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Chapter 6
• Circular Motion,
Orbits, and
Gravity
Uniform Circular Motion
Slide 6-23
For uniform circular motion, the
acceleration
A. is parallel to the velocity
B. is directed toward the center of
the circle
C. is large for a larger orbital the
same speed
D. is always due to gravity
100%
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Forces in Circular Motion
v = r
v2
a=
  2r
r
 mv 2

Fnet = ma = 
, toward center of circle 
 r

Slide 6-24
Time of Circular Motion
1
𝑓=
𝑇
The frequency is
just the inverse
of the period
Example of frequency units:
2𝜋𝑟
𝑣=
𝑇
𝑣 = 2𝜋𝑟𝑓
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rev
min
What happens when speed decreases
and radius decreases?
𝑣2
𝑎=
𝑟
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Comparing centripetal accelerations
Rank the acceleration of the following uniform
circular motions from smallest to largest:
They’re all going to have 4𝜋 2 so we
just need to find 𝑟𝑓 2 for each one
1
2
3
2𝜋𝑟𝑓
𝑎=
𝑟
4
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𝑣 = 2𝜋𝑟𝑓
𝑣2
𝑎=
𝑟
2
= 4𝜋 2 𝑟𝑓 2
5
Comparing centripetal accelerations
Rank the acceleration of the following uniform
circular motions from smallest to largest
𝑟1 𝑓12
10cm
1
rev
50
min
𝑟2 𝑓22
2
20cm
2
rev
50
min
𝑟3 𝑓32
2
20cm
3
rev
100
min
2
𝑟4 𝑓42
40cm
4
© 2015 Pearson Education, Inc.
rev
100
min
𝑣 = 2𝜋𝑟𝑓
𝑣2
𝑎=
𝑟
𝑟5 𝑓52
2
40cm
5
rev
200
min
2
Driving on a curvy road
𝑣2
𝑎=
𝑟
𝑎=0
𝑣2
𝑎=
𝑟
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Problem solving uniform circular motion
𝑣 = 2𝜋𝑟𝑓
𝑣2
𝑎=
𝑟
© 2015 Pearson Education, Inc.
Problem solving uniform circular motion
A level curve on a country road has a radius of 150 m. What is
the maximum speed at which this curve can be safely negotiated
on a rainy day when the coefficient of friction between the tires
on a car and the road is 0.40?
Slide 6-28
Problem solving uniform circular motion
A level curve on a country road has a radius of 150 m. What is the maximum
speed at which this curve can be safely negotiated on a rainy day when the
coefficient of friction between the tires on a car and the road is 0.40?
𝑓𝑠 = 𝜇𝑠 𝑁
𝑣2
𝑓𝑠 = 𝜇𝑠 𝑚𝑐 𝑔 = 𝑚𝑐
𝑟
𝑣 = 𝜇𝑠 𝑔𝑟
Slide 6-28
When a car turns a corner on a level road, which
force provides the necessary centripetal
acceleration?
76%
12%
10%
Ai
r
Gr
av
ity
re
sis
ta
n
ce
Fo
rc
e
0%
No
rm
al
sio
n
2%
Te
n
Friction
Tension
Normal Force
Air resistance
Gravity
Fr
ict
io
n
A.
B.
C.
D.
E.
Additional Questions
A coin sits on a rotating
turntable.
1. At the time shown in the
figure, which arrow gives the
direction of the coin’s
velocity?
Slide 6-45
Answer
A coin sits on a rotating
turntable.
1. At the time shown in the
figure, which arrow gives the
direction of the coin’s
velocity?
A
Slide 6-46
Additional Questions
A coin sits on a rotating
turntable.
2. At the time shown in the
figure, which arrow gives the
direction of the frictional
force on the coin?
Slide 6-47
Answer
A coin sits on a rotating
turntable.
2. At the time shown in the
figure, which arrow gives the
direction of the frictional
force on the coin?
D
Slide 6-48
Additional Questions
A coin sits on a rotating
turntable.
3. At the instant shown,
suppose the frictional force
disappeared. In what
direction would the coin
move?
