What are the characteristics of centripetal force?

Always acts perpendicular to the velocity of the particle. Changes the direction of motion but not the speed of the particle. The work done by centripetal force is always zero.

What is uniform circular motion?

Uniform circular motion occurs when a particle moves along a circular path at a constant speed. Here, the centripetal force is perpendicular to velocity, and there is no tangential force.

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Centripetal Force

When a particle moves along a curved path, a force is required to change the direction of motion of the particle. The force which is required to change the direction of motion is called centripetal force.

$F_{c} = \frac{m v ^{2}}{r}$

v = r ω

∴ F_c = mω² r

● Direction of centripetal force is always perpendicular to the velocity of particle.

● It can change the direction of motion but it can not change speed of the particle.

● Work done by centripetal force is always zero (whether particle completes the loop or not)

● If net force on the particle is perpendicular to its velocity than particle moves along a circular path with constant speed. This motion is called uniform circular motion

Note

Since circular motion is an accelerated motion, thus a pseudo force acts on the particle which is opposite to centripetal acceleration (away from the centre of circle). This pseudo force is called centrifugal force.

Centrifugal force = $\frac{m v ^{2}}{r}$

1.0Non- Uniform Circular Motion

When a particle moving in a circle and if the speed of particle increases or decreases then the motion is non-uniform circular motion.

In non- uniform circular motion ω ≠ constant

$∣ v ∣ \neq = constant$

In non-uniform circular motion particle has two acceleration :

(a) Tangential acceleration

a_T = = rate of change of speed;

$a_{T} = \frac{d v}{d t} = rate of change of speed;$

$v = \frac{d s}{d t} = speed; s = arc-length$

Tangential force F_T= ma_T

(b) Centripetal acceleration

$a_{c} = \frac{v ^{2}}{r} = ω^{2} r$

Centripetal force

$F_{c} = \frac{m v ^{2}}{r} = m ω^{2} r$

Net acceleration of the particle

$a = a_{c} + a_{T} ⟹ a = a_{c}^{2} + a_{T}^{2}$ and net force

$F = F_{c} + F_{T}$

If 'θ ' is the angle made by net force (F) with F_c, then tanθ = $\frac{F _{T}}{F _{c}}$

$θ = tan^{-} 1 \frac{F _{T}}{F _{c}} = tan^{- 1} \frac{a _{T}}{a _{c}}$

∴ θ = tan^–1

● In both uniform and non- uniform circular motion F_cis perpendicular to velocity. So work done by centripetal force will be zero in both the cases.

● In uniform circular motion F_T= 0 as a_T = 0

But in non- uniform circular motion F_T≠ 0. Thus there will be work done by tangential force in this case.

2.0 A Cyclist Making A Turn

Let a cyclist moving on a circular path of radius r bend away from the vertical by an angle θ. If R is the reaction of the ground, then R may be resolved into two components horizontal and vertical. The vertical component R cos θ balances the weight mg of the cyclist and the horizontal component R sin θ provides the necessary centripetal force for circular motion.

$\frac{m v ^{2}}{r}$ ....(i)

R cos θ = mg ....(ii)

Dividing (i) by (ii), we get

tan θ = $\frac{v ^{2}}{r g}$ ....(iii)

For less bending of cyclist, his speed v should be smaller and radius r of circular path should be greater.

3.0An Aeroplane Making a Turn

In order to make a circular turn, a plane must roll at some angle θ in such a manner that the horizontal component of the lift force L provides the necessary centripetal force for circular motion. The vertical component of the lift force balances the weight of the plane.

$L s in θ = \frac{m v ^{2}}{r} an d L cos θ = m g$

or, the angle θ should be such that, $\frac{v ^{2}}{r g}$

A car taking a turn on a level road

When a car takes a turn on a level road, the portion of the turn can be approximated by an arc of a circle of radius r (see fig). If the car makes the turn at a constant speed v, then there must be some centripetal force acting on the car. This force is generated by the friction between the tyres and the road. (car has a tendency to slip radially outward, so frictional force acts inwards) µ_S is the coefficient of static friction, N = mg is the normal reaction of the surface.

