Circular motion slides complete

Circular Motion – 5.1 to 5.4
Gianicolli - Chapter 5
Mr. Sharick

Kinematics of Uniform
Circular Motion
Section 5.1

Section 5.1 – Kinematics of
Uniform circular motion
 Uniform in this case means constant speed.
 Can an object moving with a constant
speed have acceleration?
 If so, we physicists are going to want to find a
value for it, but how?
Figure 1 a small object
moving in a circle. Note the
instantaneous velocity is always
tangent to the circular path.

Section 5.1 – Kinematics of
Uniform circular motion
 Uniform in this case means constant speed.
 Can an object moving with a constant
speed have acceleration?
 An object that changes direction can have
acceleration (changing velocity) while
maintaining a constant speed.
 This is the case for circular motion.

Section 5.1 – Vocabulary
New terms:
 Tangential velocity –
 Centripetal or radial acceleration –
 Period –
 Frequency –

Section 5.1 – Vocabulary
New terms:
 Tangential velocity – (vt) velocity directed tangent to
the circular path
 Centripetal or radial acceleration – (ac or ar)
acceleration directed toward the center of the circle
 Period – (T) the time required to make one complete
revolution
 Frequency – (f) the number of revolutions per
second

Section 5.1 – Tangential Velocity
 Tangential Veloctiy
 How do we calculate this?
– How do we calculate linear velocity?

distance
v
time


– What distance is the object going through?
distance
v
time


distance
v
time

2distance r

– The time for one revolution is called what?
distance
v
time

2distance r

– The time for one revolution is called what?
Period (T)
distance
v
time

2distance r

 Put it all together:
 Remember, the direction of the velocity is ________
to the circular path!

 Remember, the direction of the velocity is ________
to the circular path!
2distance r
v
time T

  

 Remember, the direction of the velocity is tangent to
the circular path!
2distance r
v
time T

  

Example Problem – Tangential Velocity
1.) Find the tangential velocity (in m/s) of the Earth
orbiting the sun. The distance between the sun and the
Earth is 149,600,000 km.

Example Problem – Tangential Velocity
1.) Find the tangential velocity (in m/s) of the Earth
orbiting the sun. The distance between the sun and the
Earth is 149,600,000 km.
11
11
7
2 2 (1.496 10 )
1
2 (1.496 10 )
29,700 /
r m
v
T year
m
v m s
3.16 10 s
 


 

 


Section 5.1 – Radial/Centripetal
Acceleration
Proof – radial/centripetal acceleration
-We are going to derive the equation for centripetal
acceleration of an object in uniform circular motion.

Acceleration
 We will be using two similar
triangles.
 1st triangle: ABC, made up
of sides r1, r2, and Δl
 2nd triangle: made up of
sides v1, v2, and Δv. (v1 =
v2; just different directions)
1
2

Interlude – Finding Δv
v2
Since v = v2-v1
Draw v1 + v = v2
v1

Interlude – Finding Δv
v2
Since v = v2-v1
Draw v1 + v = v2
v1
v

Acceleration
 The angle between the
radius and the tangential
velocity is always 90o.
 For this to remain true,
the angle between r1 and
r2 must be equal to the
angle between v1 and v2.
(Δθ equal in the picture)
1
2

Acceleration
 Therefore (similar
triangles have ratio of
sides that are equal)
 

v l
v r
1
2

Acceleration
 Therefore (similar
triangles have ratio of
sides that are equal)
 So
 

v l
v r
1
2
  
v
v l
r

Acceleration
 Recall the definition for acceleration:

Acceleration
 Recall the definition for acceleration:
v
a
t




Acceleration
 Recall the definition for
acceleration:
 Also, we found:
v
a
t



  
v
v l
r

Acceleration
 Recall the definition for
acceleration:
 Also, we found:
Thus:
v
a
t



  
v
v l
r
v l
a
r t




Acceleration
 Recall that the linear speed
around a circle can be given by the
change in arc length divided by
change in time.
 For small angles, arc length is
approximately the straight line
displacement.
 So, we have that

