Circular Motion Problems

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Important Information for Solving Problems on Circular Motion:

1. The second hand of a clock takes 60 seconds to complete one rotation. Its angular speed is 0.105 rad /s.
2. The minute hand of a clock takes 60 minutes = 60 x 60 seconds to complete one rotation. Its angular speed is 1.746 x 10-3 rad /s.
3. The hour hand of a clock takes 12 hours = 12 x 60 x 60 seconds to complete one rotation. Its angular speed is 1.455 x 10-4 rad /s.
4. The earth takes 24 hours to complete one rotation about its axis. The angular speed of the earth of its rotation about its axis is 7.273 x 10-5 rad /s.
5. The ratio of angular speeds of the second hand of a clock and the minute hand of a clock is 60:1.
6. The ratio of angular speeds of the minute hand of a clock and an hour hand of a clock is 12:1.
7. The ratio of angular speeds of the second hand of a clock and the hour hand of a clock is 720:1.

Notes:

1. For uniform circular motion Angular speed ω = constant, Linear speed v = constant, kinetic energy = constant, Angular momentum (L) = constant, angular acceleration α = 0 and the tangential acceleration aT = 0.
2. When a body is performing a uniform circular motion with velocity ‘v’ in a circle of radius ‘r’, then in half the revolution its velocity changes by ‘2v’.

• When a body is performing a uniform circular motion with velocity ‘v’ in a circle of radius ‘r’, then in half the revolution its displacement is ‘2r’ and distance traveled is πr.
• The angle between the radius vector and linear velocity of a particle performing circular motion is 90°. Hence their scalar product is zero.
• When a stone is tied to a string and rotated in a circle with a constant speed v, If the string is released then it flies along the tangent at a point of release on the circular path.
• If a graph is drawn by plotting tangential velocities of particles of a rigid body performing circular motion (except those on axis) on the y-axis and their distances from the axis of rotation on the x-axis, then the graph will be a straight line passing through the origin with a slope w.
• In the uniform circular motion, the magnitude of linear velocity and magnitude of linear (centripetal) acceleration remains constant but their direction changes continuously.

Kinematical Equations for Rotational Motion

Example – 1:

• Calculate the angular speed of the second hand, minute hand and hour hand of a clock.
• Solution:
• Given: For second Hand TS = 60 sec, For minute hand  TM = 60 min = 60 x 60 sec, For hour hand TH = 12 hr = 12 x 60 x 60 sec.
• To Find: Angular speed  ωS=?, ωM=?, ωH=?

Ans: Angular speed of second hand = 0.105 rad/s,

Angular speed of minute hand =1.746 x 10-3 rad/s,

Angular speed of hour hand =1.454 x 10-4 rad/s.

Example – 2:

• What is the angular velocity of the minute hand of a clock? What is the angular displacement of the minute hand in 20 minutes? If the minute hand is 5 cm long, what is the linear velocity of its tip?
• Solution:
• Given: For minute hand  TM = 60 min = 60 x 60 sec, t = 20 min = 20 x 60 sec, r = 5 cm = 5 x 10-2 m
• To Find: Angular speed ωM=?, Angular displacement  θ =?, Linear velocity = v =?

Ans: The angular speed of minute hand =1.746 x 10-3 rad/s,

The angular displacement of minute hand in 20 minutes =2.095 rad,

The linear speed of tip of minute hand = 8.73 x 10-5 m/s.

Example – 3:

• What is angular displacement of the minute hand of a clock in 25 minutes?
• Solution:
• Given: For minute hand  TM = 60 min = 60 x 60 sec, t = 25 min = 25 x 60 sec,
• To Find:  Angular displacement  θ =?

Ans: Angular displacement of minute hand = 2.618 rad

Example – 4:

• What is the angular velocity of the second hand of a clock? If the second hand is 10 cm long find linear velocity of its tip.
• Solution:
• Given: For second hand  Ts = 60  sec, r = 10 cm = 10 x 10-2 m,
• To Find:  Linear speed = v =?

