Uniform Circular Motion (UCM)

About This Presentation

Title:

Uniform Circular Motion (UCM)

Description:

Uniform Circular Motion (UCM) Try Example #2: A car is moving on a circular track of radius 0.522 km. The magnitude of its centripetal acceleration is 4.00 m/s2. Find ... – PowerPoint PPT presentation

Number of Views:176

Avg rating:3.0/5.0

Slides: 135

Provided by: Simp58

Category:

more less

Transcript and Presenter's Notes

Title: Uniform Circular Motion (UCM)

1
Uniform Circular Motion (UCM)
2
Uniform Circular Motion (UCM)

Motion in a circle at a constant speed

3
Uniform Circular Motion (UCM)

Motion in a circle at a constant speed

4
Uniform Circular Motion (UCM)

Motion in a circle at a constant speed

A
5
Uniform Circular Motion (UCM)

Motion in a circle at a constant speed
What is the direction of
the instantaneous velocity
at point A?

A
6
Uniform Circular Motion (UCM)

Motion in a circle at a constant speed
Just like projectile motion,
the direction of v is along
the tangent-line at
a point on the particles
path

vA
A
7
Uniform Circular Motion (UCM)

What is the direction of the instantaneous
velocity at point B?

vA
A
B
8
Uniform Circular Motion (UCM)

vA
A
B
vB
9
Uniform Circular Motion (UCM)

How should the length or magnitude
of vA compare with
the length or magnitude
of vB ? Why?

vA
A
B
vB
10
Uniform Circular Motion (UCM)

The length or magnitude
of vA is the same as the
length or magnitude of
vB because the speed
stays constant.

vA
A
B
vB
11
Uniform Circular Motion (UCM)

The length or magnitude
of vA is the same as the
length or magnitude of
vB because the speed
stays constant.
Even if vA vB and
the speed stays the same,
why is UCM considered to be acceleration?

vA
A
B
vB
12
Why is UCM considered to be acceleration?
Even if vA vB and the speed stays
the same, the velocity changes
direction. This means there is a non-zero
change- in velocity, ?v. Since a
?v/?t, whenever there is a non-zero
change-in velocity, there is a non-zero
acceleration.
vA
A
B
vB
13
Deriving a formula to find the magnitude of UCM
acceleration
14
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM

right
15
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
What is the direction of
the instantaneous
velocity at point S?

right
S
r
16
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
Right and a little up

vS
right
S
r
17
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
What is the direction of the
instantaneous velocity
at point F?

vS
right
S
F
r
r
18
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
Right and a little down

vS
right
vF
S
F
r
r
19
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
We want to create a
a formula for
instantaneous a ?v/?t.
What must be true
about ?t?

vS
right
vF
S
F
r
r
20
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
To calculate instantaneous
acceleration ?t must be
very small, so

vS
right
vF
S
F
r
r
21
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
To calculate instantaneous
acceleration ?t must be
very small, so
S and F are actually so
close together that the
arc distance SF is very
nearly the straight line
distance SF

vS
right
vF
S
F
r
r
22
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
To calculate instantaneous
acceleration ?t must be
very small, so
S and F are actually so
close together that the
arc distance SF is very
nearly the straight line
distance SF
We can make a triangle SFO,
but keep in mind angle O is
actually very small.

vS
right
vF
S
F
r
r
O
23
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
To calculate instantaneous
acceleration ?t must be
very small, so
S and F are actually so
close together that the
arc distance SF is very
nearly the straight line
distance SF
We can make a triangle SFO,
but keep in mind angle O is
actually very small.

vS
right
vF
S
F
r
r
O
24
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
To calculate instantaneous
acceleration ?t must be
very small, so
S and F are actually so
close together that the
arc distance SF is very
nearly the straight line
distance SF
We can make a triangle SFO,
but keep in mind angle O is
actually very small.
For ?t very small, straight line SF?d

vS
right
vF
S
?d
F
r
r
O
25
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
Since a ?v/?t , we want to find ?v
But ?v vF- vS , so...