Acceleration
change in time.
displacement.
s
v
t
s l



  

Acceleration
change in time.
displacement.
s
v
t
s l



  
v l
a
r t

 


Acceleration
change in time.
displacement.
s
v
t
s l



  
2
=
v l v s v
a
r t r t r
   
    
   

Acceleration
 Hopefully, it is clear that this
formula can only give positive
answers, and therefore only can
find the magnitude of the
acceleration.
 Acceleration is a VECTOR so… in
what direction is the
acceleration?
2

v
a
r

Finding the direction of centripetal
acceleration:
vf
Draw v = vf-vi
vi

Finding the direction of centripetal
acceleration:
tf
-vi
vf
v
Notice the direction
of v is towards the
center of the circle!
The direction of acceleration is also
towards the center of the circle!
THUS, centripetal!
v = vf-vi

Section 5.1 - Centripetal/Radial
Acceleration
 Acceleration of an object in uniform circular
motion is given by:
 Direction is toward the _______________

Section 5.1 - Centripetal/Radial
Acceleration
 Acceleration of an object in uniform circular
motion is given by:
 Direction is toward the center (centripetal)
2
c
v
a
r


Section 5.1 – Period and
Frequency
 Period and frequency are inversely
proportional.
1
T
f


Frequency
Ex: An object makes 10 revolutions in 2
seconds. What is the objects frequency and
period?

Frequency
 Ex: An object makes 10 revolutions in 2
seconds. What is the objects frequency and
period?
sec
sec
/ sec
10
5
2sec
1 1
0.2
5
rev
rev
rev
rev
f
T
f
 
  

Section 5.1 – Example 5.1 –
Acceleration of a revolving ball
A 150-g ball at the end of a string is revolving uniformly
in a horizontal circle of radius 0.600 m. The ball makes
2.00 revolutions in a second. What is the centripetal
acceleration?

Acceleration of a revolving ball
A 150-g ball at the end of a string is revolving uniformly
in a horizontal circle of radius 0.600 m. The ball makes
2.00 revolutions in a second. What is the centripetal
acceleration?
2 2
2
1/ 1/ (2.00 / ) 0.500
2 2(3.14)(0.600 )
7.54 /
(0.500 )
(7.54 / )
94.8 /
(0.600m)
c
T f rev s s
r m
v m s
T s
v m s
a m s
r

  
  
  

Interlude – a useful substitution
 We have 2 equations:
 Plugging v in the 1st into the 2nd we get:
2
c
v
a
r
2 r
v
T



Interlude – a useful substitution
 We have 2 equations:
 Plugging v in the 1st into the 2nd we get:
2
c
v
a
r
2 r
v
T


2
2
2 2
2
2
2 4
or
c
c c
r
v T
a
r r
r
a r a
T T

 
 
 
  
 
  
 

Moon’s centripetal acceleration
The moon’s nearly circular orbit about the Earth has a
radius of orbit of about 384,000km and a period T of
27.3 days. Determine the acceleration of the Moon
towards the Earth.

Moon’s centripetal acceleration
The moon’s nearly circular orbit about the Earth has a
radius of orbit of about 384,000km and a period T of
27.3 days. Determine the acceleration of the Moon
towards the Earth.
6
2 8
2
2 6 2
(27.3days)(24.0h/ day)(3600s/ h) 2.36x10
4 (39.5)(3.84x10 m)
0.00272m/s
(2.36x10 )
c
T s
r
a
T s

 
  

Dynamics of Uniform
Circular Motion
Section 5.2

Section 5.2 - DYNAMICS of
Circular Motion
Revisiting Newton’s 2nd Law
Newton stated that if a body is accelerating,
there must be a NET force acting on it, in the
direction of the acceleration.
The NET force that causes circular motion is
called the CENTRIPETAL force.
Note: This is not a new type of force, we will
treat it very much the same as we did any
NET force.