The maximum safe velocity v is

$\frac{m v ^{2}}{r} = μ_{s} N = μ_{s} m g$

$μ_{s} = \frac{v ^{2}}{r g} or v = μ_{s} r g$

It is independent of the mass of the car. The safe velocity is the same for all vehicles of larger and smaller mass.

4.0Banking of Road

If a cyclist takes a turn, he can bend from his vertical position. This is not possible in the case of car, truck or train. The tilting of the vehicle is achieved by raising the outer edge of the circular track, slightly above the inner edge. This is known as banking of curved track. The angle of inclination with the horizontal is called the angle of banking.

If driver moves with slow velocity that friction does not play any role in negotiating the turn. The various forces acting on the vehicle are :

(i) Weight of the vehicle (mg) in the downward direction.

(ii) Normal reaction (N) perpendicular to the inclined surface of the road. Resolve N in two components.

(a) N cos θ, vertically upwards which balances weight of the vehicle.

∴ N cos θ = mg .....(i)

(b) Nsinθ, in horizontal direction which provides necessary centripetal force.

∴ N sin θ = $\frac{m v ^{2}}{r}$ .....(ii)

on dividing equation (ii) by equation (i)

$\frac{N s i n θ}{N c o s θ} = \frac{\frac{m v ^{2}}{r}}{m g}$

or $tan θ = \frac{v ^{2}}{r g}$

∴ $θ = tan^{- 1} (\frac{v ^{2}}{r g})$

Where m is the mass of the vehicle, r is radius of curvature of the road, v is speed of the vehicle and θ is the banking angle $sin θ = \frac{h}{b}$ .

Here, h = height to which outer side of the road is raised.

b = width of the road.

Factors that decide the value of the angle of banking are as follows :

(i) Velocity of the vehicle

(ii) Radius of the curve

(iii) Acceleration due to gravity

Thus, there is no need of mass of the vehicle to express the value of angle of banking i.e. angle of banking ⇒ does not dependent on the mass of the vehicle.

$v^{2} = g r tan θ ∴ v = g r tan θ (maximum safe speed)$

This gives the maximum safe speed of the vehicle. In actual practice, some frictional forces are always present. So, the maximum safe velocity is always much greater than that given by the above equation. While constructing the curved track, the value of θ is calculated for fixed values of v_max and r. This explains why along the curved roads, the speed limit at which the curve is to be negotiated is clearly indicated on sign boards.

The outer side of the road is raised by h = b × θ. When θ is small, then

$tan θ \approx sin θ = \frac{h}{b}; Also tan θ = \frac{v ^{2}}{r g}$

$\frac{v ^{2}}{r g} = \frac{h}{b} or h = \frac{v ^{2}}{r g} \times b$

This gives us the height through which the outer edge is raised above the inner edge.

5.0Solved Examples

1. A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg. What is the angle of banking required to prevent wearing out of the rail?

Solution

Angle of banking = $tan^{- 1} (\frac{v ^{2}}{R g}) = tan^{- 1} (\frac{15 \times 15}{30 \times 10}) = 3 7^{\circ}$

2. At what angle should a highway be banked for cars travelling at a speed of 100 km/h if the radius of the road is 400m and no frictional forces are involved?

Solution

The banking should be done at an angle θ such that

$tan θ = \frac{v ^{2}}{r g} = \frac{\frac{250}{9} \times \frac{250}{9}}{400 \times 10} = \frac{625}{81 \times 40} = 0.19$

$θ = tan^{- 1} 0.19 \approx 0.19 radian \approx 0.19 \times 57. 3^{\circ} \approx 1 1^{\circ}$

3. An aircraft executes a horizontal loop at a speed of 720 km/h with its wing banked at 15°. Calculate the radius of the loop.

