: In vector notation, centrifugal force is expressed as $F_{cf} = -m(\omega \times (\omega \times r))$, which simplifies to $m\omega^2 r_{\perp}$ directed outward.
6. How does the effective acceleration due to gravity $g'$ vary with latitude $\lambda$ due to Earth's rotation?
Explanation: The effective acceleration due to gravity $g'$ varies with latitude $\lambda$ as $g' = g - R\omega^2 \cos^2(\lambda)$, making it minimum at the equator and maximum at the poles.
7. What is the maximum reduction in the acceleration due to gravity at the Earth's equator?
Explanation: At the equator ($\lambda = 0$), $\cos(0) = 1$, so the reduction is $\Delta g = R\omega^2$.
8. What is the critical angular velocity $\omega$ of Earth for which objects at the equator would appear weightless (float)?
Explanation: Setting apparent weight $g' = g - R\omega^2 = 0$ gives the critical angular velocity $\omega = \sqrt{\frac{g}{R}}$.
9. What is the centrifugal potential energy $V_{cf}$ associated with the centrifugal force field?
Explanation: The conservative pseudo-force field in a rotating frame has a centrifugal potential energy given by $V_{cf} = -\frac{1}{2}m\omega^2 r^2$.
10. What principle is utilized when applying the centrifugal force to solve dynamics problems in a rotating frame as static equilibrium problems?
Explanation: Centrifugal force is utilized in rotating frames to apply D'Alembert's principle, treating equilibrium problems dynamically.
11. What is the approximate angular speed $\omega$ of Earth's rotation?
Explanation: Earth's angular speed is approximately $\omega \approx 7.29 \times 10^{-5} \text{ rad/s}$.
12. In cylindrical coordinates aligned with the rotation axis, what is the expression for centrifugal force?
Explanation: In cylindrical coordinates, the centrifugal force is expressed as $F_{cf} = m\omega^2 r \hat{r}$ directed radially outward.
13. Why does an object in an inertial frame tend to move in a straight line when unconstrained, rather than experiencing an outward pull?
Explanation: Centrifugal force does not exist in an inertial frame; objects simply tend to move in a straight line due to inertia (Newton's first law).
14. What is the centrifugal acceleration $a_{cf}$ for a rigid body rotating with angular velocity $\omega$ at distance $r$ and tangential velocity $v$?
Explanation: Centrifugal acceleration is given by $a_{cf} = \omega^2 r = \frac{v^2}{r}$.
15. What condition must hold for the centrifugal force to be correctly evaluated using $F_{cf} = m\omega^2 r$?
Explanation: The formula $F_{cf} = m\omega^2 r$ applies under conditions of uniform circular motion or observations from a rotating frame.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. If the Earth were to rotate so fast that the apparent weight of a body at the equator becomes zero, what would be the length of the day?
Explanation: For zero apparent weight at the equator, $\omega = \sqrt{\frac{g}{R}}$. Taking $g \approx 9.8 \text{ m/s}^2$ and $R \approx 6400 \text{ km}$, $\omega \approx 1.24 \times 10^{-3} \text{ rad/s}$. The time period $T = \frac{2\pi}{\omega} \approx 5060 \text{ s} \approx 1.41 \text{ hours}$ (or $\approx 84 \text{ minutes}$).
2. A body of mass $m$ is placed on a rotating platform at a distance $r$ from the axis of rotation. If the coefficient of static friction is $\mu$, what is the maximum angular velocity $\omega$ for the body not to slip?
Explanation: In the rotating frame, the centrifugal force $m\omega^2 r$ is balanced by the maximum static frictional force $\mu m g$. Thus, $m\omega^2 r \le \mu mg \implies \omega \le \sqrt{\frac{\mu g}{r}}.$
3. At what latitude $\lambda$ will the effective acceleration due to gravity be equal to half of the true acceleration due to gravity, ignoring Earth's spherical bulging? (Assuming $R\omega^2 \approx 0.034 \text{ m/s}^2$)
Explanation: Since $R\omega^2 \approx 0.034 \text{ m/s}^2$ and $g \approx 9.8 \text{ m/s}^2$, the maximum reduction $R\omega^2$ is very small compared to $g$. Thus, $g'$ can never drop to $\frac{g}{2}$ solely due to centrifugal force.
