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**Example -1:**

- 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.
**Solution:****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 = ?,

F = m r ω^{2}

∴ 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 -2:

- 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
**.** **Solution:****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 = ?,

F = m r ω^{2}

∴ 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 – 3:

- 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.
**Solution:****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 = ?,

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 – 4:

- An object of mass 0.5 kg is whirled in a horizontal circle of radius 20 cm. Calculate the maximum number of revolution per minute, so that the string does not break. Breaking tension of the string is 9.86 N.
**Solution:****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 = ?

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 – 5:

- 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.
**Solution:****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 = ?

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

#### Example – 6:

- 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.
**Solution:****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 = ?

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 – 7:

- 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 – 8:

- 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?
**Solution:****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 = ?,

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 ω^{2}

∴ F = 1 x 2 x (25.12)^{2 }= 1262 N

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

#### Example – 9:

- 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.
**Solution:****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 = ?

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

**Ans: **Angular speed = 62.84 rad/s, The tension in string =.197.2 N

#### Example – 10:

- 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.
**Solution:****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 = ?

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 – 11:**

- 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.
**Solution:****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 = ?,

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 – 12:

- 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/s
^{2})

**Solution:****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 = ?

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 – 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.
**Solution:****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 = ?

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 the fixed point?
**Solution:****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 = ?

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 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.
**Solution:****Given:**Radius of earth = R = 6400 km = = 6.4 x 10^{6}m.,**To find:**Angular speed of earth = ω = ?, period of earth = T = ?- 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

The duration of the day then = 5077 s

#### 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. **Solution:****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 = ?

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

**Ans: **The velocity of the electron is 2.185 x 10^{6 }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/s
^{2}) **Solution:****Given:**Radius of circular path = r = 8 m, g = 9.8 m/s^{2},**To find:**rpm = N =?- 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:** Max. No. of revolutions per minute = 94.8

#### 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/s
^{2}) **Solution:****Given:**Radius of circle = r = 0.75 m, g = 9.8 m/s^{2}**To find:**linear speed = v = ?, angular speed = ω= ?

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

#### 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.
**Solution:****Given:**Radius of circle = r = 19.5 + 0.5 = 20 m,**To find:**velocity of car = v = ?

At the highest point, the centrifugal force and weight of 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.
**Solution:****Given:**Radius of circle = r = 30 + 0.5 = 30.5 m,**To find:**velocity of car = v = ?

** **

** **At the highest point, the centrifugal force and weight of 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.
**Solution:****Given:**Radius of hollow sphere = r = 3 m,**To find:**Minimum speed required = v = ?, Angular speed = ω = ?

At the highest point, the centrifugal force and weight of car 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.
**Solution:****Given:**Radius of sphere = r = 12.8 m,**To find:**Minimum speed = v = ?

At the highest point, the centrifugal force and weight of car 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.
**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 ω^{2}

∴ 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.
**Solution:****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 = ?,

v = r ω

∴ ω = v/r = 25/250 = 0.1 rad/s

a = r ω^{2}

∴ a = 250 x (0.1)^{2 }= 2.5 m/s^{2}

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

**Ans: **Angular speed = 0.1 rad/s, Centripetal acceleartion = 2.5 m/s^{2} radially inward

Centripetal force = 5000 N radially inward.

#### 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.
**Solution:****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**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.
**Solution:****Given:**radius of circular path = r = 0.1 m, coefficient of friction = μ = 0.25, g = 9.8 m/s^{2}**To find:**Speed of disc = v = ?

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/s
^{2}) **Solution:****Given:**radius of circular path = r = 0.25 m, coefficient of friction = μ = 0.2, g = 9.8 m/s^{2}**To find:**Angular speed of disc = ω =?

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.
**Solution:****Given:**Radius of circular path = r = 1.5 cm = 0.015 m, rpm = N = 120 r.p.m., g = 9.8 m/s^{2}**To find:**coefficient of friction = μ =?

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

**Ans: **Coefficient of friction = 0.2415

#### Example – 29:

- 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/s
^{2}) **Solution:****Given:**Radius of circular path = r = 40 m, coefficient of friction = μ = 0.3, g = 9.8 m/s^{2}**To find:**Maximum speed = v = ?

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

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

#### Example – 30:

- 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.
**Solution:****Given:**Radius of circle = r = 2 cm = 0.02 m, rpm = N = 180 r.p.m., g = 9.8 m/s^{2}**To find:**coefficient of friction = μ = ?

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

**Ans: **The coefficient of friction = 0.724

#### Example – 31:

- 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 will the coin remain on the disc if its centre is 16 cm from the axis of rotation of the disc?
**Solution:****Given:**Initial radius of circle = r_{1}= 9 cm, initial rpm = N_{1}= 60 r.p.m., final radius of circle = r_{2}= 16 cm**To find:**Final rpm = N_{2}=?

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

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

**Example – 32:**

- 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.?
**Solution:****Given:**Initial radius of circle = r_{1}= 12 cm, initial rpm = N_{1}= 45 r.p.m., Final rpm = N_{2 }= 45 r.p.m.**To find:**final radius of circle = r_{2}=?

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

**Ans: **The max. displacement from centre = 5.33 cm.

#### 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 surface of the road is 0.2.
**Solution:****Given:**radius of curve = r = 100 m, coefficient of friction = μ = 0.2, g = 9.8 m/s^{2}**To find:**maximum speed = v = ?

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.
**Solution:****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/s^{2}**To find:**radius of curve = r = ?

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.
**Solution:****Given:**Speed of the car = v = 36 km/hr = 36 x 5/18 = 10 m/s ,r = 20m, g = 9.8 m/s^{2}**To find:**coefficient of friction = μ = ?

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

**Ans: **The coefficient of friction = 0.5102

#### Next Topic: Banking of a Road

#### Previous Topic: Concept of Centripetal Force and Centrifugal Force

Science > Physics > Circular Motion > You are Here |

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