Now v = r ω = 10 x 10-2 x 0.105 = 1.05 x 10-2 m/s = 1.05 cm/s

Ans: Angular velocity of second hand = 0.105 rad/s, Linear speed of tip of second hand = 1.05 cm/s

Example – 5:

• The second hand of a watch is 1.5 cm long. Find the linear speed of a point on the second hand at a distance of 0.5 cm from the tip.
• Solution:
• Given: For second hand  Ts = 60  sec, r = 1.5 cm – 0.5 cm = 1 cm = 1 x 10-2 m,
• To Find:  Linear speed = v =?

Now v = r ω = 1 x 10-2 x 0.105 = 1.05 x 10-3 m/s = 1.05 mm/s

Ans: Linear speed of the point on the second hand = 1.05 mm/s

Example – 6:

• The extremity of the hour hand of a clock moves 1/20th as fast as the minute hand. What is the length of the hour hand if the minute hand is 10 cm long?
• Solution:
• Given: vH = 1/20 v, rM = 10 cm
• To Find: rH =?

Ans: Length of hour hand = 6 cm

Example – 7:

• Calculate angular velocity of the earth due to its spin motion.
• Solution:
• Given: For the earth = T = 24 hr = 24 x 60 x 60 sec
• To Find: Angular velocity = ω = ?

Ans: Angular speed of the earth due to its spin motion is 7.273 x 10-5 rad/s.

Example – 8:

• A turntable rotates at 100 rev/sec. Calculate its angular speed in rad/s and degrees/s.
• Solution:
• Given: n = 100 r.p.s.
• To Find: Angular speed =?

Ans: Angular speed = 10.47 rad/s, Angular speed = 600 degrees/s

Example – 9:

• 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.
• Solution:
• Given: r = 2 m, Angular speed = N = 1800 r.p.m.
• To Find: vTip =? vMid =?

Ans: The angular speed of tip of blade = 376.8 rad/s, AThe angular speed of point midway = 188.4 rad/s

Example – 10:

• A turntable has a constant angular speed of 45 r.p.m. Express this in rad per second and degrees per second. If the radius of the turntable is 0.5 m, what is the linear speed of a point on the rim?
• Solution:
• Given: N = 45 r.p.m., r = 0.5 m
• To Find: angular speed in rad/s and degrees/s, linear velocity = v = ?

Ans: Angular speed = 2.355 rad/s = 270 degrees/s Linear speed on point on rim = 2.355 m/s.

Example – 11:

• The linear velocity of a point on the rotating disc is 3 times greater than at a point on the at a distance of 8 cm from it. What is the diameter of the disc?
• Solution:
• Given: vT = 3 vP, Let rT = r, rP = (r – 8) cm
• To Find: Diameter of the disc =?

Angular velocity for both the points is the same.
Ans: Diameter of disc = 24 cm.

Example – 12:

• A disc has a diameter of one metre and rotates about an axis passing through its centre and at right angles to its plane at the rate of 120 rev/min. What is the angular and linear speed of a point on the rim and at a point halfway to the centre.
• Solution:
• Given: d = 1m , rT = r = 0.5 m, N = 120 r.p.m., for rP = r/2 = 0.25 m
• To Find: vT = ?, VP = ?

Angular velocity for both the points is the same.

The linear speed of point on the rim

vT =  rTω  = 0.5 x 12.56 = 6.28 m/s

vP =  rPω  = 0.25 x 12.56 = 3.14 m/s

Ans: Angular speed of point on rim = 12.57 rad/s, Linear peed of point on rim = 6.28 m/s,
Angular speed of point on halfway =12.57m/s, Linear speed of point on halfway = 3.14 m/s.

Example – 13:

• A body rotates in a circular path of radius 0.25 m at 240 r.p.m. Find its angular and linear speeds. If the angular speed changes to 330 r.p.m. in 10 s. Find the angular and linear accelerations.
• Solution:
• Given: r = 0.25 m, N1 = 240 r.p.m.,N2 = 330 r.p.m., t = 10 s.
• To find: α = ?, a = ?

Initial Condition

Final Condition

Now a = r α  = 0.25 x 0.942 = 0.2355 m/s2.

Ans: Initial angular velocity = 25.12 rad/s, Initial linear velocity = 6.28 m/s
Final angular velocity = 34.54 rad/s, Final linear velocity = 8.635 m/s
Angular acceleration = 0.942 rad/s2, Linear acceleration = 0.2355 m/s2.

