Example – 19:

The vertical section of a road over a bridge in the direction of its length is in the form of an arc of a circle of radius 19.5 m. Find the greatest velocity at which a car can cross the bridge without losing contact with the road at the highest point if the c.g. of the car is 0.5 m from the ground.

Given: Radius of circle = r = 19.5 + 0.5 = 20 m,

To find: velocity of car = v =?

Solution:

At the highest point, the centrifugal force and weight of the car are equal in magnitude.

Ans: The velocity of car = 14 m/s

Example – 20:

A flyover bridge is in the form of a circular arc of radius 30 m. Find the limiting speed at which a car can cross the bridge without losing contact with the road at the highest point.  Assume the centre of gravity of the car to be 0.5 m above the road.

Given: Radius of circle = r = 30 + 0.5 = 30.5 m,

To find: velocity of car = v =?

Solution:

At the highest point, the centrifugal force and weight of the car are equal in magnitude.

Ans: The velocity of car = 17.3 m/s

Example – 21:

A motorcyclist rides in a vertical circle in a hollow sphere of radius 3 m. Find the minimum speed required so that he does not lose contact with the sphere at the highest point. Also, find its angular speed.

Given: Radius of hollow sphere = r = 3 m,

To find: Minimum speed required = v = ?, Angular speed = ω = ?

Solution:

At the highest point, the centrifugal force and weight of the motorcyclist and motorcycle are equal in magnitude.

Ans: The velocity of car = 5.42 m/s, Angular speed = 1.807 rad/s.

Example – 22:

A motorcyclist rides in a vertical circle in a hollow sphere of radius 12.8 m. Find the minimum speed required so that he does not lose contact with the sphere at the highest point.

Given: Radius of sphere = r = 12.8 m,

To find: Minimum speed = v =?

Solution:

At the highest point, the centrifugal force and weight of the motorcyclist and motorcycle are equal in magnitude.

Ans: Minimum speed of motorcycle = 11.2 m/s

Example – 23:

An object of mass 100 g moves around the circumference of a circle of radius 2m with a constant angular speed of 7.5 rad/s. Compute its linear speed and force directed towards centre.

Given: mass of the body = m = 100 g = 0.1 kg, Radius of circular path = r = 2 m, Angular speed = ω = 7.5 rad/s,

To find: Linear speed = v =? Centripetal force = F = ?

Solution:

v = r ω = 2 x 7.5 = 15 m/s

F = m r ω²

∴ F = 0.1 x 2 x (7.5)² = 11.25 N

Ans: Linear speed = 15 m/s, Centripetal force = 11.25 N radially inward.

Example – 24:

A car of mass 2000 kg rounds a curve of radius 250m at 90 km/hr. Compute its angular speed, centripetal acceleration and centrifugal force.

Given: Mass of car = m = 2000 kg, Radius of circular path = r = 250 m, Linear velocity of car = v = 90 km/hr = 90 x 5/18 = 25 m/s,

To find:  Angular speed = ω =? centripetal acceleration = a = ?, centripetal force = F = ?,

Solution:

v = r ω

∴ ω = v/r = 25/250 = 0.1 rad/s

a = r ω²

∴ a = 250 x (0.1)² = 2.5 m/s²

F = m a = 2000 x 2.5 = 5000 N radially inward

Ans: Angular speed = 0.1 rad/s, Centripetal acceleration = 2.5 m/s² radially inward

The centripetal force and centrifugal force are equal in magnitude but opposite in direction.

Centrifugal force = 5000 N

Ans: Centrifugal force = 5000 N radially outward

Example – 25:

A 0.5 kg mass is tied to one end of a string and rotated in a horizontal circle of 1.25 m radius about the other end. What is the tension in the string if the period of revolution is 5 s. What is the maximum speed of rotation and the corresponding period if the string can withstand a maximum tension of 150 N.

Given: Mass of body = m = 0.5 kg, radius of curvature = r = 1.25 m, Period of revolution = T = 5 s,

To Find: Maximum speed of rotation = v =? F =? Part – II: Linear speed v =? when maximum tension = F = 150 N

Solution:

Part – I:

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

Part – II:

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

Ans: Tension in string = 0.986 N Maximum speed = 19.36 m/s. Period of revolution at maximum speed = 0.41 s

Example – 26:

A coin kept on a horizontal rotating disc has its centre at a distance of 0.1 m from the axis of rotation of the disc. If the coefficient of friction between the coin and the disc is 0.25, find the speed of the disc at which the coin would be about to slip off.

Given: radius of circular path = r = 0.1 m, coefficient of friction = μ = 0.25, g = 9.8 m/s²

To find: Speed of disc = v =?

Solution:

Necessary centripetal force is provided by the friction between the disc and coin.

Ans: The speed of the disc = 0.495 m/s.

Example – 27:

A coin kept on a horizontal rotating disc has its centre at a distance of 0.25 m from the axis of rotation of the disc. If μ = 0.2, find the angular velocity of the disc at which the coin is about to slip off.   (g = 9.8 m/s²)

Given: radius of circular path = r = 0.25 m, coefficient of friction = μ = 0.2, g = 9.8 m/s²

To find: Angular speed of disc = ω =?