Slide 6-49
Answer
A coin sits on a rotating
turntable.
3. At the instant shown,
suppose the frictional force
disappeared. In what
direction would the coin
move?
A
Slide 6-50
A ball moves CW in circular motion.
What happens when it reaches the
opening?
95%
A
B
C
D
5%
D
0%
C
B
0%
A
A.
B.
C.
D.
Problem solving uniform circular motion
In the hammer throw, an
athlete spins a heavy mass
in a circle at the end of a
chain, then lets go of the
chain. For male athletes,
the “hammer” is a mass of
7.3 kg at the end of a 1.2 m
chain.
𝑣2
𝑚
𝑟
A world-class thrower can get the hammer up to a speed of 29
m/s. If an athlete swings the mass in a horizontal circle centered
on the handle he uses to hold the chain, what is the tension in the
chain?
2
m
29 s
𝑣
𝑇=𝑚
= 7.3kg
𝑟
1.2m
2
Slide 6-29
Problem solving uniform circular motion
A car of mass 1500 kg goes over a
hill at a speed of 20 m/s. The shape
of the hill is approximately circular,
with a radius of 60 m, as in the figure
at right. When the car is at the
highest point of the hill,
a. What is the force of gravity on
the car?
b. What is the normal force of the
road on the car at this point?
𝐹 = 𝑚𝑔
Slide 6-30
Problem solving uniform circular motion
A car of mass 1500 kg goes over a hill
at a speed of 20 m/s. The shape of the
hill is approximately circular, with a
radius of 60 m, as in the figure at right.
When the car is at the highest point of
the hill,
a. What is the force of gravity on the
car?
b. What is the normal force of the
road on the car at this point?
𝑣2
𝐹𝑦 = 𝑁 + 𝑚𝑔 = −𝑚
𝑟
𝑣2
𝑁 = −𝑚
+𝑔
𝑟
Slide 6-30
Maximum walking speed
vmax  gr
Slide 6-31
Example Problem
A roller coaster car goes through a vertical loop at a constant
speed. For positions A to E, rank order the:
• centripetal acceleration
• normal force
• apparent weight
𝑣2
𝑎=
𝑟
Slide 6-32
Example Problem
A roller coaster car goes through a vertical loop at a constant
speed. For positions A to E, rank order the:
• centripetal acceleration
• normal force
• apparent weight
Slide 6-32
Example Problem
A roller coaster car goes through a vertical loop at a constant
speed. For positions A to E, rank order the:
• centripetal acceleration
• normal force
• apparent weight
Slide 6-32
Net Force causes centripetal acceleration
𝐹 = 𝐹net
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Net Force causes centripetal acceleration
© 2015 Pearson Education, Inc.
Friction force as a Net Force
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Problem solving circular motion with Net Force
𝑣2
𝑓𝑠 = 𝜇𝑠 𝑚𝑐 𝑔 = 𝑚𝑐
𝑟
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𝑣 = 𝜇𝑠 𝑔𝑟
A car travels around a corner with the
cruise control on. Which vector is the
car’s centripetal acceleration?
A
B
C
D
E
94%
E
2%
D
2%
C
2%
B
0%
A
A.
B.
C.
D.
E.
A car travels around a corner while speeding up.
Which vector is the net force on the car?
A
B
C
D
E
84%
6%
E
C
B
0%
2%
D
8%
A
A.
B.
C.
D.
E.
A car travels around a corner while speeding up.
Which vector is the force of friction from the
road on the car?
A
B
C
D
E
96%
D
E
2%
0%
C
0%
B
2%
A
A.
B.
C.
D.
E.
Fictitious Forces in circular motion
Feels like something is
pushing you out of
the car
© 2015 Pearson Education, Inc.
A car travels around a corner at a constant
speed. Which vector is the fictitious force on the
car?
0%
0%
D
E
0%
B
0%
C
A
B
C
D
E
A
A.
B.
C.
D.
E.
100%
Simulating Gravity
This one guy you knew from a while back says that in
order to avoid loss of bone density on his trip to mars so,
the ship must be spinning to simulate gravity. How fast
must the craft below spin to simulate gravity?