Solution

Speed, $v = \frac{720 \times 1000}{3600} = 200 m/s$

And tan θ = tan 15° = 0.2679

$tan θ = \frac{v ^{2}}{r g} or r = \frac{v ^{2}}{g t a n θ}$

m = 15235.7 m = 15.24 km.

$r = \frac{200 \times 200}{9.8 \times 0.2679} = 15235.7 m = 15.24 km .$

4. A string breaks under a tension of 10 kg-wt. If a string is used to revolve a body of mass
1.2 g in a horizontal circle of radius 50 cm, what is the maximum speed with which a body can be revolved? When a body is revolving at maximum speed, what is its time period? (g = 9.8 m/s²)

Solution

Mass of a body = m = 1.2 g = 1.2 × 10^–3 kg,

radius of circular path = r = 50 cm = 0.5 m,

Centripetal force = F = 10 kg wt = 10 × 9.8 N.

The necessary centripetal force acting on a body is given by tension in the string.

Maximum speed = 22.1 m/s, Period of revolution at maximum speed = 0.016 s

5. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration is varying with time t as a_c = k² rt², where k is a constant. Determine the power delivered to particle by the forces acting on it.

Solution

If v is instantaneous velocity, centripetal acceleration

∴ $a_{c} = \frac{v ^{2}}{r} = \frac{v ^{2}}{r} = k^{2} r t^{2} ∴ v = k r t$

In circular motion work done by centripetal force is always zero & work is done only by tangential force.

Tangent acceleration

$a_{T} = \frac{d v}{d t} = \frac{d}{d t} (k r t) = k r$

∴ Tangential force F_T = ma_T = mkr

Power P = F_Tv = (mkr) (krt) = mk²r²t

6.0Also Read

Galaxies	Moon	Mirrors
Eclipse	Solar Eclipse	Magnets
Star	Rainbow	Velocity

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Centripetal Force

When a particle moves along a curved path, a force is required to change the direction of motion of the particle. The force which is required to change the direction of motion is called centripetal force.

$F_{c} = \frac{m v ^{2}}{r}$

v = r ω

∴ F_c = mω² r

● Direction of centripetal force is always perpendicular to the velocity of particle.

● It can change the direction of motion but it can not change speed of the particle.

● Work done by centripetal force is always zero (whether particle completes the loop or not)

● If net force on the particle is perpendicular to its velocity than particle moves along a circular path with constant speed. This motion is called uniform circular motion

Note

Since circular motion is an accelerated motion, thus a pseudo force acts on the particle which is opposite to centripetal acceleration (away from the centre of circle). This pseudo force is called centrifugal force.

Centrifugal force = $\frac{m v ^{2}}{r}$

1.0Non- Uniform Circular Motion

When a particle moving in a circle and if the speed of particle increases or decreases then the motion is non-uniform circular motion.

In non- uniform circular motion ω ≠ constant

$∣ v ∣ \neq = constant$

In non-uniform circular motion particle has two acceleration :

(a) Tangential acceleration

a_T = = rate of change of speed;

$a_{T} = \frac{d v}{d t} = rate of change of speed;$

$v = \frac{d s}{d t} = speed; s = arc-length$

Tangential force F_T= ma_T

(b) Centripetal acceleration

$a_{c} = \frac{v ^{2}}{r} = ω^{2} r$

Centripetal force

$F_{c} = \frac{m v ^{2}}{r} = m ω^{2} r$

Net acceleration of the particle

$a = a_{c} + a_{T} ⟹ a = a_{c}^{2} + a_{T}^{2}$ and net force

$F = F_{c} + F_{T}$

If 'θ ' is the angle made by net force (F) with F_c, then tanθ = $\frac{F _{T}}{F _{c}}$

$θ = tan^{-} 1 \frac{F _{T}}{F _{c}} = tan^{- 1} \frac{a _{T}}{a _{c}}$

∴ θ = tan^–1

● In both uniform and non- uniform circular motion F_cis perpendicular to velocity. So work done by centripetal force will be zero in both the cases.