Solution:

Necessary centripetal force is provided by the friction between the disc and coin.

Ans: The angular speed of the disc = 7.84 rad/s.

Example – 28:

A coin just remains on a disc rotating at 120 r.p.m. when kept at a distance of 1.5 cm from the axis of rotation. Find the coefficient of friction between the coin and the disc.

Given: Radius of circular path = r = 1.5 cm = 0.015 m, rpm = N = 120 r.p.m., g = 9.8 m/s²

To find: coefficient of friction = μ =?

Solution:

Necessary centripetal force is provided by the friction between the disc and coin.

Ans: Coefficient of friction = 0.2415

Example – 29:

A coin just remains on a disc rotating at a steady rate of 180 rev/min if kept at a distance of 2 cm from the axis of rotation. Find the coefficient of friction between the coin and the disc.

Given: Radius of circle = r = 2 cm = 0.02 m, rpm = N = 180 r.p.m., g = 9.8 m/s²

To find: coefficient of friction = μ =?

Solution:

Necessary centripetal force is provided by the friction between the disc and coin.

Ans: The coefficient of friction = 0.724

Example – 30:

A coin kept with its centre at a distance of 9 cm from the axis of rotation of a disc starts slipping off when the disc speed reaches 60 r.p.m. Up to what speed the coin will remain on the disc if its centre is 16 cm from the axis of rotation of the disc?

Given: Initial radius of circle = r₁ = 9 cm, initial rpm = N₁= 60 r.p.m., final radius of circle = r₂ = 16 cm

To find: Final rpm = N₂ =?

Solution:

Necessary centripetal force is provided by the friction between the disc and coin.

Ans: The required angular speed = 45 r.p.m.

Example – 31:

A coin placed on a turntable rotating at 30 r.p.m. revolves with the table without slipping provided it is not more than 12 cm away from the axis. How far from the axis can the coin be placed so that it revolves with the turntable without slipping if the speed of rotation is 45 r.p.m.?

Given: Initial radius of circle = r₁ = 12 cm, initial rpm = N₁= 45 r.p.m., Final rpm = N₂ = 45 r.p.m.

To find: final radius of circle = r₂ =?

Solution:

Necessary centripetal force is provided by the friction between the disc and coin.

Ans: The maximum displacement from centre  = 5.33 cm.

Example – 32:

A coin is placed at a distance of 10 cm from the centre of a turntable of radius 1m just begins to slip, when the turntable rotates at a speed of 90 r.p.m. Calculate the coefficient of static friction between the coin and the turntable.

Given: Radius of circle = r = 10 cm = 0.1 m, speed of rotation = N = 90 r.p.m.

To Find: Coefficient of Friction = m =?

Solution:

The necessary centripetal force is provided by the friction between the turntable and coin

Ans: The coefficient of friction is 0.9066

Example – 32:

With what maximum speed a car be safely driven along a curve of radius 40 m on a horizontal road if the coefficient of friction between the car tyres and the road surface is 0.3? (g = 9.8 m/s²)

Given: Radius of circular path = r = 40 m, coefficient of friction = μ = 0.3, g = 9.8 m/s²

To find: Maximum speed = v =?

Solution:

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: The speed of the car = 10.84 m/s.

Example – 33:

Find the maximum speed at which a car can be safely driven along a curve of 100m radius. The coefficient of friction between tyres and the surface of the road is 0.2.

Given: radius of curve = r = 100 m, coefficient of friction = μ = 0.2, g = 9.8 m/s²

To find: maximum speed = v =?

Solution:

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: The speed of the car = 14 m/s.

Example – 34:

A car travelling at 18 km/h just rounds a curve without skidding. If the road is plain and the coefficient of friction between the road surface and the tyres is 0.25, find the radius of the curve.

Given: Speed of car = v = 18 km/hr = 18 x 5/18 = 5 m/s, coefficient of friction = μ = 0.25, g = 9.8 m/s²

To find: radius of curve = r =?

Solution:

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: The radius of road = 10.2 m.

Example – 35:

A car can just go around a curve of 20 m radius without skidding when travelling at 36 km/h. If the road is plain, find the coefficient of friction between the road surface and tyres.

Given: Speed of the car = v = 36 km/hr = 36 x 5/18 = 10 m/s, r = 20m,  g = 9.8 m/s²

To find: coefficient of friction = μ =?

Solution:  

Necessary centripetal force is provided by the friction between the road and tyres.

Ans: The coefficient of friction = 0.5102

Example – 04:

A flat curve highway has a radius of curvature 400 m. A car rounds the curve at a speed of 32 m/s. What is the minimum value of the coefficient of friction that will prevent the car from sliding?

Given: Radius of curvature = r = 400 m, speed of car = 32 m/s, g = 9.8 m/s².

To Find: Coefficient of Friction = m =?

