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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. 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?
**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

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 – 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.

mg = m r ω^{2}

∴ g = r ω^{2}

∴ ω^{2 }= 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/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

Science > Physics > Circular Motion > You are Here |

Physics |
Chemistry |
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Mathematics |