Centripetal force is the net inward force directed toward the center of curvature that is necessary for an object to move along a curved path. It is not a fundamental force of nature, but a descriptive role fulfilled by tension, gravity, friction, or the normal force.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Understanding centripetal force formula, uniform circular motion, banking of roads, and conical pendulum kinematics.
PGT extension
Non-uniform circular motion, tangential and radial components of acceleration, rotating reference frames, and fictitious centrifugal force.
Prerequisites
Newton's Laws of Motion
Kinematics of Circular Motion
Vector Algebra and Components
Definition and Direction
Centripetal force always acts orthogonally to the instantaneous velocity vector of the particle.
It points radially inward toward the instantaneous center of the circular path.
It changes the direction of the velocity vector continuously without necessarily changing its magnitude in uniform circular motion.
Origin of Centripetal Force
Centripetal force is a resultant force, supplied by existing physical interactions such as gravitational force, tension, friction, or electrostatic force.
It does not do any work on the particle in uniform circular motion because the displacement is always perpendicular to the force.
The absence of a sufficient centripetal force results in the particle moving along a path tangential to the circle due to inertia.
Uniform vs Non-Uniform Circular Motion
In uniform circular motion, the tangential acceleration is zero, and the net force is purely centripetal (radial).
In non-uniform circular motion, both tangential acceleration (due to changing speed) and radial/centripetal acceleration (due to changing direction) coexist.
The total acceleration vector is the vector sum of the radial acceleration and tangential acceleration.
Centrifugal Force in Rotating Frames
In a non-inertial rotating frame of reference, an outward pseudo force called the centrifugal force appears to act on masses.
The magnitude of the centrifugal force is equal to mω2r, directed radially outward from the axis of rotation.
It is introduced to apply Newton's laws of motion within accelerated rotating frames.
Derivations
Derivation of Centripetal Acceleration and Force
Conditions: Particle moving with uniform speed v along a circular path of radius r.
Consider position vectors r(t) and r(t + dt) subtending a small angle d$\theta = (v dt)/r$.
Express velocity vector as $v = v u_{t}$, where ut is the unit tangent vector.
Find the change in velocity vector dv over time dt, noting that the direction of dv points toward the center of the circle.
Take the limit as dt approaches zero to obtain magnitude of acceleration ac = v (dθ/dt) = v2/r.
Multiply mass m by acceleration ac to obtain the centripetal force Fc = m v2/r.
Result: $F_c = \frac{m v^2}{r} = m \omega^2 r$
Derivation of Optimum Speed for Banking of Roads
Conditions: Vehicle of mass m moving on a road banked at angle θ with coefficient of static friction μ.
Identify forces acting on the vehicle: normal force N perpendicular to the road, gravitational force mg downward, and frictional force f along the inclined plane.
Resolve forces into horizontal (radial) and vertical components.
Equate vertical components: N cos(θ) = mg + f sin(θ).
Equate horizontal components to the centripetal force: N sin(θ) + f cos(θ) = \frac{m v^2}{r}.
Substitute maximum static friction $f = \muN$ to find the maximum safe speed.
m = mass, $\omega = angular velocity of reference frame$, r = radial distance Newton (N)
Non-inertial rotating frame
Minimum Speed at Highest Point of Vertical Loop
$$v_{min} = \sqrt{r g}$$
r = radius of loop, g = gravitational acceleration m/s
String or track remains taut/engaged at top
Graphs and experiments
Centripetal Force versus Angular Velocity Squared
Draw: A straight line passing through the origin on a Cartesian plot with F_c on the y-axis and ω2 on the x-axis.
Interpret: The slope of the straight line equals the product of mass and radius (mr), verifying the direct proportionality F_c ∝ ω2.
Exam use: Interpreting experimental data plots in objective questions.
Conical Pendulum Experiment
Draw: A schematic showing a mass suspended by a string tracing out a horizontal circle, with resolved components of tension T.
Interpret: Demonstrates the dynamic equilibrium where the vertical component of tension balances weight, and the horizontal component provides the centripetal force.
Exam use: Deriving periodic time formulas as a function of string length and angle.
Radial vs Tangential Acceleration in Non-Uniform Circular Motion
Draw: A vector diagram at a point on a curved trajectory showing a radial vector pointing inward and a tangential vector perpendicular to it.
Interpret: Highlights that centripetal force only changes direction, whereas tangential force changes the kinetic energy and speed of the particle.
Exam use: Distinguishing force components in complex mechanics problems.
Solved numericals
TGT core
A stone of mass 0.5 kg is tied to the end of a string 1.5 m long and whirled in a horizontal circle at a constant speed of 3 m/s. Calculate the tension in the string.