Solution:

The necessary centripetal force is provided by the friction between the road and tyres of the car

Ans: The coefficient of friction is 0.2612

Example – 36:

A rotor has a diameter of 4.0 m. The rotor is rotated about the central vertical axis. The occupant remains pinned against the wall. When the linear velocity of the drum is 8 m/s. Compute the coefficient of static friction between the wall of the rotor and the clothing of occupant. Also, calculate the angular velocity of the drum. How many revolutions will it make in a minute?

Given: diameter = d = 4 m, radius = r = 4/2 = 2 m, linear speed = v =.8 m/s , g = 9.8 m/s² ,

To find:  Coefficient of friction = μ = ?, Angular velocity = ω = ?, rpm =  N = ?

Solution:

As the occupant remains pinned against the wall, his weight is equal to the frictional force.

Frictional force = Weight of a body

Ans: Coefficient of friction = 0.3063, Angular velocity = 4 rad/s, Number of revolutions per minute = 38.21.

Previous Topic: More Problems on Centripetal Force

Next Topic: Concept of Banking of Road

Science > Physics > Circular Motion > Numerical Problems on Centripetal Force – 02

The post Numerical Problems on Centripetal Force – 02 appeared first on The Fact Factor.

Example – 01:

A 0.5 kg mass is rotated in a horizontal circle of radius 20 cm.  Calculate the centripetal force acting on it, if its angular speed of revolution is 0.8 rad /s.

Given: Mass of a body = m = 0.5 kg, radius of circular path = r = 20cm = 0.2 m, Angular speed = ω = 0.8 rad/s,

To find: Centripetal force = F =?

Solution:

F = m r ω²

∴ F = 0.5 x 0.2 x (0.8)²

∴ F = 0.5 x 0.2 x (0.64) = 0.064 N

Ans: Centripetal force = 0.064 N acting radially inward.

Example – 02:

An object of mass 0.5 kg is whirled in a horizontal circle of radius 20 cm. Calculate centripetal force acting on it, if its angular speed of revolution is 0.6 rad/s.

Given: Mass of a body = m = 0.5 kg, radius of circular path = r = 20cm = 0.2 m, Angular speed = ω = 0.6 rad/s,

To find: Centripetal force = F =?

Solution:

F = m r ω²

∴ F = 0.5 x 0.2 x (0.6)²

∴ F = 0.5 x 0.2 x (0.36) = 0.036 N

Ans: Centripetal force = 0.036 N acting radially inward.

Example – 03:

A one kg mass tied at the end of the string 0. 5 m long is whirled in a horizontal circle, the other end of the string being fixed. The breaking tension in the string is 50 N. Find the greatest speed that can be given to the mass.

Given: Mass of a body = m = 1 kg, radius of circular path = r = 0.5 m, Centripetal force  = F = 50 N.

To find: Greatest speed = v =?

Solution:

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

Ans: The greatest speed can be given to mass = 5 m/s

Example – 04:

2 kg mass is tied to a string at one end and rotated in a horizontal circle of radius 0.8 m about the other end. If the breaking tension in the string is 250 N, find the maximum speed at which mass can be rotated.

Given: Mass of a body = m = 2 kg, radius of circular path = r = 0.8 m, Centripetal force  = F = 250 N.

To find: Maximum speed = v =?

Solution:

Ans: The greatest speed can be given to mass =10 m/s

Example – 05:

An object of mass 0.5 kg is whirled in a horizontal circle of radius 20 cm. Calculate the maximum number of revolutions per minute, so that the string does not break. The breaking tension of the string is 9.86 N.

Given: Mass of a body = m = 0.5 kg, radius of circular path = r = 20 cm = 0.2 m, Centripetal force  = F = 9.86 N.

To find: Maximum rpm = N =?

Solution:

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

Ans: Max. No. of revolutions = 94.87 r.p.m.

Example – 06:

A mass of 5 kg is tied at the end of the string 1.2 m long rotates in a horizontal circle. If the breaking tension in the string is 300 N, find the maximum number of rotations per minute the mass can make.

Given: Mass of a body = m = 5 kg, radius of circular path = r = 1.2 m, Centripetal force  = F = 300 N.

To find: Maximum rpm = N =?

Solution:

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

Ans: Max. No. of revolutions per minute = 67.56

Example – 07:

A stone is tied to a string 50 cm long and rotated uniformly in a horizontal circle about the other end. If the string can support a maximum tension ten times the weight of the stone, find the maximum number of revolutions per second the string can make before it breaks.

Given: Mass of a body = m, radius of circular path = r = 50 cm = 0.5 m, Centripetal force  = 10 mg.

To find: Maximum number of revolutions per second = n =?

Solution:

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

Ans: Max. No. of revolutions per second = 2.23

Example – 08:

A certain string breaks under a tension of 45 kg-wt. A mass of 100 g is attached to one end of a piece of this string 500 cm long and rotated in a horizontal circle. Neglecting the effect of gravity, find the greatest number of revolutions which, the sting can make without breaking.

Solution:

Given: Mass of a body = m = 100g = 0.1 kg, radius of circular path = r = 500 cm = 5 m, Centripetal force  = F = 9.86 N.