vS
right
vF
S
?d
F
r
r
O
26
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
Since a ?v/?t , we want to find ?v
But ?v vF- vS , so..
?v vF (- vS)

vS
right
vF
S
?d
F
r
r
O
27
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
Since a ?v/?t , we want to find ?v
But ?v vF- vS , so..
?v vF (- vS)
Now let's arrange tip-to-tail

vS
right
vF
S
?d
F
r
r
O
28
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
Since a ?v/?t , we want to find ?v
But ?v vF- vS , so..
?v vF (- vS)
Now let's arrange tip-to-tail

vS
right
vF
S
?d
F
r
r
vF
O
29
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
Since a ?v/?t , we want to find ?v
But ?v vF- vS , so..
?v vF (- vS)
Now let's arrange tip-to-tail

vS
right
vF
S
?d
F
r
r
vF
O
-vS
30
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
Since a ?v/?t , we want to find ?v
But ?v vF- vS , so..
?v vF (- vS)
Now let's arrange tip-to-tail

vS
right
vF
S
?d
F
r
r
vF
?v
O
-vS
31
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
Since a ?v/?t , we want to find ?v
But ?v vF- vS , so..
?v vF (- vS)
Now let's arrange tip-to-tail
Note ?v occurs between S and F,
which is the top of the circle.

vS
right
vF
S
?d
F
r
r
vF
?v
O
-vS
32
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
Since a ?v/?t , we want to find ?v
But ?v vF- vS , so..
?v vF (- vS)
Now let's arrange tip-to-tail
Note ?v occurs between S and F,
which is the top of the circle. Therefore
?v has a direction towards the centre of the
circle.

vS
right
vF
S
?d
F
r
r
vF
?v
O
-vS
33
Deriving a formula to find the magnitude of UCM
acceleration
up

Consider a particle undergoing UCM
Since a ?v/?t , we want to find ?v
But ?v vF- vS , so..
?v vF (- vS)
Now let's arrange tip-to-tail
Note ?v occurs between S and F,
which is the top of the circle. Therefore
?v has a direction towards the centre of the
circle.
The instantaneous acceleration is also toward the
centre. Why?

vS
right
vF
S
?d
F
r
r
vF
?v
O
-vS
34
Deriving a formula to find the magnitude of UCM
acceleration
up

Let's change our ?v vector diagram to scalars
only
How does I-vS compare with vF?

vF
?v
vS
right
vF
-vS
S
?d
F
r
r
O
35
Deriving a formula to find the magnitude of UCM
acceleration
up

Let's change our ?v vector diagram to scalars
only
Remember I-vS vF
because the speed is constant.

vF
?v
vS
right
vF
-vS
S
?d
F
r
r
O
36
Deriving a formula to find the magnitude of UCM
acceleration
up

Let's change our ?v vector diagram to scalars
only
Remember I-vS vF
because the speed is constant.
Let's call I-vS vF v (speed)

vF
?v
vS
right
vF
-vS
S
?d
F
r
r
O
37
Deriving a formula to find the magnitude of UCM
acceleration
up

Let's change our ?v vector diagram to scalars
only
Remember I-vS vF
because the speed is constant.
Let's call I-vS vF v (speed)
Now we can change our ?v vector
diagram to a triangle with magnitudes
only like this...

vF
vF
?v
vS
right
vF
-vS
S
?d
F
r
r
O
38
Deriving a formula to find the magnitude of UCM
acceleration
up

Let's change our ?v vector diagram to scalars
only
Remember I-vS vF
because the speed is constant.
Let's call I-vS vF v (speed)
Now we can change our ?v vector
diagram to a triangle with magnitudes
only like this...

vF
vF
?v
vS
right
vF
-vS
S
?d
F
r
r
O
v
?v
v
39
Deriving a formula to find the magnitude of UCM
acceleration
v

We now have two similar
triangles. Later for homework,
see if you can prove it.

