# Physics – Circular Motion: Textbook Unsolved Problems

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#### Example – 01:

• Calculate the angular velocity and linear velocity of a tip of the minute hand of length 10 cm?
• Given: Period of minute hand = T = 60 min = 60 x 60 s, length of minute hand = r = 10 cm = 0.1 m
• To Find: Angular velocity = ω = ?, linear velocity = ?
• Solution:

Angular velocity is given by The linear velocity of the tip is given by

v = r w = 0.1 x 1.745 x 10-3 = 1.745 x 10-4 m/s

Ans: The angular speed and linear speed of the tip of the minute hand are 1.745 x 10-3 rad/s and  1.745 x 10-4 m/s respectively.

#### Example – 02:

• Propeller blades of an aeroplane are 2 m long. When the propeller is rotating at 1800 rev/min, compute the tangential velocity of the tip of the blade. Also, find the tangential velocity at a point on blade midway between tips and axis.
• Given: r = 2 m, Angular speed = N = 1800 r.p.m.
• To Find: vTip =? vMid =?
• Solution: Ans: The angular speed of the tip of blade = 376.8 rad/s, The angular speed of point midway = 188.4 rad/s

#### Example – 03:

• 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 ω2

∴   a = 250 x (0.1)= 2.5 m/s2

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

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

Centripetal force = 5000 N radially inward.

#### Example – 04:

• A bucket containing water is tied to one end of a rope and rotated about the other end in a vertical circle at arm’s length. Find the minimum speed at the top to ensure that no water spills out. Also, find the angular speed. (g = 9.8 m/s2) take, r = 0.75 m
• 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

#### Example – 05:

• A motorcyclist at a speed of 5 m/s is describing a circle of radius 25 m. Find his inclination with the vertical. What is the value of the coefficient of friction between the road and tyres?
• Solution:
• Given: Speed of motor cycle = v = 5 m/s, radius of circle = r = 25 m,  g = 9.8 m/s2 ,
• To find:  Angle of inclination = θ= ?,  Coefficient of friction = μ = ? Necessary centripetal force is provided by the friction between the road and tyres. Ans: Angle made with the vertical = 5°49’, The coefficient of friction = 0.1020

#### Example – 06:

• A stone weighing 1 kg is whirled in a vertical circle at the end of a rope of length 0.5 m. Find the tension in the string at a) lowest position b) mid position c) highest position
• Solution:
• Given: Radius of circle = r =  0.5 m,  mass of the body = m = 1 kg, g = 9.8 m/s2.
• To find: Tension at lowest point = TL = ?, Tension when string is horizontal = TM = ?, Tension at topmost point = TH = ?. Ans: Tension at the lowest position = 58.8 N,

Tension  at mid position  = 29.4 N,

Tension at the highest position = 0 N

#### Example – 07:

• An object of mass 0.5 kg attached to a string of length 0.5 m is whirled in a vertical circle at constant angular speed if the maximum tension in the string is 5 kg wt. Calculate the speed of the object and maximum number of revolutions it can complete in one minute
• Given: Mass of object = m = 0.5 kg, Radius of circle = r = 0.5 m, Tension in the string = T = 5 kg wt = 5 x 9.8 N.
• To Find: Speed of the object = ?, the maximum number of revolutions per minute = N =?
• Solution:

In vertical circle maximum tension is at the lowermost point  Ans: Speed of object = 6.64 m/s and maximum number of revolutions = 126.7 r.p.m.

