Centripetal Force Formula with Examples

Centripetal force is a force that operates on a body travelling in a circular direction and is directed towards the centre around which the body is moving. When an item moves at a constant speed around a circular route, it encounters an accelerating centripetal force towards the centre.

What is Centripetal Force?

In the uniform circular motion of a particle, the centripetal force is the force on the particle which at every instant points radially towards the centre of the circle and produces the centripetal acceleration to move the particle in its circular path.

Centripetal Force Formula

F_c = m × v²/r

Where,

• F_c is the centripetal force
• m is mass
• v is velocity
• r is the radius of the path

Derivation of Formula

The force of a moving object is

F = ma

PQ + QS = PS

-v₁ + v₂ = Δv

Δv = v₂ – v₁

The PQS and AOB triangles are similar. Therefore,

Δv/AB = v/r

AB = vΔt

Δv/v×Δt = v/r

Δv/Δt = v²/r

a = v²/r

Now, 

F_c = ma

Therefore, F_c = m × v²/r

Examples of Centripetal Force

1. When a metal bob tied to one end of a string is whirled around a circle the tension in the string is centripetal force. If the hold at the Centre is released, the bob flies away tangentially.
2. For the motion of the moon or a satellite around the earth, the gravitational force provides the centripetal force.
3. We know that the electrons revolve around the nucleus in nearly circular orbits. The electrostatic force between the electrons and the nucleus provides the centripetal force.
4. When a pendulum is pulled away from its equilibrium position and is pushed tangentially, it moves along a circle. The horizontal component of tension T provides the centripetal force.  
5. For an Earth-satellite in a circular orbit, the centri- petal force is the gravitational force exerted by the Earth on the satellite.
6. In the Bohr atom, the centripetal force on an elec- tron in circular orbit around the nucleus is the attractive Coulomb force of the nucleus.
7. When an object tied at the end of a string is revolved in a horizontal circle, the centripetal force is the tension in the string.
8. When a car takes a turn in a circular arc on a horizontal road with constant speed, the force of static friction between the car tyres and road surface is the centripetal force.

Sample Problems

Problem 1: A van of 1,250 Kg is travelling at 50.0 m/s and covers a curve of the radius of 200 m. calculate the centripetal force.

Solution

Mass = 1,250 Kg

Radius = 200 m

Velocity = 50.0 m/s

We know that ,

Centripetal force (F_c)= mv²/r

= 1250 × (50)² / 200

= 15,625 N

Problem 2: A car of mass 1000 kg is moving in a circular path with a velocity of 1.8 km/h. If the radius of a circular path is 10 m, then what is the centripetal force acting on a car?

Solution

Mass of a car, m = 1000 kg

Velocity of a car, v = 1.8 km/h = 0.5 m/s

Radius of a circular path, r = 10 m

Using the equation of centripetal force,

F_c = m v² /r

F_c = [1000 × (0.5)²] /10

F_c = 250 /10

F^c = 25 N

Therefore, the centripetal force acting on a car is 25 N.

Problem 3: One boy of mass 30 kg is enjoying a merry-go-round ride. A boy sitting on a merry-go-round rotates in a circular path of radius 6 m with a velocity of 2 m/s. Calculate the centripetal force acting on a boy.

Solution

Mass of a boy, m = 30 kg

Radius of a circular path, r = 6 m

Velocity of a boy, v = 2 m/s

Using the equation of centripetal force,

F_c = mv² / r

F_c = [30 × (2)²] /6

F_c = 120 /6

F_c = 20 N

Therefore, the centripetal force acting on a boy is 20 N.

Problem 4: Calculate the centripetal force acting on a 2500 kg truck moving with a velocity of 1.2 km/h in a circular path of a radius of 15 m.

Solution

Mass of a truck, m = 2500 kg

Velocity of a truck, v = 1.2 km/h = 0.33 m/s

Radius of a circular path, r = 15 m

Using the equation of centripetal force,

F_c = mv²/ r

F_c = [2500 × (0.33)²] /15

F_c = 272.25 /15

F_c = 18.15 N

Therefore, the centripetal force acting on a truck is 18.15 N

Problem 5: What is the centripetal force acting on a sports bike of mass 140 kg moving in a circular path of radius 25 m with a velocity of 2.5 km/h?

Solution

Mass of a sports bike, m = 140 kg

Radius of a circular path, r = 25 m

Velocity of a sports bike, v = 2.5 km/h = 0.69 m/s

Using the equation of centripetal force,

F_c = mv²/r

F_c = [140 × (0.69)²] /25

F_c = 66.654 /25

F_c = 2.6 N

Therefore, the centripetal force acting on a sports bike is 2.6 N.