Example – 14:

• A disc is rotating in a horizontal plane about a vertical axis passing through its centre at 150 r.p.m. When accelerated its speed increases to 1050 r.p.m. in 3 s. What is the angular acceleration caused assuming it to be uniform? What will be the angular velocity of the disc in r.p.m. after 2 more seconds? How many rotations does it make during this time and what is the angular displacement?
• Solution:
• Given: N1 =150 r.p.m.,N2=1050 r.p.m., t = 3 s.
• To Find: α = ?, N = ? after 2 sec, = ?. No. of revolutions = ?

Ans: Angular acceleration =31.42 rad/s2, Angualr velocity after 2 sec = 1650 r.p.m.
No. of rotations in these 2 sec = 45

Example – 15:

• The angular acceleration of a body rotating about a given axis is 1 rad/s2. Through what angle does it rotates during the time in which its angular velocity increases from 5 rad/s to 15 rad/s.
• Solution:
• To Find: θ = ?.

Ans: Angle traced = 100 rad.

Example – 16:

• A disc rotating about an axis passing through its centre and right angles to its plane has its angular speed reduced 50 r.p.s. to 25 r.p.s. at 5 s. How much time does it take and how many revolutions does it make during this time? How much time does it take and how many more revolutions does it make before coming to rest.
• Solution:
• Given: n1 = 50 r.p.s., n2 = 15 r.p.s., t = 5 s.

ω1 = 2πn1  = 2π x 50 = 100π rad/s

ω2 = 2πn2  = 2π x 15 = 30π rad/s

Ans: No. of revolutions made in 5 sec =187.5, Time taken by disc before coming to rest = 5 s,
No. of revolution made before coming to rest = 62.5.

Example – 17:

• The angular velocity of a disc rotating in a horizontal plane about a vertical axis passing through its centre increases from 600 r.p.m. to 3000 r.p.m. in 5 s. Find the angular acceleration of the disc assuming it to be constant. What are the initial and final angular velocities of the disc? What is the angular displacement and number of revolutions made by the disc during this time? Find the linear velocity of a point on the rim of the disc if its radius is 0.5 m.
• Solution:
• Given: N1 = 600 r.p.m.,N2 = 3000 r.p.m., t = 5s, r = 0.5 m
• To Find: α = ?, No. of revolutions = ?, velocity of point on rim = ?

Ans: Initial angular velocity = 62.8 rad/s, Final angular velocity = 314 rad/s
No. of revolutions = 150, Linear velocity of point on rim = 157 m/s.

Example – 18:

• A satellite revolves around the earth in a circular orbit of radius 7000 km. If the period of revolution is 2 hours, calculate its angular speed, linear speed, and centripetal acceleration.
• Solution:
• Given: r = 7000 km = 7000 x 103 m, T = 2 hr = 2 x 60 x 60 s = 7200 s
• To Find: ω = ?, v = ?, acp = ?

Ans: Angular speed = 8.72 x 10-4 rad/s, Linear speed 6.10 km/s,
Centripetal acceleration = 5.32 m/s2.

Example – 19:

• Find the speed at which the point on the equator move as the earth rotates about its axis. Take radius of the earth as 6400 km.
• Solution:
• Given: r = 6400 km = 6400 x 103 m, T = 24 hr = 24 x 60 x 60 s
• To find: = ?, v = ?, acp = ?

Ans: Linear speed of point on equator = 465.1 m/s

Example – 20:

• The tangential acceleration of a particle moving in a circular path of radius 5 cm is 2 m/s2. The angular velocity of the particle increases from 10 rad/s to 20 rad/s during this time. Find the duration of time and number of revolutions completed during this time.
• Solution:
• Given: a= 2 m/s2, r = 5 cm = 5 x10-2 m, ω1 = 10 rad/s, ω2 = 20 rad/s
• To Find: t = ?, No. of revolutions = ?,

Ans: Time interval = 0.25 s, No. of rotations = 0.6.

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One Comment

1. Rajkumar patil

Nice
Very nice
Zakas