4. Which of the following statements correctly distinguishes between centrifugal and centripetal forces regarding work done on a particle moving in a uniform circular path?
Explanation: In uniform circular motion, centripetal force is perpendicular to displacement in the inertial frame (zero work). In the rotating frame, the particle is at rest, so displacement is zero, and thus net work done by all forces (including centrifugal) in that frame is also zero.
5. If a particle of mass $2 \text{ kg}$ is rotated in a horizontal plane at a radius of $0.5 \text{ m}$ with an angular speed of $10 \text{ rad/s}$, what is the centrifugal force experienced in the rotating frame?
Explanation: Using $F_{cf} = m\omega^2 r$, we have $F_{cf} = 2 \times (10)^2 \times 0.5 = 2 \times 100 \times 0.5 = 100 \text{ N}$.
6. What is the primary reason why centrifugal force is termed a 'pseudo-force'?
Explanation: Pseudo-forces do not arise from physical interactions (like gravity or electromagnetism); they emerge mathematically when transforming equations of motion to an accelerated or rotating coordinate system.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. In which frame of reference does the centrifugal force explicitly appear in the equation of motion?
Explanation: Centrifugal force is a pseudo-force that appears exclusively when observations are made from a non-inertial rotating frame of reference.
2. What is the ratio of centrifugal forces acting on two masses $m_1$ and $m_2$ rotating at the same radius with the same angular velocity?
Explanation: Since $F_{cf} = m\omega^2 r$, with $\omega$ and $r$ constant, $F_{cf}$ is directly proportional to mass $m$. Thus, the ratio is $\frac{m_1}{m_2}$.
3. How does the apparent acceleration due to gravity change as one moves from the poles to the equator?
Explanation: Effective gravity $g' = g - R\omega^2 \cos^2(\lambda)$ is maximum at the poles ($\lambda = 90^\circ$) and minimum at the equator ($\lambda = 0^\circ$). Moving from poles to equator decreases $g'$.
4. If the angular speed of rotation of a centrifuge is doubled, what happens to the centrifugal force on a sample inside it?
Explanation: Centrifugal force is proportional to $\omega^2$ ($F_{cf} = m\omega^2 r$). Doubling $\omega$ increases the force by a factor of $2^2 = 4$.
5. What is the work done by the centrifugal force on a body displaced radially outward by $\Delta r$ in a rotating frame?
Explanation: Since $V_{cf} = -\frac{1}{2}m\omega^2 r^2$, the work done by the conservative centrifugal pseudo-force is negative of change in potential energy: $W = -\Delta V_{cf} = \int_{r_1}^{r_2} m\omega^2 r dr = \frac{1}{2}m\omega^2 (r_2^2 - r_1^2)$.
6. Which of the following phenomena is explained by the existence of centrifugal force in Earth's rotating frame?
Explanation: The outward centrifugal force due to Earth's rotation causes the equatorial regions to bulge outward, resulting in an oblate spheroid shape.
7. Which force balances the centrifugal force in a rotating frame when an object is in static equilibrium on a rotating turntable?
Explanation: In a rotating frame, static equilibrium of an object on a turntable requires the outward centrifugal force to be balanced by real inward forces like static friction or tension.
8. What is the dimension of centrifugal force?
Explanation: Centrifugal force is a force, so its dimensional formula is same as force: $[MLT^{-2}]$.
9. In the vector expression $F_{cf} = -m(\omega \times (\omega \times r))$, what does the term $(\omega \times r)$ represent?
Explanation: The cross product of angular velocity and position vector, $\omega \times r$, represents the linear velocity vector $v$ of the point.
10. If an observer inside a rotating space station experiences simulated gravity via centrifugal force, towards which direction must the floor of the station be curved?
Explanation: To experience outward centrifugal force acting 'downwards' against the floor, the floor must be curved such that the normal reaction is directed inward toward the axis of rotation (convex relative to outside, concave when viewed from inside facing the outer rim).