Identify given parameters: m = 0.5 kg, r = 1.5 m, $v = 3 m/s$.
Apply the centripetal force formula: $F_c = m * (v^2) / r$.
A particle of mass 0.2 kg moves in a circle of radius 1.0 m with an angular velocity that increases according to $\omega = 2t rad/s$. Find the total force acting on the particle at t = 1 s.
Calculate total force $F_net = m * a_net = 0.2 * 2\sqrt{}5 = 0.4\sqrt{}5 N$.
Answer: 0.4√5 N (approx 0.894 N)
TGT core
Find the maximum speed at which a car can safely negotiate a unbanked circular curve of radius 50 m if the coefficient of friction between the tires and the road is 0.6. (Take $g = 10 m/s^2)$
Identify given parameters: r = 50 m, $\mu = 0.6$, $g = 10 m/s^2$.
Set maximum friction equal to centripetal force: μm$g = m * (v^2) / r$.
Cancel mass m from both sides: $v^2 = \mu * g * r$.
Substitute values: $v^2 = 0.6 * 10 * 50 = 300$.
Take square root: $v = \sqrt{}300 = 10\sqrt{}3 m/s \approx 17.32 m/s$.
Answer: 10√3 m/s (approx 17.32 m/s)
Assumptions and limitations
Centripetal force formulas assume an inertial frame of reference unless a pseudo centrifugal force is explicitly incorporated in rotating frames.
For macroscopic objects, macroscopic friction and normal reactions are treated as continuous rather than discrete molecular interactions.
Common misconceptions
Common error: Centrifugal force is a real outward force exerted by the center on the revolving body.
Correct: Centrifugal force is a fictitious force experienced only in a non-inertial rotating frame of reference.
In an inertial frame, only inward centripetal forces act; outward forces do not exist physically on the mass.
Common error: Centripetal force does work on an object in uniform circular motion.
Correct: Centripetal force does zero work on an object in uniform circular motion.
Work is the dot product of force and displacement. Since the centripetal force is perpendicular to the displacement vector at every instant, the work done is zero.
Common error: Centripetal force is a distinct, new type of fundamental force like gravity or electromagnetism.
Correct: Centripetal force is a functional label given to any net inward force acting on a circular path.
Gravity, tension, or friction individually or collectively serve as the centripetal force depending on the physical context.
Quick revision
Centripetal force always points radially inward and has a magnitude of m v^2 / r or m ω^2 r.
Centripetal Force (Velocity form): $F_c = \frac{m v^2}{r}$
Centripetal Force (Angular form): $F_c = m \omega^2 r$
Must remember
Centripetal force always acts orthogonally to the instantaneous velocity vector and points radially inward.
Centripetal force does zero work on a particle in uniform circular motion because force and displacement are perpendicular.
In uniform circular motion, tangential acceleration is zero, whereas in non-uniform circular motion, both radial and tangential accelerations coexist.
Centrifugal force is a pseudo force of magnitude $m \omega^2 r$ introduced in non-inertial rotating frames.
The minimum speed at the highest point of a vertical loop for a string or track to remain taut is $v_{min} = \sqrt{r g}$.
Glossary
Centripetal Force ($F_c$)A net inward force directed toward the center of curvature that changes the direction of velocity in circular motion, given by $F_c = \frac{m v^2}{r}$.
Centrifugal Force ($F_{cf}$)An outward fictitious force of magnitude $m \omega^2 r$ experienced in a non-inertial rotating reference frame.
Banking Angle ($\theta$)The angle at which a road surface is tilted relative to the horizontal to provide necessary centripetal force without relying solely on friction.
Tangential Acceleration ($a_t$)The component of acceleration parallel to the velocity vector, responsible for changing the magnitude of speed in non-uniform circular motion.
Uniform vs Non-Uniform Circular Motion
Parameter
Uniform Circular Motion
Non-Uniform Circular Motion
Tangential Acceleration ($a_t$)
Zero ($a_t = 0$)
Non-zero ($a_t \neq 0$)
Speed ($v$)
Constant
Changing with time
Net Acceleration ($a_{net}$)
Only radial ($a_c = \frac{v^2}{r}$)
Vector sum of radial and tangential ($a_{net} = \sqrt{a_c^2 + a_t^2}$
Work Done by Net Force
Zero
Non-zero (due to tangential force component)
Centripetal vs Centrifugal Force
Property
Centripetal Force
Centrifugal Force
Nature
Real resultant physical force
Fictitious / Pseudo force
Frame of Reference
Inertial frame
Non-inertial rotating frame
Direction
Radially inward
Radially outward
Magnitude
$\frac{m v^2}{r}$ or $m \omega^2 r$
$m \omega^2 r$
Practice Quiz
Choose an option to reveal the explanation.