#### Example – 08:

• A motor van weighing 4400 kg rounds a level curve of radius 200 m on the unbanked road at 60 km/hr. What should be the minimum value of the coefficient of friction to prevent skidding? At what angle the road should be banked for this velocity?
• Solution:
• Given: Mass of vehicle = m = 4400 kg, velocity of vehicle = v = 60 km/hr = 60 x 5/18 = 16.67 m/s , r = 200m,  g = 9.8 m/s2
• To Find:  Coefficient of friction = μ = ?, Angle of banking = θ = ?,

Necessary centripetal force is provided by the friction between the road and tyres. Ans: The coefficient of friction = 0.1418, The angle of banking = 8°4’

#### Example – 09:

• A string of length 0.5 m carries a bob of mass 100 g and with a period 1.41 s. Calculate the angle of inclination of the string with vertical and the tension in the string.
• Given: length of string = l = 0.5 m, mass of bob = m = 100 g = 0.1 kg, period = T = 1.414 s,
• To Find: Angle of inclination = q – ?, and Tension in string = F = ?
• Solution:

The time period of a simple pendulum is given by The tension in the string is given by Ans: Angle of inclination of string = 9°19′

Tension in string = 0.993 N

#### Example – 10:

• A pilot of mass 50 kg in a jet aircraft while executing a loop the loop with a constant speed of 250 m/s. If the radius of the circle is 5 km, compute the force exerted by seat on the pilot a) at the top of loop b) at the bottom loop.
• Solution:
• Given: mass of piolot = m = 50kg, radius of circle = r = 5 km = 5000 m,  velocity of plane = v = 250 m/s  ,
• To find: force at the top of loop = FT = ?, Force at bottom of loop = FB = ?,.
• At the topmost point, the centrifugal force acts vertically upward and the weight of the body acts vertically downward. Thus the net force on the pilot by the seat is given by • At the bottom-most point, both the centrifugal force and the weight of the body act vertically downward. Thus the net force on the pilot by the seat is given by Ans: The force exerted by the seat on the pilot at the topmost point is 135 N.

The force exerted by the seat on the pilot at the bottom-most point is 1115 N.

Example – 11:

• A ball is released from a height h along the slope and at the end of a slope moves along a circular track of radius R without falling vertically downwards. Determine the height h in terms of R. • Solution:
• As the ball is released from a height along the slope and moves along a circular track thus it is looping a loop or it has sufficient velocity at the lowest point to loop a loop.

At lowest point, v = √5gr

By the law of conservation of energy

Potential energy at P = Kinetic energy at Q #### Example – 12:

• A block of mass 1 kg is released from point P on a frictionless track which ends in a quarter-circular track of radius 2m at the bottom. What is the magnitude of radial acceleration and total acceleration of the block when it arrives at Q? • Given: Height of point P above datum = 6 m, Radius of circular track = 2 m
• To find: the magnitude of radial acceleration = aR = ? and magnitude of total acceleration = a =?
• Solution:

By the principle of conservation of energy

Total Energy at P = Total energy at Q The magnitude of radial acceleration is given by At point Q the tangential acceleration is acceleration due to gravity Ans: Magnitude of radial acceleration = 39.2 m/s2 and the magnitude of total acceleration = 40.4 m/s2.

#### Example – 13:

• A circular race course track has a radius of 500 m and is banked to 10°. If the coefficient of friction between the road and tyre is 0.25. Compute (i) the maximum speed to avoid slipping (ii) optimum speed to avoid wear and tear of the tyres.
• Solution:
• Given: radius of curve = r = 500m,  Angle of banking = θ = 10°, coefficient of friction = μ = 0.25,  g = 9.8 m/s2 ,
• To find: safe velocity = v = ?, to avoid wear and tear v = ?

Maximum sped to avoid slipping. Maximum speed to avoid wear and tear Ans: The maximum velocity to avoid skidding = 46.74 m/s,

The maximum velocity to avoid wear and tear of tyres. = 29.39 m/s.

#### Example – 14:

• 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
• Solution:
• Given: r = 1.5 cm = 1.5 x 10-2 m, For hour hand TH = 12 hr = 12 x 60 x 60 sec
• To Find: Angular velocity = ω = ?, linear velocity = v = ?, angular acceleration = α = ?, radial acceleration = ar = ?, tangential acceleration = aT =?, linear acceleration a = ? 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 aT = 0

Radial acceleration is given by  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.

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