● In uniform circular motion F_T= 0 as a_T = 0

But in non- uniform circular motion F_T≠ 0. Thus there will be work done by tangential force in this case.

2.0 A Cyclist Making A Turn

Let a cyclist moving on a circular path of radius r bend away from the vertical by an angle θ. If R is the reaction of the ground, then R may be resolved into two components horizontal and vertical. The vertical component R cos θ balances the weight mg of the cyclist and the horizontal component R sin θ provides the necessary centripetal force for circular motion.

$\frac{m v ^{2}}{r}$ ....(i)

R cos θ = mg ....(ii)

Dividing (i) by (ii), we get

tan θ = $\frac{v ^{2}}{r g}$ ....(iii)

For less bending of cyclist, his speed v should be smaller and radius r of circular path should be greater.

3.0An Aeroplane Making a Turn

In order to make a circular turn, a plane must roll at some angle θ in such a manner that the horizontal component of the lift force L provides the necessary centripetal force for circular motion. The vertical component of the lift force balances the weight of the plane.

$L s in θ = \frac{m v ^{2}}{r} an d L cos θ = m g$

or, the angle θ should be such that, $\frac{v ^{2}}{r g}$

A car taking a turn on a level road

When a car takes a turn on a level road, the portion of the turn can be approximated by an arc of a circle of radius r (see fig). If the car makes the turn at a constant speed v, then there must be some centripetal force acting on the car. This force is generated by the friction between the tyres and the road. (car has a tendency to slip radially outward, so frictional force acts inwards) µ_S is the coefficient of static friction, N = mg is the normal reaction of the surface.

The maximum safe velocity v is

$\frac{m v ^{2}}{r} = μ_{s} N = μ_{s} m g$

$μ_{s} = \frac{v ^{2}}{r g} or v = μ_{s} r g$

It is independent of the mass of the car. The safe velocity is the same for all vehicles of larger and smaller mass.

4.0Banking of Road

If a cyclist takes a turn, he can bend from his vertical position. This is not possible in the case of car, truck or train. The tilting of the vehicle is achieved by raising the outer edge of the circular track, slightly above the inner edge. This is known as banking of curved track. The angle of inclination with the horizontal is called the angle of banking.

If driver moves with slow velocity that friction does not play any role in negotiating the turn. The various forces acting on the vehicle are :

(i) Weight of the vehicle (mg) in the downward direction.

(ii) Normal reaction (N) perpendicular to the inclined surface of the road. Resolve N in two components.

(a) N cos θ, vertically upwards which balances weight of the vehicle.

∴ N cos θ = mg .....(i)

(b) Nsinθ, in horizontal direction which provides necessary centripetal force.

∴ N sin θ = $\frac{m v ^{2}}{r}$ .....(ii)

on dividing equation (ii) by equation (i)

$\frac{N s i n θ}{N c o s θ} = \frac{\frac{m v ^{2}}{r}}{m g}$

or $tan θ = \frac{v ^{2}}{r g}$

∴ $θ = tan^{- 1} (\frac{v ^{2}}{r g})$

Where m is the mass of the vehicle, r is radius of curvature of the road, v is speed of the vehicle and θ is the banking angle $sin θ = \frac{h}{b}$ .

Here, h = height to which outer side of the road is raised.

b = width of the road.

Factors that decide the value of the angle of banking are as follows :

(i) Velocity of the vehicle

(ii) Radius of the curve

(iii) Acceleration due to gravity

Thus, there is no need of mass of the vehicle to express the value of angle of banking i.e. angle of banking ⇒ does not dependent on the mass of the vehicle.

$v^{2} = g r tan θ ∴ v = g r tan θ (maximum safe speed)$

This gives the maximum safe speed of the vehicle. In actual practice, some frictional forces are always present. So, the maximum safe velocity is always much greater than that given by the above equation. While constructing the curved track, the value of θ is calculated for fixed values of v_max and r. This explains why along the curved roads, the speed limit at which the curve is to be negotiated is clearly indicated on sign boards.

