Numerical Problems: Circular Motion

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For uniform circular motion Angular speed ω, Linear speed t, kinetic energy, Angular momentum (L) are constant. The angular acceleration α and the tangential acceleration a_T are zero.

Problems Based on Hands of Clock:

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 ratio of angular speeds of the second hand of a clock and the minute hand of a clock is 60:1.
5. The ratio of angular speeds of the minute hand of a clock and an hour hand of a clock is 12:1.
6. The ratio of angular speeds of the second hand of a clock and the hour hand of a clock is 720:1.

Example – 01:

Calculate the angular speed of the second hand, minute hand and hour hand of a clock.

Given: For second Hand T_S = 60 sec, For minute hand T_M = 60 min = 60 x 60 sec, For hour hand T_H = 12 hr = 12 x 60 x 60 sec.

To Find: Angular speed ω_S=? ω_M=? ω_H=?

Solution: 

The tips of the second hand, the minute hand, and the hour hand perform the uniform circular motion.

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 – 02:

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?

Given: For minute hand  T_M = 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 ω_M=?, Angular displacement  θ =?, Linear velocity = v =?

Solution: 

The tip of the minute hand performs the uniform circular motion.

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 the tip of minute hand = 8.73 x 10^-5 m/s.

Example – 03:

What is the angular displacement of the minute hand of a clock in 25 minutes?

Given: For minute hand  T_M = 60 min = 60 x 60 sec, t = 25 min = 25 x 60 sec,

To Find:  Angular displacement  θ =?

Solution: 

The tip of the minute hand performs the uniform circular motion.

Ans: Angular displacement of minute hand = 2.618 rad

Example – 04:

What is the angular velocity of the second hand of a clock? If the second hand is 10 cm long find the linear velocity of its tip.

Given: For second hand  T_s = 60  sec, r = 10 cm = 10 x 10^-2 m,

To Find:  Linear speed = v =?

Solution: 

The tip of the second hand performs the uniform circular motion.

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 – 05:

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.

Given: For second hand  T_s = 60  sec, r = 1.5 cm – 0.5 cm = 1 cm = 1 x 10^-2 m,

To Find:  Linear speed = v =?

Solution:

The tip of the second hand performs the uniform circular motion.

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 – 06:

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?

Given: v_H = 1/20 v_M , r_M = 10 cm

To Find: r_H =?

Solution:

Ans: Length of hour hand = 6 cm

Example – 07:

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

Given: r = 1.5 cm = 1.5 x 10^-2 m, For hour hand T_H = 12 hr = 12 x 60 x 60 sec

To Find: Angular velocity = ω = ?, linear velocity = v = ?, angular acceleration = α = ?, radial acceleration = a_r = ?, tangential acceleration = a_T =?, linear acceleration a = ?

Solution:

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

∴  Angular acceleration = α = 0 and tangential acceleartion = a_T = 0

Radial acceleration is given by

a_r = v² /r = ( 2.181 x 10^-6 ) ² / (1.5 x 10^-2) = 3.171 x 10^-10 m/s²

a² = a_r ² + a_T ²

as a_T = 0, a = a_r = 3.171 x 10^-10 m/s²

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.

Problems Based on Earth:

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.

Example – 08:

Calculate the angular velocity of the earth due to its spin motion.

Given: For the earth = T = 24 hr = 24 x 60 x 60 sec

To Find: Angular velocity = ω = ?

Solution:

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

Problems Based on Rotating Discs:

Example – 09:

A turntable rotates at 100 rev/sec. Calculate its angular speed in rad/s and degrees/s.

Given: n = 100 r.p.s.

To Find: Angular speed =?

Solution:

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

Example – 10:

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 the blade midway between tips and axis.

Given: r = 2 m, Angular speed = N = 1800 r.p.m.

To Find: v_Tip =? v_Mid =?

Solution:

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

Example – 11:

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?

Given: N = 45 r.p.m., r = 0.5 m

To Find: angular speed in rad/s and degrees/s, linear velocity = v = ?

Solution:

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

Example – 12:

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?

Given: v_T = 3 v_P, Let r_T = r, r_P = (r – 8) cm

To Find: Diameter of the disc =?

Solution:

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

Example – 13:

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.

Given: d = 1m , r_T = r = 0.5 m, N = 120 r.p.m., for r_P = r/2 = 0.25 m

To Find: v_T = ?, V_P = ?

Solution:

Angular velocity for both the points is the same.

The linear speed of point on the rim

v_T =  r_Tω  = 0.5 x 12.56 = 6.28 m/s

v_P =  r_Pω  = 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.

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