Problems on Centripetal Force 01

Physics Chemistry Biology Mathematics
Science > Physics > Circular MotionYou are Here

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 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 -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 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 – 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 ω2

∴  F = 1 x 2 x (25.12)= 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/s2)
  • 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 106 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 10m/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)
  • Solution:
  • Given:  Radius of circular path = r = 8 m, g = 9.8 m/s2,
  • 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 ω2

∴   g = r ω2

∴   ω=   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)
  • Solution:
  • Given: Radius of circle = r = 0.75 m, g = 9.8 m/s2
  • 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 MotionYou are Here
Physics Chemistry Biology Mathematics

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