Section 5.2 – Dynamics of Uniform
Circular Motion
Centripetal or radial force – the force required to keep
an object in circular motion is as follows:
What is the direction of this force?
2 2
2
4
c c
mv m r
F ma
r T

  

Section 5.2 – Centripetal vs.
Centrifugal
 Centri from the Latin Centrum: the middle point of a
circle, the center.
 Petal from the Latin Peto – to seek, strive after,
endeavor to obtain.
 Fugal from the Latin Fugito – to flee, to fly from,
avoid or shun.
 By looking at the direction of the change in velocity,
you should be able to see the direction of the
acceleration.
 Is it Centripetal or Centrifugal?

Centrifugal
 Consider a person swinging a ball on the end
of a string around her head:
 A ___________ (center seeking) force causes
circular motion NOT a __________ (center
fleeing) force.
 The force that is acting on the ball causing
circular motion is inward (centripetal).
 The force acting on the hand is outward, and
NOT causing circular motion.
 A centrifugal force is a “fictitious” or “pseudo”
force caused by viewing the situation from an
accelerating reference frame.

Centrifugal
 Consider a person swinging a ball on the end
of a string around her head:
 A centripetal (center seeking) force causes
circular motion NOT a centrifugal (center
fleeing) force.
 The force that is acting on the ball causing
circular motion is inward (centripetal).
 The force acting on the hand is outward, and
NOT causing circular motion.
 A centrifugal force is a “fictitious” or “pseudo”
force caused by viewing the situation from an
accelerating reference frame.

Section 5.2 - Centrifugal force – the
fictitious force
Wahoo!
I’m stuck to the wall!
I feel like I’m being pressed against
the wall.
Force on person due
to the wall
Fictitious force to explain zero
acceleration

Section 5.2 - Why is it said that
centrifugal “fictitious” force is a
misconception, and NOT real?
Because to an
observer in an inertial
reference frame
“fictitious” force does
not exist.

Section 5.2 - Centrifugal forces are
not real!
 Remember, Newton’s laws only apply in inertial
reference frames.
 Centrifugal (fictitious) force is a non-Newtonian, and
is caused by acceleration and is only real to an
observer in the accelerating reference frame.
 In the picture above (a) represents what would
happen if centrifugal forces were real. (b) Shows
what does happen in an inertial frame of reference.

Section 5.2 – Solving Dynamics
Problems for Uniform Circular Motion

Section 5.2 – Example 5.3 – Force
on revolving ball (horizontal)
Estimate the force a person must exert on a string
attached to a 0.150 kg ball to make the ball revolve in a
horizontal circle of radius 0.600m, making 2.00
revolutions per second.
mg
θ
T

Tetherball
The game of tetherball is played with a ball tied to a
pole with a string. When the ball is struck, it whirls
around the pole as shown. In what direction is the
acceleration of the ball, and what causes the
acceleration.
q

Section 5.2 – Extra Example – Conical
Pendulum (yellow packet page 9)
A particle of mass, m, is suspended from a string of length, L, and
travels at a constant speed, v in a horizontal circle if radius, r. The
string makes an angle, q, given by sin q = r/L, as shown in the
figure. Find the tension in the string and the speed of the particle.
q

Section 5.2 – Example 5.5 – Revolving
ball (vertical circle)
A 0.150 kg ball on the end of a 1.10 m long cord is swung in a
vertical circle. a.) Determine the minimum speed the ball must
have at the top of its arc so that it continues moving in a circle.

Section 5.2 – Example 5.5 – Revolving
ball (vertical circle)
A 0.150 kg ball on the end of a 1.10 m long cord is swung in a
vertical circle. b.) Calculate the tension in the cord at the bottom of
the arc assuming the ball is moving at twice the speed of part (a).

A Car Rounding a
Curve
Section 5.3

Section 5.3 – A Car Rounding a
Curve
 One example of centripetal acceleration
occurs when a car rounds a curve.
 In this situation, what do you feel?
 What is causing the car to move in circular
motion?

Curve
 You feel like you are being thrown _______, but in
reality, you tend to move in a _______ _____as the
car begins moving in a curve. The force you feel is
the car pushing you!
 The car is moving
because of _________
between the tires
and the road!

