# Physics – Circular Motion: Past Board Exam Questions

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### March – 2013

#### Question – 01:

• A car of mass 1500 kg rounds a curve of radius 250 m at 90 km/h. Calculate the centripetal force acting on it. (2 M)
• Given: mass of car = m = 1500 kg,radius of curve =  r = 250 m, speed of car = v = 90 km/h = 90 x 5/18 = 25 m/s
• To find: Centripetal force = F = ?
• Solution:

Centripetal force = F = mv2/r = 1500 x (25)2/ 250 = 3750 N

Ans: Centripetal force = 3750 N

#### Question – 02:

• A particle of mass ‘m’ just completes the vertical circular motion. Derive the expression for the difference in tensions at the highest and the lowest points. (3M)
• Consider a small body of mass ‘m’ attached to one end of a string and whirled in a vertical circle of radius ‘r’. In this case, the acceleration of the body increases as it goes down the vertical circle and decreases when goes up the vertical circle. Hence the speed of the body changes continuously. It is maximum at the bottommost position and minimum at the uppermost position of the vertical circle. Hence the motion of the body is not uniform circular motion. Irrespective of the position of the particle on the circle, the weight ‘mg’ always acts vertically downward. • Let ‘v’ be the velocity of the body at any point P on the vertical circle. Let L be the lowest point of the vertical circle. Let ‘h’ be the height of point P above point L. let ‘u’ be the velocity of the body at L. By the law of conservation of energy

Energy at point P = Energy at point L This is an expression for the velocity of a particle at any point performing a circular motion in a vertical circle.

Consider the centripetal force at point P Substituting in equation (2) This is the expression for the tension in the string.

When the body is at the lowermost position i.e. body is at L) (h = 0) When the body is at the uppermost position i.e. body is at H (h = 2r) Thus the difference in tensions at the two positions • Thus the tension in the string at the lowest point L is greater than the tension at the highest point H by six times the weight of the body.

### October – 2013

#### Question – 01:

• A racing car completes 5 rounds of a circular track in 2 minutes. Find the radius of the track if the car has a uniform centripetal acceleration of π2 ms-2. (2 M)
• Given: ac = π2 ms-2, t = 2 minutes = 2 x 60 = 120 s, Number of rounds = n = 5
• To find: r = ?
• Solution:

The velocity of the car is given by Ans: The radius of the circular track is 144 m

#### Question – 02:

• In a conical pendulum, a string of length 120 cm is fixed at a rigid support and carries a mass of 150 g at its free end. The mass is revolved in a horizontal circle of radius 0.2 m around the vertical axis. Calculate the tension in the string. (3 M)
• Given: Length of pendulum = l = 120 cm = 1.2 m, mass = m = 150 g = 0.150 kg , radius of circle = r = 0.2 m
• To find: θ =?

### From figure For equilibrium
Total upward force = Total downward force T = 1.49 N
Ans: Tension in the string is 1.49 N

### March – 2014

#### Question – 01:

• For a particle performing uniform circular motion v. = ω x r  Obtain an expression for linear acceleration of particle performing the non-uniform circular motion. (2 M)

The relation between linear velocity and angular velocity In vector form is written as • Thus the acceleration of the body performing non-uniform 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 #### Question – 02:

• A stone of mass 1 kg is whirled in a horizontal circle attached at one end of 1m long string. If a string makes an angle of 30° with the vertical. calculate the centripetal force acting on it. (3 M)
• Given: Length of string = l = 1 m, mass of stone = m = 1 kg, angle with vertical = θ = 30°
• To find: Centripetal force = F =?
• Solution:

### For equilibrium
Total upward force = Total downward force Ans: Centripetal force is 5.66 N

### October – 2014

#### Question – 01:

• Draw a diagram showing all components of forces acting on a vehicle moving on a curved bank road. Write the necessary equation for maximum safe speed and state the significance of each term involved in it. (3 M)

The various forces acting on a  vehicle moving on a curved bank road The free body diagram of a car is as follows The equation for maximum safe speed on the banked road is Where r = radius of circular path

g = acceleration due to gravity

θ = angle of banking

μs = coefficient of static friction

#### Question – 02:

• A stone of mass 5 kg, tied to one end of a rope of a length 0.8 m, is whirled in a vertical circle. Find the minimum velocity at the highest point and at the midway point. (2 M)
• Given: Mass of stone = 5 g, radius of circle = 0.8 m
• To Find: Minimum velocity at highest point and midway point = ?
• Solution:

Minimum velocity at highest point is given by

v = √g r  = √9.8 x 0.8   =  √7.84  = 2.8 m/s

Minimum velocity at midway point is given by

v = √3g r  = √3 x 9.8 x 0.8   =  √23.52  = 4.85 m/s

Ans: Minimum velocity at highest point is 2.8 m/s and

Minimum velocity at midway point is 4.85 m/s

### March – 2015

#### Question – 01:

• In circular motion assuming , obtain an expression for the resultant acceleration of a particle in terms of tangential and radial component. (2 M)

The relation between linear velocity and angular velocity In vector form is written as • Thus the acceleration of the body performing non-uniform 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 #### Question – 02:

• The period of a conical pendulum in terms of its length (l), semi-vertical angle (θ) and acceleration due to gravity (g) is ………
a) b) c) d) (1 M)

#### Question – 03:

• The spin dryer of a washing machine rotating at 15 r.p.s slows down to 5 r.p.s after making 50 revolutions. Find the angular acceleration. (2 M)
• Given: Initial angular speed = n1 = 15 r.p.s., final angular speed = n2 = 5 r.p.s., No. of revolutions = 50
• To Find: Angular acceleration = α = ?