?v
v
S
?d
F
r
r
O
40
Deriving a formula to find the magnitude of UCM
acceleration
v

We now have two similar
triangles. Later for homework,
see if you can prove it.
If these triangles are similar,
then what is
?v ?
v

?v
v
S
?d
F
r
r
O
41
Deriving a formula to find the magnitude of UCM
acceleration
v

We now have two similar
triangles. Later for homework,
see if you can prove it.
?v ?d
v r

?v
v
S
?d
F
r
r
O
42
Deriving a formula to find the magnitude of UCM
acceleration

We now have two similar
triangles. Later for homework,
see if you can prove it.
?v ?d
v r
?v ?

v
?v
v
S
?d
F
r
r
O
43
Deriving a formula to find the magnitude of UCM
acceleration

We now have two similar
triangles. Later for homework,
see if you can prove it.
?v ?d
v r
?v ?d v
r

v
?v
v
S
?d
F
r
r
O
44
Deriving a formula to find the magnitude of UCM
acceleration

We now have two similar
triangles. Later for homework,
see if you can prove it.
?v ?d
v r
?v ?d v
r
But a ?v
?t

v
?v
v
S
?d
F
r
r
O
45
Deriving a formula to find the magnitude of UCM
acceleration

We now have two similar
triangles. Later for homework,
see if you can prove it.
?v ?d
v r
?v ?d v
r
But a ?v ?
?t

v
?v
v
S
?d
F
r
r
O
46
Deriving a formula to find the magnitude of UCM
acceleration

We now have two similar
triangles. Later for homework,
see if you can prove it.
?v ?d
v r
?v ?d v
r
But a ?v ?d v
?t ?t r

v
?v
v
S
?d
F
r
r
O
47
Deriving a formula to find the magnitude of UCM
acceleration

We now have two similar
triangles. Later for homework,
see if you can prove it.
?v ?d
v r
?v ?d v
r
But a ?v ?d v
?t ?t r
But what is

v
?v
v
S
?d
F
r
r
O
48
Deriving a formula to find the magnitude of UCM
acceleration

We now have two similar
triangles. Later for homework,
see if you can prove it.
?v ?d
v r
?v ?d v
r
But a ?v ?d v
?t ?t r
But what is
So a ?

v
?v
v
S
?d
F
r
r
O
49
Deriving a formula to find the magnitude of UCM
acceleration

We now have two similar
triangles. Later for homework,
see if you can prove it.
?v ?d
v r
?v ?d v
r
But a ?v ?d v
?t ?t r
But what is
So a v v
r

v
?v
v
S
?d
F
r
r
O
50
Deriving a formula to find the magnitude of UCM
acceleration

We now have two similar
triangles. Later for homework,
see if you can prove it.
?v ?d
v r
?v ?d v
r
But a ?v ?d v
?t ?t r
But what is
So a v v or a ?
r

v
?v
v
S
?d
F
r
r
O
51
Deriving a formula to find the magnitude of UCM
acceleration

We now have two similar
triangles. Later for homework,
see if you can prove it.
?v ?d
v r
?v ?d v
r
But a ?v ?d v
?t ?t r
But what is
So a v v or a v2
r r

v
?v
v
S
?d
F
r
r
O
52
Short-Cut Formula for the magnitude of the
Instantaneous Acceleration of UCM
53
Short-Cut Formula for the magnitude of the
Instantaneous Acceleration of UCM

We have derived a v2
r
where v is the speed and r is the radius

54
Short-Cut Formula for the magnitude of the
Instantaneous Acceleration of UCM

We have derived a v2 Memorize this!
r
where v is the speed and r is the radius

55
Short-Cut Formula for the magnitude of the
Instantaneous Acceleration of UCM

We have derived a v2 Memorize this!
r
where v is the speed and r is the radius
As we determined earlier, for UCM, ?v is always
directed toward the __________ of the _________.

56
Short-Cut Formula for the magnitude of the
Instantaneous Acceleration of UCM

We have derived a v2 Memorize this!
r
where v is the speed and r is the radius
As we determined earlier, for UCM, ?v is always
directed toward the center of the circle.

57
Short-Cut Formula for the magnitude of the
Instantaneous Acceleration of UCM

We have derived a v2 Memorize this!
r
where v is the speed and r is the radius
As we determined earlier, for UCM, ?v is always
directed toward the center of the circle.
But a and ?v are always in the ___________
direction.

58
Short-Cut Formula for the magnitude of the
Instantaneous Acceleration of UCM

We have derived a v2 Memorize this!
r
where v is the speed and r is the radius
As we determined earlier, for UCM, ?v is always
directed toward the center of the circle.
But a and ?v are always in the same direction.
Why?