To find: Maximum rps = n =?,

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

Ans: Max. No. of revolutions per second = 4.73

Example – 09:

The breaking tension of a string is 80 kg.-wt. A mass of 1 kg is attached to the string and rotated in a horizontal circle on a horizontal surface of radius 2 m. Find the maximum number of revolutions made without breaking.

Given: Mass of a body = m = 1 kg, radius of circular path = r = 2 m, Centripetal force = F = 80 kg wt = 80 x 9.8 N.

To find: Maximum rps = n =?

Solution:

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

Ans: Max. No. of revolutions per second = 3.15

Example – 10:

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 resolved? When a body is revolving at maximum speed, what is its period? (g = 9.8 m/s²)

Given: Mass of a body = m = 1.2 g = 1.2 x 10^-3 kg, radius of circular path = r = 50 cm = 0.5 m, Centripetal force = F = 10 kg wt = 10 x 9.8 N.

To find: Maximum speed = v =? Period = T = ?

Solution:

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

Ans: Max. Speed = 202.1 m/s, Period of revolution at maximum speed = 0.016 s

Example – 11:

A spherical body of mass 1 kg and diameter 2 cm rotates in a horizontal circle at the end of a string 1.99 m. long. What is the tension in the string when the speed of rotation is 6 revolutions in 1.5 s?

Given: Mass of the body = m = 1 kg, diameter of sphere = d = 2cm = 0.02 m. Radius of sphere = 0.01 m, radius of circular path = r = 1.99 + 0.01 = 2m, No. of revolutions = 6, time taken t = 1.5 s.

To find: Tension in string = F =?

Solution:

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

n = 6/1.5 = 4 rad/s

ω = 2πn = 2 x 3.14 x 4 = 25.12 rad/s

F = m r ω²

∴ F = 1 x 2 x (25.12)² = 1262 N

Ans: The tension in the string = 1262 N radially inward.

Example – 12:

A spherical bob of diameter 3 cm having a mass 100 g is attached to the end of a string of length 48.5 cm. Find the angular velocity and the tension in the string, if the bob is rotated at a speed of 600 r.p.m. in a horizontal circle. If the same bob is now whirled in a vertical circle of the same radius, what will be the difference in the tensions at the lowest point and the highest point?

Given: Mass of bob = m = 100 g = 0.1 kg, Radius of circular path = r = 48.5 cm + 1.5 cm = 50 cm = 0.5 m, rpm = N = 600 r.p.m.,

To Find: Angular velocity = ω=? Tension in the string = F = ?

Solution:

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

The difference in the tensions at the lowest point and the highest point

= 6mg = 6 x 0.1 x 9.8 = 5.88 N

Ans: Angular speed = 62.84 rad/s, The tension in string =.197.2 N, The difference in the tensions at the lowest point and the highest point is 5.88 N

Example – 13:

A body of mass 2 Kg is tied to the end of a string of length 1.5 m and revolved about the other end (kept fixed) in a horizontal circle. If it makes 300 rev/min, calculate the linear velocity, the acceleration and the force acting upon the body.

Given: Mass of the body = m = 2 kg, radius of circular path = r = 1.5 m, Revolutions per minute = N = 300 r.p.m.,

To find: Linear speed = v =? Acceleration = a = ?, Force = F = ?

Solution:

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

Ans: Linear speed of body = 47.13 m/s; acceleration of body = 1480 m/s2;  Force acting on body = 2958N radially inward

Example – 14:

A body of mass 20 g rests on a smooth horizontal plane. The body is tied by a light inextensible string 80 cm long to a fixed point in the plane. Find the tension in the string if the body is rotated in a circular path at 30 rev/min. What is the force experienced by a fixed point?

Given:  Mass of the body = m = 20 g = 0.020 kg, radius of circular path = r = 80 cm = 0.8 m, rpm = N = 30 r.p.m.,

To find: Tension in string = F =?

Solution:

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

Ans: Tension in the string = 0158 N radially inward. The fixed point experiences a force of 0.158 N Radially outward.

Example – 15: 

How fast should the earth rotate about it axis so that the apparent weight of a body at the equator be zero? How long would a day be then? Take the radius of the earth = 6400 km.

Given:  Radius of earth = R = 6400 km = = 6.4 x 10⁶ m.,

To find: Angular speed of earth = ω =?, period of earth = T = ?

Solution:

As the apparent weight of the body is zero. The centrifugal force and the weight of the body are equal in magnitude. Let m be the mass of the body.

Ans: The angular speed of earth then = 1.24 x 10^-3 rad/s

T = 2π/ ω = (2 x 3.142)/( 1.24 x 10^-3)  = 5077 s

Ans: The duration of the day then = 5077 s

Example – 02:

An object of mass 100 g moves around the circumference of a circle of radius 2m with a constant angular speed of 7.5 rad/s. Compute its linear speed and force directed towards centre.

Solution:

Given: mass of thre body = m = 100 g = 0.1  kg, Radius of circular path = r = 2 m,  Angulae speed = ω = 7.5 rad/s,

To find: Linear speed = v  = ?, Centripetal force = F = ?

v = r ω = 2 x 7.5 = 15 m/s

F = m r ω²

∴  F = 0.1 x 2 x (7.5)² = 11.25 N

Ans: Linear speed = 15 m/s, Centripetal force = 11.25 N radially inward.