𝑣
𝐹𝑔 = 𝑁
𝑣2
𝑚𝑔 = 𝑚
𝑟
𝑁
𝑣 = 𝑔𝑟
© 2015 Pearson Education, Inc.
Tilted roads in favor of circular motion
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Example Problem: Over the Top
A handful of professional skaters have taken a skateboard
through an inverted loop in a full pipe. For a typical pipe with
diameter 14 ft, what is the minimum speed the skater must have
at the very top of the loop?
Slide 6-33
Example Problem: Over the Top
A handful of professional skaters have taken a skateboard
through an inverted loop in a full pipe. For a typical pipe with
diameter 14 ft, what is the minimum speed the skater must have
at the very top of the loop?
𝑣2
𝐹 = 𝐹𝑔 = 𝑚
𝑟
𝑁=0
𝐹𝑔
𝑣2
−𝑚𝑔 = −𝑚
𝑟
𝑣=
𝑔𝑟
ft
help: 𝑔 = 32 2
s
Slide 6-33
Keeping riders safe in circular motion
© 2015 Pearson Education, Inc.
When a ball on the end of a string is
swung in a vertical circle, the ball’s
acceleration is because
100%
Th
e
0%
er
at
io
n
ar
e.
..
cc
el
re
ct
io
n
is
n
ba
ll
Th
e
ee
d
an
d
di
is
ch
an
gin
g
0%
sp
di
re
ct
io
n
Th
e
Th
e
sp
ee
d
is
ch
a
ng
in
g
0%
ot
a
A. The speed is changing
B. The direction is changing
C. The speed and direction are
changing
D. The ball is not acceleration
The Force of Gravity
Slide 6-35
The inverse-square law
𝑦
𝑥
𝑔 versus 𝐺 in Newton’s law of gravitation
𝑔 is a value unique to the mass
that’s pulling
𝐺 is the same throughout the
whole universe!
© 2015 Pearson Education, Inc.
Example Problem
A typical bowling ball is spherical, weighs 16 pounds, and has a
diameter of 8.5 in. Suppose two bowling balls are right next to
each other in the rack. What is the gravitational force between
the two—magnitude and direction? What is the magnitude and
direction of the force of gravity on a 60 kg person?
Slide 6-36
A 60 kg person stands on each of the following
planets. Ranks order her weight on the three
bodies, from highest to lowest
19%
4%
C>
A>
B
0%
C>
B>
A
B>
C
>A
0%
B>
A>
C
A>B>C
B>A>C
B>C>A
C>B>A
C>A>B
A>
B>
C
A.
B.
C.
D.
E.
77%
Which vector pair matches the car’s
motion over the hill?
94%
2%
C
B
4%
A
A. A
B. B
C. C
© 2015 Pearson Education, Inc.
The Force of Gravity
𝑣=
© 2015 Pearson Education, Inc.
𝐺𝑀e
~17,000mph
𝑟e
Example Problem: Orbital Motion
Phobos is one of two small moons that
orbit Mars. Phobos is a very small
moon, and has correspondingly small
gravity—it varies, but a typical value is
about 6 mm/s2. Phobos isn’t quite
round, but it has an average radius of
about 11 km. What would be the orbital
speed around Phobos, assuming it was
round with gravity and radius as noted?
𝑣=
𝐺𝑀
=
𝑟e
3
𝑚
.006
−11
24 kg
6.67 ∙ 10
∙
5.97
∙
10
kg ∙ s2 9.8
6.4 ∙ 106 m
Slide 6-34
A satellite orbits the earth. A space shuttle crew
is sent to boost the satellite into a higher orbit.
Which of these quantities increases?
94%
0%
Pe
rio
et
d
al
ac
ce
tio
le
na
ra
lf
t io
or
n
ce
of
th
e
ea
rth
nt
rip
Gr
av
ita
Ce
ed
ar
s
pe
ed
4%
2%
0%
An
gu
l
Speed
Angular speed
Period
Centripetal acceleration
Gravitational force of the
earth
Sp
e
A.
B.
C.
D.
E.
Summary
Slide 6-41
Summary
Slide 6-42