59
Short-Cut Formula for the magnitude of the
Instantaneous Acceleration of UCM

We have derived a v2 Memorize this!
r
where v is the speed and r is the radius
As we determined earlier, for UCM, ?v is always
directed toward the center of the circle.
But a and ?v are always in the same direction.
This is because a ?v/?t , and division of a
vector by a positive scalar always results in a
new vector in the same direction as the original.

60
Short-Cut Formula for the magnitude of the
Instantaneous Acceleration of UCM

We have derived a v2 Memorize this!
r
where v is the speed and r is the radius
As we determined earlier, for UCM, ?v is always
directed toward the center of the circle.
But a and ?v are always in the same direction.
This is because a ?v/?t , and division of a
vector by a positive scalar always results in a
new vector in the same direction as the original.
Therefore the direction of the instantaneous
acceleration for UCM is toward the ______ of the
______.

61
Short-Cut Formula for the magnitude of the
Instantaneous Acceleration of UCM

We have derived a v2 Memorize this!
r
where v is the speed and r is the radius
As we determined earlier, for UCM, ?v is always
directed toward the center of the circle.
But a and ?v are always in the same direction.
This is because a ?v/?t , and division of a
vector by a positive scalar always results in a
new vector in the same direction as the original.
Therefore the direction of the instantaneous
acceleration for UCM is toward the center of the
circle.

62
A New Name for the Instantaneous Acceleration of
UCM
63
A New Name for the Instantaneous Acceleration of
UCM

A Latin adjective is used by physicists to help
them remember that the direction of the
instantaneous acceleration for UCM is always
directed towards the centre of the circle. Do
you know the name of this Latin adjective?

64
A New Name for the Instantaneous Acceleration of
UCM

A Latin adjective is used by physicists to help
them remember that the direction of the
instantaneous acceleration for UCM is always
directed towards the centre of the circle.
Instantaneous UCM acceleration is called
centripetal acceleration from the Latin petare
to seek and centri the centre

65
A New Name for the Instantaneous Acceleration of
UCM

A Latin adjective is used by physicists to help
them remember that the direction of the
instantaneous acceleration for UCM is always
directed towards the centre of the circle.
Instantaneous UCM acceleration is called
centripetal acceleration from the Latin petare
to seek and centri the centre
The instantaneous acceleration for UCM will
henceforth be referred to as centripetal
acceleration with new symbol ac . The formula
for the magnitude of the centripetal acceleration
is now modified as ac v2 / r

66
A New Name for the Instantaneous Acceleration of
UCM

A Latin adjective is used by physicists to help
them remember that the direction of the
instantaneous acceleration for UCM is always
directed towards the centre of the circle.
Instantaneous UCM acceleration is called
centripetal acceleration from the Latin petare
to seek and centri the centre
The instantaneous acceleration for UCM will
henceforth be referred to as centripetal
acceleration with new symbol ac . The formula
for the magnitude of the centripetal acceleration
is now modified as ac v2 / r Memorize
please!

67
Other Formulas for UCM centripetal acceleration
68
Other Formulas for UCM centripetal acceleration

We can use ac v2 / r if we know the speed and
radius of UCM. But suppose we don't know the
speed, but know the time for the UCM to complete
one revolution or rotation. This special time is
called _____________ or the time to complete one
cycle and it has the symbol _____.

69
Other Formulas for UCM centripetal acceleration

We can use ac v2 / r if we know the speed and
radius of UCM. But suppose we don't know the
speed, but know the time for the UCM to complete
one revolution or rotation. This special time is
called period or the time to complete one cycle
and it has the symbol T.

70
Other Formulas for UCM centripetal acceleration

We can use ac v2 / r if we know the speed and
radius of UCM. But suppose we don't know the
speed, but know the time for the UCM to complete
one revolution or rotation. This special time is
called period or the time to complete one cycle
and it has the symbol T. For example, the earth
going around the sun is approximately UCM with a
period of
T _________days.

71
Other Formulas for UCM centripetal acceleration

We can use ac v2 / r if we know the speed and
radius of UCM. But suppose we don't know the
speed, but know the time for the UCM to complete
one revolution or rotation. This special time is
called period or the time to complete one cycle
and it has the symbol T. For example, the earth
going around the sun is approximately UCM with a
period of
T 365.25 days.