Curve
 You feel like you are being thrown outward, but in
reality, you tend to move in a straight line as the car
begins moving in a curve. The force you feel is the
car pushing you!
 The car is moving
because of friction
between the tires
and the road!

Section 5.3 - Example 5.7 – Skidding
on a curve (yellow packet page 7)
A 1000 kg car rounds a curve on a flat road of
radius 50 m at a speed of 50 km/hr (14 m/s).
Will the car make the turn, or will it skid, if: (a)
the pavement is dry and the coefficient of
static friction is ms = 0.60?

Section 5.3 - Example 5.7 – Skidding
on a curve (yellow packet page 7)
A 1000 kg car rounds a curve on a flat road of
radius 50 m at a speed of 50 km/hr (14 m/s).
Will the car make the turn, or will it skid, if: (b)
the pavement is icy and ms = 0.25?

Section 5.3 – Example 5.8 – Banking angle
w/o friction(yellow packet page 8)
For a car traveling with a speed, v, around a curve
of radius, r. (a) Determine a formula for the angle
at which a road should be banked so that no
friction is required to keep the car on the road.

Section 5.3 – Example 5.8 – Banking angle
w/o friction (yellow packet page 8)
For a car traveling with a speed, v, around a curve
of radius, r. (b) What is this angle for a curve of
radius 50 m at a designed speed of 50 km/h?

Section 5.3 – Banking angles with
friction (not in book)
Two cases:
- Faster than ideal speed
- Slower than ideal speed
What happens in each case?

V > V_ideal
 The car will want to move __
the slope, so that friction is
pointing ______ the slope
(______ motion)
 The math is much more
complicated with friction!

V > V_ideal
 The car will want to move up
the slope, so that friction is
pointing down the slope
(opposing motion)
 The math is much more
complicated with friction!

Circular motion slides complete

V < V_ideal
 The car will want to move
______ the slope, so that
friction is pointing ___ the
slope (________ motion)

V < V_ideal
 The car will want to move
down the slope, so that
friction is pointing up the
slope (opposing motion)

Nonuniform Circular
Motion
Section 5.4
total

Section 5.4 - Nonuniform Circular
Motion
 Combining the new with the old!
 Remember acceleration is a vector!
(in relation to the velocity)
Total tangential radial
Total
a a a
OR
a a a
 
 
total

Section 5.4 – Tangential vs centripetal
acceleration
 For tangential acceleration (caused by tangent force)
 For centripetal acceleration (caused by centripetal
force)
For a car, where would each of these forces come
from?

Section 5.4 – Tangential vs centripetal
acceleration
 For tangential acceleration (caused by tangent force)
vf=vi + at
 For centripetal acceleration (caused by centripetal
force)
a = v2/r
 For a car, where would each of these forces come
from?

Section 5.4 – Example 5.9 – Two
components of acceleration
A racing car starts from rest in the pit area and
accelerates at a uniform rate to a speed of 35 m/s in 11
s, moving on a circular track of radius 500 m. Assuming
constant tangential acceleration, find (a) the tangential
acceleration.

Section 5.4 – Example 5.9 – Two
components of acceleration
A racing car starts from rest in the pit area and
accelerates at a uniform rate to a speed of 35 m/s in 11
s, moving on a circular track of radius 500 m. Assuming
constant tangential acceleration, find (b) the centripetal
acceleration when the speed is 30 m/s.

Section 5.4 – Example Problem: Nonuniform
circular motion (from yellow packet page 10)
A particle moves clockwise in a circle of radius 1.0 m. It
starts at rest at the origin at time, t=0. Its speed
increases at a constant rate of 1.6 m/s2.
a.) How long does it take to travel half way around the
circle (point A to Point B)?
A B
r =1.0 m

b.) What is its speed at point B?
c.) What is the direction of its velocity at that time?
A B
r =1.0 m

d.) What is the centripetal (radial) acceleration?
e.) What is the tangential (linear) acceleration?
A B
r =1.0 m

f.) What is the magnitude and direction of the total
acceleration at point B?
A B
r =1.0 m

Circular motion slides complete

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Circular motion slides complete