1. What is the angle between the centripetal force and the instantaneous velocity vector of a particle in uniform circular motion?
Explanation: Centripetal force always acts radially inward along the radius, whereas the instantaneous velocity vector is tangent to the circular path. The angle between radius and tangent is $90^\circ$.
2. How much work is done by the centripetal force on a particle moving in a uniform circular motion?
Explanation: Work is defined as the dot product of force and displacement ($W = \vec{F} \cdot \vec{d}$). Since centripetal force is perpendicular to the displacement at every instant, $\cos(90^\circ) = 0$, making the work done zero.
3. If the radius of a circular path is doubled while keeping the linear speed constant, the centripetal force becomes:
Explanation: The centripetal force is given by $F_c = \frac{m v^2}{r}$. If $r$ is doubled and $v$ remains constant, $F_c$ becomes half of its original value.
4. Which of the following forces can act as a centripetal force for a planet orbiting the Sun?
Explanation: The gravitational pull between the Sun and the planet provides the necessary centripetal force required for orbital motion.
5. In non-uniform circular motion, the total acceleration vector of the particle is given by:
Explanation: The radial acceleration ($a_c$) and tangential acceleration ($a_t$) are mutually perpendicular. Hence, the net acceleration is the vector sum magnitude $\sqrt{a_c^2 + a_t^2}$.
6. What is the expression for the minimum speed required at the highest point of a vertical loop of radius $r$ so that a string remains taut?
Explanation: At the highest point, if the string just remains taut, tension becomes zero, and weight alone provides the centripetal force: $mg = \frac{m v^2}{r}$, yielding $v_{min} = \sqrt{r g}$.
7. What is the nature of the centrifugal force experienced in a rotating frame of reference?
Explanation: Centrifugal force is a fictitious or pseudo force introduced to apply Newton's laws of motion correctly in non-inertial rotating frames.
8. For a vehicle taking a turn on a flat unbanked circular road, what provides the necessary centripetal force?
Explanation: On a flat circular road, the frictional force directed toward the center of the curve supplies the required centripetal force.
9. If a particle moves in a circle of radius $r$ with an angular velocity $\omega$, its centripetal acceleration is:
Explanation: The kinematic relation for centripetal acceleration in terms of angular velocity is $a_c = \omega^2 r$.
10. What is the banking angle $\theta$ formula for a road of radius $r$ and speed $v$ in the absence of friction?
Explanation: Balancing vertical and horizontal components for a vehicle on a banked road without friction gives $\tan \theta = \frac{v^2}{r g}$.
11. In a conical pendulum, if the length of the string is $L$ and the semi-vertical angle is $\theta$, the radius of the horizontal circular path is:
Explanation: From geometry, the radius of the circular path swept by the bob is $r = L \sin \theta$.
12. What happens to a body if the sufficient centripetal force acting on it suddenly vanishes while in circular motion?
Explanation: Due to inertia, in the absence of a net inward centripetal force, the body continues to move along the straight-line path tangential to the circle.
13. What is the magnitude of the centrifugal force in a rotating frame with angular velocity $\omega$ for a mass $m$ at distance $r$?
Explanation: The magnitude of the centrifugal force is given by $F_{cf} = m \omega^2 r$.
14. In uniform circular motion, what is the value of the tangential component of acceleration ($a_t$)?
Explanation: In uniform circular motion, the speed remains constant, meaning the rate of change of speed (tangential acceleration) is zero ($a_t = 0$).
15. Which physical interaction supplies the centripetal force for an electron revolving around the nucleus in a hydrogen atom?
Explanation: The attractive electrostatic force (Coulomb force) between the positively charged nucleus and the negatively charged electron provides the centripetal force.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. A stone of mass $m$ is tied to a string of length $L$ and rotated in a vertical circle. What is the difference in tension between the lowest point and the highest point of the circle?
Explanation: At the lowest point, $T_1 - mg = \frac{m v_1^2}{L}$. At the highest point, $T_2 + mg = \frac{m v_2^2}{L}$. Using work-energy theorem, $v_1^2 - v_2^2 = 4gL$. Subtracting the equations gives $T_1 - T_2 = 6mg$.
2. A car is moving with speed $v$ on a banked circular road of radius $r$ and banking angle $\theta$. If the coefficient of static friction between the tires and the road is $\mu_s$, what is the maximum safe speed without skidding?