Angular displacement = θ = No.of revolutions x 2π

∴ Angular displacement = θ =50 x 2π = 100π rad

Initial angular speed = ω1 = 2πn = 2 x π x 15 = 30π rad/s

Final angular speed = ω2 = 2πn = 2 x π x 5 = 30π rad/s

We have   ω22 = ω12  + 2αθ

∴   (10π)2 = (30π)2  + 2 x α x 100π

∴   100π2 = 900π2  + α x 200π

∴   – 800π2 = α x 200π

∴   α = – 800π2 / 200π = 4π rad/s2

∴   α =  4 x 3.142 = 12.57  rad/s2

Ans: Angular acceleration is 12.57  rad/s2

### October – 2015

#### Question – 01:

• Draw neat labelled diagram of conical pendulum showing all forces acting on it. State the expression for the periodic time in terms of length. (2M)

#### Question – 02:

• A stone of mass 100 g attached to a string of length 50 cm is whirled in a vertical circle by giving velocity at the lowest point at 7 m/s. Find the velocity at the highest point. (3M)

### March – 2016

#### Question – 01:

• With usual notations prove that (2M)

#### Question – 02:

• Define linear S.H.M. and show that S.H.M. is a projection of U.C.M. on any diameter. (3M)

### July – 2016

#### Question – 01:

• The difference in tensions in the string at the lowest and highest points in the path of the particle of mass ‘m’ performing vertical circular motion is ….
a) 2mg b) 4mg c) 6mg d) 8mg (1M)

#### Question – 02:

• Draw a neat labelled diagram showing the various forces and their components acting on a vehicle moving along a curved banked road. (2M)

#### Question – 03:

• A stone of mass 2 kg is whirled in a horizontal circle attached at the end of 1.5 m long string. If the string makes an angle of 30O with the vertical, compute its period.

### March – 2017

#### Question – 01:

• Explain the concept of centripetal force. (2 M)
• Centripetal force is a force which is acting on a body performing circular motion and is acting along the radius of the circular path and directed towards the centre of the circle.
• 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.
• Without it, the circular motion is not possible.
• It is real force and it arises in an inertial frame of reference.

#### Question – 02:

• A vehicle is moving on a circular track whose surface is inclined towards the horizon at an angle of 10°. The maximum velocity with which it can move safely is 36 km / hr. Calculate the length of the circular track. [π = 3.142] (3M)
• Given: angle of banking = θ = 10°, the maximum velocity of car = v = 36 kmph = 36 x 5/18 = 10 m/s, g = 9.8 m/s2.
• To find: length of circular tack =?
• Solution:

maximum speed on the banked road is given by

v = √r g tanθ

∴ v2 = r g tanθ

∴  r = v2/g tanθ = (10)2/ (9.8 x tan 10°) = 100 /(9.8 x 0.6484)

∴  r =  15.74 m

Length of circular track = circumference = 2πr = 2 x 3.142 x 15.74 = 98.9 m

Ans: Length of circular track is 98.9 m

### July – 2017

#### Question – 01:

• Draw neat, labelled diagram showing different forces acting on a vehicle moving along a

#### Question – 02:

• A body of mass ‘m’ performs uniform circular motion along a circular path of radius ‘r’ with
velocity ‘v’. If its angular momentum is L, then the centripetal force acting on it is _______.

### March – 2018

#### Question – 01:

• A particle of mass m performs vertical motion in a circle of radius r. Its potential energy at the highest point is _______. (g is acceleration due to gravity)
(A) 2mgr        (B) mgr    (C) 0   (D) 3 mgr  (1 M)
• A particle of mass m performs vertical motion in a circle of radius r. Its potential energy at the highest point is 2mgr.

#### Question – 02:

• Distinguish between centripetal and centrifugal force. (2M)
 Centripetal Force Centrifugal Force Centripetal force is a force which is acting on a body performing circular motion and is acting along the radius of the circular path and directed towards the centre of the circle. The imaginary (pseudo) 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 as centrifugal force. It is a real force. It is an imaginary force or a pseudo force. It arises in an inertial frame of reference. It is experienced in non – inertial frame of reference. It is always directed towards the centre of the circular path. It is always directed away from the centre of the circle along the radius. Without it, the circular motion is not possible. Centrifugal force doesn’t have an independent existence. Example: The moon or a Satellite revolves around the earth in circular orbit. Necessary centripetal force is provided by the gravitational force of attraction between satellite and earth. Example: When moving 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

#### Question – 03:

• A flat curve on a highway has a radius of curvature 400 m. A car goes around a curve at a speed of 32 m/s. What is the minimum value of coefficient of friction that will prevent the car from sliding? (g = 9.8 m/s2)  (2M).
• Given: radius of curve = r = 400 m, speed of car = v = 32 m/s, g = 9.8 m/s2
• To find: Coefficient of friction = μ = ?
• Solution:

The safe velocity on unbanked road is given by

v = √μ r g

μ = v2/rg = (32)2/(400 x 9.8) = 0.2612

Ans: The minimum value of coefficient of friction is 0.2612

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