The outer side of the road is raised by h = b × θ. When θ is small, then

$tan θ \approx sin θ = \frac{h}{b}; Also tan θ = \frac{v ^{2}}{r g}$

$\frac{v ^{2}}{r g} = \frac{h}{b} or h = \frac{v ^{2}}{r g} \times b$

This gives us the height through which the outer edge is raised above the inner edge.

5.0Solved Examples

1. A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg. What is the angle of banking required to prevent wearing out of the rail?

Solution

Angle of banking = $tan^{- 1} (\frac{v ^{2}}{R g}) = tan^{- 1} (\frac{15 \times 15}{30 \times 10}) = 3 7^{\circ}$

2. At what angle should a highway be banked for cars travelling at a speed of 100 km/h if the radius of the road is 400m and no frictional forces are involved?

Solution

The banking should be done at an angle θ such that

$tan θ = \frac{v ^{2}}{r g} = \frac{\frac{250}{9} \times \frac{250}{9}}{400 \times 10} = \frac{625}{81 \times 40} = 0.19$

$θ = tan^{- 1} 0.19 \approx 0.19 radian \approx 0.19 \times 57. 3^{\circ} \approx 1 1^{\circ}$

3. An aircraft executes a horizontal loop at a speed of 720 km/h with its wing banked at 15°. Calculate the radius of the loop.

Solution

Speed, $v = \frac{720 \times 1000}{3600} = 200 m/s$

And tan θ = tan 15° = 0.2679

$tan θ = \frac{v ^{2}}{r g} or r = \frac{v ^{2}}{g t a n θ}$

m = 15235.7 m = 15.24 km.

$r = \frac{200 \times 200}{9.8 \times 0.2679} = 15235.7 m = 15.24 km .$

4. A string breaks under a tension of 10 kg-wt. If a string is used to revolve a body of mass
1.2 g in a horizontal circle of radius 50 cm, what is the maximum speed with which a body can be revolved? When a body is revolving at maximum speed, what is its time period? (g = 9.8 m/s²)

Solution

Mass of a body = m = 1.2 g = 1.2 × 10^–3 kg,

radius of circular path = r = 50 cm = 0.5 m,

Centripetal force = F = 10 kg wt = 10 × 9.8 N.

The necessary centripetal force acting on a body is given by tension in the string.

Maximum speed = 22.1 m/s, Period of revolution at maximum speed = 0.016 s

5. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration is varying with time t as a_c = k² rt², where k is a constant. Determine the power delivered to particle by the forces acting on it.

Solution

If v is instantaneous velocity, centripetal acceleration

∴ $a_{c} = \frac{v ^{2}}{r} = \frac{v ^{2}}{r} = k^{2} r t^{2} ∴ v = k r t$

In circular motion work done by centripetal force is always zero & work is done only by tangential force.

Tangent acceleration

$a_{T} = \frac{d v}{d t} = \frac{d}{d t} (k r t) = k r$

∴ Tangential force F_T = ma_T = mkr

Power P = F_Tv = (mkr) (krt) = mk²r²t

6.0Also Read

Galaxies	Moon	Mirrors
Eclipse	Solar Eclipse	Magnets
Star	Rainbow	Velocity

Frequently Asked Questions

What are the characteristics of centripetal force?

What is uniform circular motion?

Frequently Asked Questions

What are the characteristics of centripetal force?

What is uniform circular motion?

Join ALLEN!

Centripetal Force

1.0Non- Uniform Circular Motion

2.0 A Cyclist Making A Turn

3.0An Aeroplane Making a Turn

A car taking a turn on a level road

4.0Banking of Road

5.0Solved Examples

6.0Also Read

Table of Contents

Join ALLEN!

Centripetal Force

1.0Non- Uniform Circular Motion

2.0 A Cyclist Making A Turn

3.0An Aeroplane Making a Turn

A car taking a turn on a level road

4.0Banking of Road

5.0Solved Examples

6.0Also Read

Table of Contents