Explanation: The standard expression for maximum safe speed on a banked road with friction is $v_{max} = \sqrt{r g \left(\frac{\mu_s + \tan \theta}{1 - \mu_s \tan \theta}\right)}$.
3. A particle moves along a circle of radius $1\text{ m}$ with a uniform speed of $2\text{ m/s}$. What is the magnitude of its net acceleration?
Explanation: For uniform circular motion, tangential acceleration is zero. Net acceleration is purely centripetal: $a_c = \frac{v^2}{r} = \frac{2^2}{1} = 4\text{ m/s}^2$.
4. A conical pendulum has a string of length $L$ and makes an angle $\theta$ with the vertical. The time period $T$ of revolution of the conical pendulum is given by:
Explanation: For a conical pendulum, $T = 2\pi \sqrt{\frac{L \cos \theta}{g}}$ derived from $T \cos \theta = mg$ and $T \sin \theta = m \omega^2 L \sin \theta$.
5. If a particle's speed in circular motion of radius $r$ increases with time as $v = kt$ (where $k$ is a constant), the magnitude of its total acceleration at time $t$ is:
Explanation: Here, tangential acceleration $a_t = \frac{dv}{dt} = k$, and radial acceleration $a_c = \frac{v^2}{r} = \frac{k^2 t^2}{r}$. Total acceleration is $\sqrt{a_c^2 + a_t^2} = \sqrt{\left(\frac{k^2 t^2}{r}\right)^2 + k^2}$.
6. A body of mass $m$ is rotated in a horizontal circle of radius $r$ by a string. If the maximum tension the string can withstand is $T_{max}$, what is the maximum angular velocity with which the body can be rotated?
Explanation: Tension provides the centripetal force: $T_{max} = m \omega_{max}^2 r$. Solving for $\omega_{max}$ gives $\sqrt{\frac{T_{max}}{m r}}$.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Which of the following statements about centripetal force is correct?
Explanation: Centripetal force is a resultant force directed radially inward. It is not a new fundamental force and does zero work in uniform circular motion.
2. In non-uniform circular motion, the tangential acceleration arises due to:
Explanation: Tangential acceleration is responsible for changing the speed (magnitude of velocity) of the particle along the circular path.
3. A coin placed on a rotating turntable does not slip if the centrifugal force is:
Explanation: In the rotating frame, the tendency of the coin to fly outward due to centrifugal force is counterbalanced by static friction. It remains at rest if $m \omega^2 r \le f_{s,max}$.
4. If both the mass and speed of a particle in circular motion are doubled while the radius remains constant, by what factor does the centripetal force change?
Explanation: Formula $F_c = \frac{m v^2}{r}$. Doubling mass ($2m$) and doubling speed ($(2v)^2 = 4v^2$) results in $2 \times 4 = 8$ times the original centripetal force.
5. What is the tension at the lowest point of a vertical loop of radius $r$ for a particle of mass $m$ just able to complete the loop?
Explanation: At the lowest point, $T - mg = \frac{m v_{low}^2}{r}$. Using $v_{low}^2 = 5gr$, we get $T - mg = 5mg \implies T = 6mg$ ? Wait, let's check: $v_{low}^2 = 5gr$, so $\frac{m(5gr)}{r} = 5mg$. Then $T = 5mg + mg = 6mg$. Ah, let's verify option: $6mg$ is correct. Let's make sure correct index matches $6mg$ (which is index 2).
6. Why do passengers in a car feel pushed outward when the car takes a sharp turn?
Explanation: Passengers tend to continue straight due to inertia, and relative to the accelerating car frame, they experience an outward pseudo force (centrifugal force).
7. The dimension of centripetal force is:
Explanation: Centripetal force is a force, so its dimensional formula is $[\text{mass}] \times [\text{acceleration}] = [M][L T^{-2}] = [M L T^{-2}].
8. If a particle moves in a circle of radius $r$ with constant frequency $f$ (revolutions per second), its centripetal acceleration is:
Explanation: Since $\omega = 2 \pi f$, centripetal acceleration $a_c = \omega^2 r = (2\pi f)^2 r = 4 \pi^2 f^2 r$.
9. What is the primary purpose of banking a race track curve?
Explanation: Banking tilts the normal reaction, providing a horizontal component that assists or fully supplies the required centripetal force, reducing wear and skid risk.
10. Which acceleration component is always present in any type of circular motion (uniform or non-uniform)?
Explanation: Circular motion inherently requires a change in the direction of velocity at every instant, which necessitates a non-zero radial (centripetal) acceleration.