72
Other Formulas for UCM centripetal acceleration

We can use ac v2 / r if we know the speed and
radius of UCM. But suppose we don't know the
speed, but know the time for the UCM to complete
one revolution or rotation. This special time is
called period or the time to complete one cycle
and it has the symbol T. For example, the earth
going around the sun is approximately UCM with a
period of
T 365.25 days.
We can quickly derive a formula for ac in terms
of r and T.

73
Deriving a Centripetal Acceleration Formula in
terms of T and r
74
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t

75
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t
For an object undergoing UCM, what formula can we
use to calculate the distance ?d once around the
circumference of a circle if we know the radius
r?

76
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t
?d 2pr

77
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t
?d 2pr
What do we call the time interval ?t for an
object to complete one cycle or revolution?

78
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t
?d 2pr
?t T

79
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t Eq 1
?d 2pr Eq 2
?t T Eq 3

80
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t Eq 1
?d 2pr Eq 2
?t T Eq 3
Without simplifying, sub equations 2 and 3 into
1

81
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t Eq 1
?d 2pr Eq 2
?t T Eq 3
Without simplifying, sub equations 2 and 3 into
1
v 2pr/T Eq 4

82
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t Eq 1
?d 2pr Eq 2
?t T Eq 3
Without simplifying, sub equations 2 and 3 into
1
v 2pr/T Eq 4 We know ac v2 / r Eq 5

83
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t Eq 1
?d 2pr Eq 2
?t T Eq 3
Without simplifying, sub equations 2 and 3 into
1
v 2pr/T Eq 4 We know ac v2 / r Eq 5
Without simplifying, sub Eq 4 into Eq5

84
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t Eq 1
?d 2pr Eq 2
?t T Eq 3
Without simplifying, sub equations 2 and 3 into
1
v 2pr/T Eq 4 We know ac v2 / r Eq 5
Without simplifying, sub Eq 4 into Eq5
ac (2pr/T)2 / r

85
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t Eq 1
?d 2pr Eq 2
?t T Eq 3
Without simplifying, sub equations 2 and 3 into
1
v 2pr/T Eq 4 We know ac v2 / r Eq 5
Without simplifying, sub Eq 4 into Eq5
ac (2pr/T)2 / r With pencil, simplify to a
two-story expression

86
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t Eq 1
?d 2pr Eq 2
?t T Eq 3
Without simplifying, sub equations 2 and 3 into
1
v 2pr/T Eq 4 We know ac v2 / r Eq 5
Without simplifying, sub Eq 4 into Eq5
ac (2pr/T)2 / r With pencil, simplify to a
two-story expression ac 4p2r / T2

87
Deriving a Centripetal Acceleration Formula in
terms of T and r

The basic constant speed formula is
v ?d/?t Eq 1
?d 2pr Eq 2
?t T Eq 3
Without simplifying, sub equations 2 and 3 into
1
v 2pr/T Eq 4 We know ac v2 / r Eq 5
Without simplifying, sub Eq 4 into Eq5
ac (2pr/T)2 / r With pencil, simplify to a
two-story expression ac 4p2r / T2 Memorize
please!

88
Deriving a Centripetal Acceleration Formula in
terms of f and r
89
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) What is f
called in physics?

90
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) f is called
frequency.

91
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) f is called
frequency. In SI, the unit of frequency is
cycles/s or 1/s or s-1 or the ________.

92
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) f is called
frequency. In SI, the unit of frequency is
cycles/s or 1/s or s-1 or the Hertz (Hz).

93
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) f is called
frequency. In SI, the unit of frequency is
cycles/s or 1/s or s-1 or the Hertz (Hz).
How are frequency and period related in an
equation?