Example – 16:

An electron of mass 9 x 10^-31 kg is revolving in a stable orbit of radius 5.37 x 10^-11 m.  If the electrostatic force of attraction between electron and proton is 8 x 10^-8 N. Find the velocity of the electron.

Given: Mass of electron = m = 9 x 10^-31 kg, r = 5.37 x 10^-11 m, F = 8 x 10^-8 N

To find: velocity of electron = v =?

Solution:

The necessary centripetal force is given by electrostatic attraction between an electron and a proton.

Ans: The velocity of the electron is 2.185 x 10⁶ m/s

Example – 17:

A bucket containing water is tied to one end of a rope 8 m long and rotated about the other end in a vertical circle. Find the minimum number of rotations per minute in order that water in the bucket may not spill? (g = 9.8 m/s²)

Given:  Radius of circular path = r = 8 m, g = 9.8 m/s²,

To find: rpm = N =?

Solution:

At the highest point, the centrifugal force and the weight of water and bucket are equal in magnitude. Let m be the mass of the water and bucket.

mg = m r ω²

∴   g = r ω²

∴   ω² =   g /r = 9.8/8 = 1.225

∴   ω=   1.107 rad/s

ω= 2πN / 60

∴   N = 60ω/2π = (60 x 1.107) / (2 x 3.142) = 10.57

Ans: Max. No. of revolutions per minute = 10.57

Example – 18:

A bucket containing water is tied to one end of a rope 0.75 m long and rotated about the other end in a vertical circle. Find the speed in order that water in the bucket may not spill? Also, find the angular speed. (g = 9.8 m/s²)

Given: Radius of circle = r = 0.75 m, g = 9.8 m/s²

To find: linear speed = v =? angular speed = ω= ?

Solution:

At the highest point, the centrifugal force and the weight of water and bucket are equal in magnitude.

Let m be the mass of the water and bucket.

Ans: The linear velocity of bucket = 2.711 m/s, Angular speed of bucket = 3.615 rad/s

Previous Topic: Theory of Centripetal Force and Centrifugal Force

Next Topic: More Problems on Centripetal Force

Science > Physics > Circular Motion > Numerical Problems on Centripetal Force – 01

The post Numerical Problems on Centripetal Force – 01 appeared first on The Fact Factor.

In this article, we shall study the concept of centripetal force and centrifugal force, their expressions, and examples.

Centripetal Force:

A force which acts on a body performing circular motion and acts along the radius of the circular path and directed towards the centre of the path is called centripetal force.

Example: When a stone tied to one end of a string is whirled horizontally in a circle, there is an inward force exerted by the string on the stone called tension. The tension in the string provides necessary centripetal force for circular motion.

Without centripetal force, circular motion is not possible. If centripetal force vanishes at some instant, then the body ceases to move in the circular path and flies along the tangent at that point to the circular path.

On the surface of the earth, the centripetal force is maximum at the equator and zero on the poles.

Expression For Centripetal Force:

If ‘m’ is the mass of the particle performing circular motion then the magnitude of the centripetal force is given by

Centripetal force = mass  x  centripetal acceleration.

It is a vector quantity and always directed towards the centre of the circle. In vector form

Characteristics of Centripetal Force:

• It is a real force because it is provided by gravitational, electromagnetic or nuclear interaction.
• It arises in an inertial frame of reference. The inertial reference frame is that frame of reference which is moving with uniform velocity w.r.t. earth.
• It is always directed towards the centre of the circular path.
• Without it, the circular motion is not possible.
• The work done by the centripetal force is zero because the displacement of the particle (tangential) is perpendicular to the direction of the centripetal force (radial) i.e. there is no displacement in the direction of the centripetal force.
• The torque produced by the centripetal force at the center of the circular path is zero.
• The direction of the centripetal force does not depend on the sense of rotation of the body.

Notes:

The kinetic energy of a body performing a circular motion in terms of centripetal force is

K.E. = F.r/2

The linear momentum of a body performing a circular motion in terms of centripetal force is

p = F.r / v

Change in the velocity of a body performing U.C.M. and when it moves through angle θ is

dv = 2 sin(θ/2)

Examples of Centripetal Force:

• When a stone tied to one end of a string is whirled horizontally, there is an inward force exerted by the string on the stone called tension. This force provides necessary centripetal force for circular motion.

• Satellite revolves around the earth in a circular orbit. Necessary centripetal force is provided by the gravitational force of attraction between satellite and earth.

• When a vehicle moves around a horizontal circular road, the centripetal force for circular motion is provided by the frictional force between the road and the wheels.

• In an atom, an electron moves around the nucleus in an orbit. The centripetal force required for the motion of the electron is provided by the electrostatic force of attraction between the negatively charged electron and positively protons.

• When a coin is kept on rotating disc necessary centripetal force is provided by friction between the coin and the surface of the rotating disc.

• In a circus, when the motorcyclist moves in a vertical circle in the sphere of death, at the highest point of the vertical circle, the necessary centripetal force is provided by the weight of the motorcycle and motorcyclist.

