Diploma sem 2 applied science physics-unit 4-chap-2 circular motion

CIRCULAR MOTION
Course: Diploma
Subject: Applied Science Physics
Unit: IV
Chapter: II

Circular motion
 Uniform circular motion is the motion of an
object traveling at a constant speed on a circular
path.
 Let T be the time it
takes for the object
to travel once
around the circle.
 T=period of the
circular motion
2 r
T

 

Angular velocity
 the angular velocity is defined as the rate of
change of angular displacement and is a vector
quantity which specifies the angular speed
(rotational speed) of an object and the axis about
which the object is rotating.
 Angular velocity can be expressed as :
 ω = θ / t
 where
ω= angular velocity (rad/s)
θ = angular distance (rad)
t = time (s)
rad=radians

Frequency and Period
 If we assume an object is continuously rotating, then
another way to look at rotational motion is to
examine the period of rotation, T.
 Measurable in units of time (milliseconds, second,
hours, years, eons…) the period is how much time is
takes to make one complete rotation.
 The frequency, f, of an object is actually the inverse
of the period of rotation.

Frequency and Period
T=1/f
and
f =1/T
 The metric unit for frequency is Hertz (Hz), where 1
Hertz = 1 cycle/second. You are probably familiar
with the term Hertz.

Relation between angular velocity, period
and frequency
 Angular frequency, f, is defined as the number
of circular revolutions in a given time interval. It
is commonly measured in units of Hertz (Hz),
where 1 Hz = 1 s–1. For example, the second
hand on a clock completes one revolution every
60 seconds and therefore has an angular
frequency of 1 /60 Hz.
 The relationship between frequency and angular
velocity is:
 Where,ω=angular velocity
f=angular frequency
f
2




Now, Angular period,
Therefore, relation between angular velocity &
angular period is,
1
T
f

1
f &
T
f
2
1
T 2
2
T


 
 







Angular Acceleration
Angular acceleration is the rate of change in angular
velocity. (Radians per sec per sec.)
The angular acceleration can also be found from the
change in frequency, as follows:
2 ( )
2
f
Since f
t

  

 
2
Angular acceleration (rad/s )
t






Centripetal Acceleration and Angular
Velocity
 The angular velocity and the linear velocity are
related (v = ωr)
 The centripetal acceleration can also be related to
the angular velocity
2 2 2
2
2
c
2 r 1
v & f
T T
v 2 fr & 2 f
v r
v r
v 1
a & f
t t 2
v r
a
2 2
a 2 a r
 
  
 

  
  
  

  




 
 
 

Centripetal Force
 We should remember
from that motion is
always caused by some
kind of force.
 The same is true for
circular motion
 Centripetal Force is any
force that causes an
object to follow a
circular path.

Centripetal Force
 Centripetal force can be thought of as a
“center seeking” force.
 Centripetal force is always directed toward
the center

Measuring Centripetal Force
 We know that Force is always F=m(a)
 Centripetal force is always measured as
follows---->

Try it!
 A car is rounding a curve at 65 m/s. The
mass of the car is 1500 kg. The centripetal
force of the car going around the curve is
28 N. What is the radius of the curve?

Centrifugal Force
 Just like we have an inward seeking force with
circular motion, we can also have an outward
seeking motion.
 This outward seeking motion is called
Centrifugal force.

Centrifugal Force
 Centrifugal force can be seen if we tie a string to
a bucket and fill the bucket with water.
 The force holding the water in the bucket and
keeping it from falling out is Centrifugal Force.
 Centrifugal force is measured the same way we
measure centripetal force
 The only difference is the fact that our direction
is outward and not inward.

Example 1: A rope is wrapped many times around a drum
of radius 50 cm.
How many revolutions of the drum are required to raise a
bucket to a height of 20 m?
h = 20 m
RƟ = 40 rad
Now, 1 rev = 2π rad
Ɵ = 6.37 rev
 
1 rev
40 rad
2 rad
 
  
 


20 m
0.50 m
s
R
 

Example 2: A bicycle tire has a radius of 25 cm. If the
wheel makes 400 rev, how far will the bike have traveled?
Ɵ = 2513 rad
s = Ɵ R = 2513 rad (0.25 m)
s = 628 m
 
2 rad
400 rev
1 rev
 
  
 



Example 3: A rope is wrapped many times around a drum
of radius 20 cm. What is the angular velocity of the drum
if it lifts the bucket to 10 m in 5 s?
h = 10 m
R
ω = 10.0 rad/s
ω = =Ɵ
t
50 rad
5 s
Ɵ = 50 rad
10 m
0.20 m
s
R
 

Example 4: In the previous example, what is the
frequency of revolution for the drum? Recall that
ω = 10.0 rad/s.
h = 10 m
R
f = 95.5 rpm
2 or
2
f f 

 

10.0 rad/s
1.59 rev/s
2 rad/rev
f  

Or, since 60 s = 1 min:
rev 60 s rev
1.59 95.5
1 min min
f
s
 
  
 

Example 5: The block is lifted from rest until the
angular velocity of the drum is 16 rad/s after a
time of 4 s. What is the average angular
acceleration?
h = 20 m
R
a = 4.00 rad/s2
f o f
or
t t

 
  
 
2
16 rad/s rad
4.00
4 s s
 

REFERENCE BOOKS AUTHOR/PUBLICATION
ENGINEERING PHYSICS
B.L.THERAJA, S. CHAND
PUBLISHERS

References
1. http://www.engineeringtoolbox.com/moti
on-formulas-d_941.html
2. http://kleins-
klasses.wikispaces.com/file/view/Centrip
etal+and+Centrifugal+Forces.ppt

Diploma sem 2 applied science physics-unit 4-chap-2 circular motion

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Diploma sem 2 applied science physics-unit 4-chap-2 circular motion