94
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) f is called
frequency. In SI, the unit of frequency is
cycles/s or 1/s or s-1 or the Hertz (Hz).
Frequency and period are reciprocals of each
other. So
f 1/T

95
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) f is called
frequency. In SI, the unit of frequency is
cycles/s or 1/s or s-1 or the Hertz (Hz).
Frequency and period are reciprocals of each
other. So
f 1/T Eq 6

96
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) f is called
frequency. In SI, the unit of frequency is
cycles/s or 1/s or s-1 or the Hertz (Hz).
Frequency and period are reciprocals of each
other. So
f 1/T Eq 6 But we know ac 4p2r / T2

97
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) f is called
frequency. In SI, the unit of frequency is
cycles/s or 1/s or s-1 or the Hertz (Hz).
Frequency and period are reciprocals of each
other. So
f 1/T Eq 6 But we know ac 4p2r / T2
We can write ac 4p2r
T2

98
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) f is called
frequency. In SI, the unit of frequency is
cycles/s or 1/s or s-1 or the Hertz (Hz).
Frequency and period are reciprocals of each
other. So
f 1/T Eq 6 But we know ac 4p2r / T2
We can write ac 4p2r or 4p2r (1/T2)
T2

99
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) f is called
frequency. In SI, the unit of frequency is
cycles/s or 1/s or s-1 or the Hertz (Hz).
Frequency and period are reciprocals of each
other. So
f 1/T Eq 6 But we know ac 4p2r / T2
We can write ac 4p2r or 4p2r (1/T2)
T2
How can we express ac in terms of frequency?

100
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) f is called
frequency. In SI, the unit of frequency is
cycles/s or 1/s or s-1 or the Hertz (Hz).
Frequency and period are reciprocals of each
other. So
f 1/T Eq 6 But we know ac 4p2r / T2
We can write ac 4p2r or 4p2r (1/T2)
T2
ac 4p2r f2

101
Deriving a Centripetal Acceleration Formula in
terms of f and r

For UCM. sometimes we don't know v or T, but we
know f , the number of revolutions or complete
circles a particle goes through per unit time.
(usually in units of seconds) f is called
frequency. In SI, the unit of frequency is
cycles/s or 1/s or s-1 or the Hertz (Hz).
Frequency and period are reciprocals of each
other. So
f 1/T Eq 6 But we know ac 4p2r / T2
We can write ac 4p2r or 4p2r (1/T2)
T2
ac 4p2r f2 Memorize please!

102
Review of the Circle Three for UCM
103
Review of the Circle Three for UCM

For UCM, what is the formula for the magnitude of
the centripetal acceleration in terms of speed
and radius?

104
Review of the Circle Three for UCM

ac v2 / r UCM circle equation 1

105
Review of the Circle Three for UCM

ac v2 / r UCM circle equation 1
For UCM, what is the formula for the magnitude of
the centripetal acceleration in terms of period
and radius?

106
Review of the Circle Three for UCM

ac v2 / r UCM circle equation 1
ac 4p2r / T2 UCM circle equation 2

107
Review of the Circle Three for UCM

ac v2 / r UCM circle equation 1
ac 4p2r / T2 UCM circle equation 2
For UCM, what is the formula for the magnitude of
the centripetal acceleration in terms of
frequency and radius?

108
Review of the Circle Three for UCM

ac v2 / r UCM circle equation 1
ac 4p2r / T2 UCM circle equation 2
ac 4p2r f2 UCM circle equation 3

109
Review of the Circle Three for UCM

ac v2 / r UCM circle equation 1
ac 4p2r / T2 UCM circle equation 2
ac 4p2r f2 UCM circle equation 3
The direction of ac is always directed towards
the __________ of the _________.

110
Review of the Circle Three for UCM

ac v2 / r UCM circle equation 1
ac 4p2r / T2 UCM circle equation 2
ac 4p2r f2 UCM circle equation 3
The direction of ac is always directed towards
the centre of the circle.

111
Review of the Circle Three for UCM

ac v2 / r UCM circle equation 1
ac 4p2r / T2 UCM circle equation 2
ac 4p2r f2 UCM circle equation 3
The direction of ac is always directed towards
the centre of the circle.
If we want acceleration in units of m/s2, we must
substitute ________ units into the formula.

112
Review of the Circle Three for UCM

ac v2 / r UCM circle equation 1
ac 4p2r / T2 UCM circle equation 2
ac 4p2r f2 UCM circle equation 3
The direction of ac is always directed towards
the centre of the circle.
If we want acceleration in units of m/s2, we must
substitute MKS units into the formula.

113
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?
114
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given

115
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days

116
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days
r 384,000 km

117
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days X 24 h/day X 3600s/h ?
r 384,000 km

118
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days X 24 h/day X 3600s/h
2.36X106s
r 384,000 km

119
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days X 24 h/day X 3600s/h
2.36X106s
r 384,000 km X 1000 m/km ?