• When a body is performing looping a loop, at the highest point of the vertical circle, the necessary centripetal force is provided by the weight of the body. For looping a loop the speed of motorcyclist at the lowest point should be greater than √5gr  and at the highest point, the speed should be greater than √gr.

Centrifugal Force:

The imaginary force which acts on the particle performing a circular motion in the direction away from the centre along the radius of the circular path having the same magnitude as that of centripetal force is called centrifugal force.

Example: When moving a car takes a turn along a horizontal curved road, persons in the car experience a force in the outward direction. This force is centrifugal force

It has the same magnitude and opposite direction to that of centripetal force.

It should be noted that the centrifugal force is not real force’ since it does not arise due to either gravitational or electrostatic or nuclear interaction.

Centrifugal Force is a Pseudo Force:

• Centrifugal force is not real force since it does not arise due to either gravitational or electrostatic or nuclear interaction.
• Centrifugal force has no independent existence. It comes to play with the action of the centripetal force. It does not arise due to either gravitational or electrostatic or nuclear interaction.
• Centrifugal force acts in the non-inertial reference frame
  Hence centripetal force is a pseudo force.

Theoretical Explanation:

Case – I:

Let us consider body P is tied to a string and lying on a frictionless turntable and performing uniform circular motion along with turntable in a clockwise sense as shown. The necessary centripetal force is provided by the tension in the string. Let A be an observer standing between the centre of the table and the particle at P. Let P₁, P₂, P₃ be different positions of the particle on the circular path. Let A₁, A₂, A₃ be corresponding positions of observer A. Observer B is standing near table B outside the turntable. As observer B is stationary w.r.t. the earth he is an inertial frame of reference, while observer A is acted upon by centripetal acceleration. Thus observer A is in acceleration w.r.t. earth, observer A is a non-inertial frame of reference.

Let us check the opinion of observer B about the motion of particle P. His observation is that he is stationary and the particle P is in a circular motion and a centripetal force acts on it. The necessary centripetal force is provided by the tension in the string. There is a centripetal acceleration and is provided by centripetal force. Thus his observations by B are in accordance with Newton’s laws of motion.

Now let us check observation of observer A. He says that he is stationary and every instant particle is in front of me at a constant distance, hence particle is stationary. There is tension in the string. Thus force is acting on particle directed towards centre but particle is stationary. By Newton’s second law is F = ma. Thus if there is a force, the state of motion of the body should change. But as per observer A, the state of motion of the body is not changing in spite of the action of the force. Thus Newton’s second law of motion fails. To correct A we have to assume some imaginary force (centrifugal force) which is equal to the tension in the string acts radially outward i.e. opposite and equal to centripetal force. Under the action of these equal and opposite forces acting on the body, the body does not change its change of state of motion.

We can see that the concept of centrifugal force (imaginary) is not required by B but is required by observer A who is the noninertial frame of reference.

Case- II:

Now let us assume the string breaks when the particle is at position P₁. The particle is thrown off with uniform velocity along the tangent to the circular path at position P₁.

Let us check the new opinion of observer B about the motion of particle P. His observation is that I am he is stationary and the body is moving along a straight line with uniform velocity and no force acts on it. It is in accordance with Newton’s first law of motion.

Now let us check new observation of observer A. He says that he is stationary and the body P moves outward with constant acceleration but no force acts on it. Thus Newton’s second law of motion fails. To correct A we have to assume some imaginary force (centrifugal force) which is equal to the tension in the string acts radially outward. Under the action of this force, the body moves radially outward with acceleration.

Thus in both the case we have to assume the existence of imaginary or pseudo force to correct observer A, which is non-inertial reference frame..

Centripetal Force and Centrifugal Force Do Not Constitute Action-Reaction Pair:

A statement ” A particle moving uniformly along a circle experiences a force directed towards the centre (centripetal force) and an equal and opposite force directed away from the centre (centrifugal force). The two forces keep the particle in equilibrium” is completely wrong.

Because the particle in a circular motion is not in equilibrium because a net force (centripetal force) acts towards the centre which is required to change the direction of particle continuously. Centripetal force is a real force.

The centrifugal force is an imaginary force is required by an observer moving with the particle. The observer is a non-inertial reference frame and for the observer, the particle is at rest.

Centripetal force is required for circular motion. Centrifugal force does not have an independent existence. It comes into play when the centripetal force starts acting. Thus the centripetal force and centrifugal force do not constitute the action-reaction pair.

Characteristics of Centrifugal Force:

• It is an imaginary force or pseudo force.
• It is experienced in non – inertial frame of reference.
• The magnitude of the centripetal force is equal to that of the centripetal force.
• It is always directed away from the centre of the circle along the radius
• it is directed opposite to the centripetal force.
• Centrifugal force doesn’t have an independent existence.

Examples of Centrifugal Force:

• When a moving car along a horizontal curved road takes a turn, persons in the car experience a force in an outward direction. This force is centrifugal force.

• When the horizontal merry go round rotates about the vertical axis the chairs are pulled out due to centrifugal force.