120
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days X 24 h/day X 3600s/h
2.36X106s
r 384,000 km X 1000 m/km
3.84X108m

121
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days X 24 h/day X 3600s/h
2.36X106s
r 384,000 km X 1000 m/km
3.84X108m
Unknown
?

122
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days X 24 h/day X 3600s/h
2.36X106s
r 384,000 km X 1000 m/km
3.84X108m
Unknown
ac

123
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days X 24 h/day X 3600s/h
2.36X106s
r 384,000 km X 1000 m/km
3.84X108m
Formula Unknown ac

124
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days X 24 h/day X 3600s/h
2.36X106s
r 384,000 km X 1000 m/km
3.84X108m
Formula Unknown ac
ac 4p2r / T2

125
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days X 24 h/day X 3600s/h
2.36X106s
r 384,000 km X 1000 m/km
3.84X108m
Formula Unknown ac
ac 4p2r / T2
Sub

126
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days X 24 h/day X 3600s/h
2.36X106s
r 384,000 km X 1000 m/km
3.84X108m
Formula Unknown ac
ac 4p2r / T2
Sub ac 4p2(3.84X108) / (2.36X106)2

127
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days X 24 h/day X 3600s/h
2.36X106s
r 384,000 km X 1000 m/km
3.84X108m
Formula Unknown ac
ac 4p2r / T2
Sub ac 4p2(3.84X108) / (2.36X106)2
ac 2.72 X 10-3 m/s2 toward the
centre of the earth

128
Example 1 The moon goes around the earth in
approximately UCM every 27.3 days with respect to
the stars. (called a sidereal month). If the
centre-to-centre distance from earth to moon is
384,000 km, what is the moon's centripetal
acceleration about the earth in m/s/s?

Given T 27.3 days X 24 h/day X 3600s/h
2.36X106s
r 384,000 km X 1000 m/km
3.84X108m
Formula Unknown ac
ac 4p2r / T2
Sub ac 4p2(3.84X108) / (2.36X106)2
ac 2.72 X 10-3 m/s2 toward the
centre of the earth
Note the moon is accelerating with a very
tiny acceleration compared to
projectiles near the earth's surface.

129
Try Example 2 A car is moving on a circular
track of radius 0.522 km. The magnitude of its
centripetal acceleration is 4.00 m/s2. Find the
speed of the car in km/h.

130
Try Example 2 A car is moving on a circular
track of radius 0.522 km. The magnitude of its
centripetal acceleration is 4.00 m/s2. Find the
speed of the car in km/h.

Given
r 0.522 km 0.522 km X 1000 m/km 522 m
ac 4.00 m/s2
Unknown v ?
Formula ac v2 / r or v2 r ac or v
(rac)1/2
Sub v (522 X 4.00)1/2
45.7 m/s X 3.6
165 km/h too fast!

131
Example 3The planet Mercury moves in an
approximately circular path around the sun at an
average distance of 5.8 X 1010 m, accelerating
centripetally at 0.040 m/s2. What is the period
of revolution about the sun in days?
132
Example 3The planet Mercury moves in an
approximately circular path around the sun at an
average distance of 5.8 X 1010 m, accelerating
centripetally at 0.040 m/s2. What is the period
of revolution about the sun in days?

Given r 5.8 X 1010 m ac 0.040 m/s2
Unknown T ?
Formula ac 4p2r / T2 or T (4p2r / ac)1/2
Sub T (4p2 (5.8 X 1010 )/ 0.040 )1/2
7.6 X 106 s
7.6 X 106 s X 1 hour/3600 s X 1 day/24 hours
88 days

133
Example 4 A stone is whirled in UCM on a
smooth sheet of ice. The stone traces out a
circle of diameter 3.0 m as it accelerates
centripetally at 93.0 m/s2. Find the frequency
of rotation in Hertz and in rpm (revolutions per
minute)
134
Example 4 A stone is whirled in UCM on a
smooth sheet of ice. The stone traces out a
circle of diameter 3.0 m as it accelerates
centripetally at 93.0 m/s2. Find the frequency
of rotation in Hertz and in rpm (revolutions per
minute)