• When a stone is whirled in a circle, we feel that stone is pulling our hand because of centrifugal force.

• The earth is flattened at the poles and bulged at the equator because the centrifugal force acting on the particles on the equator is maximum.

• The drier of a washing machine acts on the principle of centrifugal force. Water particles from wet clothes are thrown outward due to centrifugal force acting on them. Drier in the washing machine consists of a cylindrical vessel with perforated walls. As the cylindrical vessel is rotated fast, centrifugal force acts on the water particles of wet clothes. Under the action of this centrifugal force, water particles are forced out of the perforations, thereby drying of the clothes.

• When a pilot moves the plane in looping a loop, he does not fall down because his weight is balanced by the centrifugal force acting on it.

• A coin kept slightly away from the center of rotating gramophone disc slips off towards the edge of the disc at a particular speed. This is due to centrifugal force acting on the coin.

• A bucket full of water is rotated in a vertical circle at a particular speed, so that water does not fall. This is because the weight of the water is balanced by centrifugal force.

• The centrifuge is a device which is used for separating heavier particles and light particles and works on the principle of centrifugal force. In the centrifuge, the tube containing liquid along with the suspended particles is whirled in a horizontal circle. Dense particles are acted upon by a centrifugal force. Hence they get accumulated at the bottom, which is outside while rotating. This is because buoyant force towards the center is greater for the lighter particle.

• Centrifugal governor works on the principle of centrifugal force. When the speed of the vehicle increases beyond a certain set limit, the fly balls under the action of centrifugal force, fly away from the axis of rotation. A bell and crank arrangement attached to it reduces the flow of fuel to the engine. Thus the speed of the vehicle is governed.

• Centrifugal pump works on this principle. It sucks water through a pipe from the reservoir and throws outward under the action of centrifugal action in the involute casing and thus produces a draft and water is lifted up.

• When wheels are rotating mud stuck to the wheels and then under the action of centrifugal force are thrown tangentially away from the wheels. These mud particles tarnish the vehicle and spray the mud on the following vehicles. At the same time, these mud article may degrade the performance of the brake. When mudguards are used the mud particles are stopped dead and fall down under gravity.

Distinguishing Between Centripetal Force and Centrifugal Force:

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Next Topic: Numerical Problems on Centripetal Force

Science > Physics > Circular Motion > Centripetal and Centrifugal Force

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In this article, we shall study the concept of centripetal acceleration and expression for it. Centripetal acceleration is also called radial acceleration.

Accelerated Motion:

Velocity is a vector quantity which has magnitude as well as direction. Either magnitude or direction or both change the motion is nonuniform motion

• Case – I: When direction remains the same, but magnitude changes e.g. Motion under gravity.
• Case – II: When magnitude remains the same, but the direction changes continuously e.g. Uniform circular motion:
• Case – III:  When both the magnitude and direction changes continuously e.g. Projectile motion

The acceleration of the body performing circular motion which is directed towards the centre of the circular path along the radius is called a radial acceleration or centripetal acceleration.

Characteristics of Centripetal Acceleration:

• It is the acceleration of a particle performing the circular motion, which is directed towards the centre of the circular path along the radius.
• It is always directed towards the centre of the circular path along the radius.
• The direction of centripetal acceleration changes continuously.
• For U.C.M. the magnitude of the centripetal acceleration is constant.
• It is denoted by the letter ‘a’. Its S.I. unit is metre per square second (m s^-2). Its dimensions are [MºL¹T ^-2].
• The magnitude of centripetal acceleration is given by

The Expression for Centripetal Acceleration

Geometric Method:

The magnitude of the velocity of a particle performing uniform circular motion is constant but its direction changes constantly. Hence the particle in circular motion has linear acceleration.

Let us consider a particle performing uniform circular motion with a linear velocity of magnitude ‘v’ and angular velocity of magnitude ‘ω’ along a circle of radius ‘r’ with centre O in an anticlockwise sense (moving from initial position A to final position B)as shown in the figure.

Triangle QBR is drawn for the velocities of the particles at A and B.

Now by definition,

The triangles AOB and RBQ are similar. Hence ∠ QBR = δθ
For smaller angular displacement δθ,

Substituting in equation (1), we get

This is the expression for the acceleration of particle performing the uniform circular motion.

This acceleration is directed towards the centre of the circular path along the radius. This acceleration is called radial acceleration or centripetal acceleration. In vector form, centripetal acceleration can be given as

The negative sign indicates that centripetal acceleration is oppositely directed to that of radius vector i.e. directed towards the centre of the circle along the radius.

Note: the alternate formula is

Centripetal acceleration is also called radial acceleration. It always acts along the radius towards the centre of the circular path. The angle between radius vector and centripetal acceleration is π radian or 180°.

Cartesian Method or Calculus Method:

Let us consider a particle performing uniform circular motion with a linear velocity of magnitude v and angular velocity of magnitude ω along a circle of radius r with centre O in an anticlockwise sense. Let the particle moves from A to P, in time ‘t’ Such that ∠ POA = θ.

But for uniform Circular motion θ = ω t. Thus, ∠ POA = ω t

Let us draw seg PM perpendicular to seg OA. Thus, ∠ POM = ω t

The radius vector at time t at P is given by

Substituting these values in equation (1)

 The linear velocity of a particle can be obtained by differentiating equation (2) w.r.t. time t.

The linear acceleration of a particle can be obtained by differentiating equation (3) w.r.t. time t.

From equations (1) and (4) we have

Thus the magnitude of the acceleration is v²/r and its direction is along the radius and the negative sign indicates that it is opposite to the radius vector i.e. the acceleration is directed towards the centre of the circular path. This acceleration is called the centripetal acceleration.

Relation between linear velocity (v) and angular velocity (ω) by calculus method:

From the equation (3) we get the instantaneous velocity as

Thus the linear velocity of a particle performing U.C.M. is radius times its angular velocity.

Angle between linear velocity (v) and radius vector by calculus method:

From the equations (2) and (3) we have

Thus the scalar or dot product of the velocity of the particle performing U.C.M. and the radius vector is zero. Hence the angle between the velocity of the particle performing U.C.M. and the radius vector.

Acceleration of a Body Performing Circular Motion has Two Components:

The relation between linear velocity and angular velocity In vector form is written as

Thus the acceleration of the body performing circular motion has two components. one along the radius of the circular path towards the centre and is called centripetal acceleration and another tangential component. The net acceleration of the body is given by

Numerical Problems:

Example – 1:

The length of an hour hand of a wristwatch is 1.5 cm. Find the magnitudes of following w.r.t. tip of the hour hand a) angular velocity b) linear velocity c) angular acceleration d) radial acceleration e) tangential acceleration f) linear acceleration

Given: r = 1.5 cm = 1.5 x 10^-2 m, For hour hand T_H = 12 hr = 12 x 60 x 60 sec

To Find: Angular velocity = ω = ?, linear velocity = v = ?, angular acceleration = α = ?, radial acceleration = a_r = ?, tangential acceleration = a_T =?, linear acceleration a = ?

Solution:

Now v = r ω = 1.5 x 10^-2 x 1.454 x 10^-4 = 2.181 x 10^-6 m/s

Tip of hour hand performs uniform circular motion

∴  α = 0 and a_T = 0

Radial acceleration is given by

Now linear acceleration

Ans: Angular speed = 1.454x 10^-4 rad/s, Linear speed =2.181 x 10^-6 m/s,
Angular acceleration = 0, Radial acceleration = 3.171 x 10^-10 m/s².
Tangential acceleration = 0, Linear acceleartion = 3.171 x 10^-10 m/s².

Example – 2:

To simulate the acceleration of high-speed fighter plane, astronauts are spun at the end of a long rotating beam of radius 5 m. Find the angular velocity required to generate a centripetal acceleration 3 times the acceleration due to gravity.

Given:  r = 5 m, a= 3g, g = 9.8 m/s².

To find:  ω =?

Solution:

Ans: Required angular velocity = 2.425 rad/s

Example – 3:

To simulate the acceleration of large rockets, astronauts are spun at the end of a long rotating beam of radius 9.8 m. Find the angular velocity required to generate a centripetal acceleration 8 times the acceleration due to gravity.

Given:  r = 9.8 m, a= 8g, g = 9.8 m/s².

To find:  ω = ?

Solution:

Ans: Required angular velocity = 2.828 rad/s

Example – 4:

A body of mass 2 Kg is tied to the end of a string of length 1.5 m and revolved around the other end (kept fixed) in a horizontal circle. If it makes 300 rev/min, calculate the linear velocity, the acceleration and the force acting upon the body.

Given:  m = 2 kg,  r = 1.5 m,  N = 300 r.p.m.,

To find: v = ?, a = ?, F = ?

Solution:

Centripetal force F = ma = 2 x 1479 = 2958 N

Ans: Linear speed of body = 47.13 m/s, Acceleration of body = 1480 m/s²,

Force acting on body = 2958 N radially inward.

Example – 5:

The tangential acceleration of a body performing circular motion is 29.48 m/s² and its linear acceleration is 52.3 m/s². Find its radial acceleration

Given: a_T = 29.48 m/s² and a = 52.3 m/s².

To Find: a_r =?

Solution:

Ans: The radial acceleration is 43.1 m/s²

Example – 6:

A particle is revolving in a circle. Its angular speed increases from 2 rad/s to 40 rad/s in 19 s. The radius of a circle is 20 cm. Compare the ratios of centripetal acceleration to tangential acceleration, at end of 19s.

Given: ω₁ = 2 rad/s, ω₂ = 40 rad/s, t = 38 s, r = 20 cm = 0.2 m.

To Find: a_r : a_t =?

Solution:

Now a_t = r a = 0.2 x 2 = 0.4 m/s²

Centripetal acceleration at end of 19 s

Ar = r ω ² = 0.2 x (40)² = 320 m/s²

Ans: the required ratio a_r : a_t =320: 0.4 = 800:1

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Next Topic: Concept of Centripetal and Centrifugal Force

Science > Physics > Circular Motion > Centripetal Acceleration

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