Unit 6 Part 2
Circular Motion
and Force
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
1
Center Seeking Acceleration
 An object traveling in a circular pattern always has an acceleration acting to the
center of its rotation (to the center of the circle traced out by the object).
 Which way does the acceleration act on the vertically swinging ball (left) and on
the man (right) on the Ferris Wheel when they are at the top of their rotation?
 Which way does the acceleration act on the vertically swinging ball (left) and on
the man (right) on the Ferris Wheel when they are at the bottom of their rotation?
a
a
a
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
a
2
FBD – Ferris Wheel
 Which way is the Normal Force acting on the man (when he is sitting in the car on





a seat) at the top of his rotation (NT)?
Which way is the weight acting on the man at the top?
Which way is the acceleration acting on the man at the top of the rotation?
Which way is the Normal Force acting on the man when he is at the bottom of his
rotation (NB)?
Which way is the weight acting on the man at the bottom?
Which way is the acceleration acting on the man at the bottom of the rotation?
NT
NB
a
a
W
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
W
3
Apparent Weight






Daniel Boone rides in a roller coaster car without a seat belt.
If the car is moving slowly, then will he be safe when he reaches the top of the hill? Explain.
Will he remain safe as the speed of the car increases (medium)? Explain.
Will he appear to weigh more or less as his speed over the top increases? Explain.
If he is going very, very fast, then what will happen to him when he reaches the top? Why?
If he was sitting on a scale at the moment he was ejected, then how much would the scale say
he weighed once he was ejected?
 No matter his speed, which way is he accelerating at the top of the hill?
a
Slow
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Medium
Fast
4
Apparent Weight
 What forces are acting on Daniel Boone while he is sitting in the car?
 Would you expect the upward Normal Force (the force the seat exerts
onto him) to increase or decrease as his speed over the top of the hill
increases? Explain.
 What would the value of the Normal Force equal when the car is moving
very, very fast resulting in the rider being ejected from the car?
NT
a
W
Run
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
5
Apparent Weight
 What trend did we observe in how much the man appeared to weigh (his apparent weight) as
the speed of the car over the top increased?
 We observed that his apparent weight is reduced as his speed increases.
 What conclusion did we draw in regards to the magnitude of the Normal Force as the speed of
the car increased?
 The Normal forces is reduced as his speed increases.
 What similarities exist in these two trends?
 Given this observed trend, what relationship do you suppose exists between the apparent
weight and the Normal Force?
 They are the same thing.
 If the car continues to speed up as it goes over the top of the hill, then what value will the
Normal Force (apparent weight) eventually reach?
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
6
Apparent Weight




Daniel Boone is now going through a valley on the coaster.
What forces are acting on him at the bottom of the valley?
In which direction is the acceleration acting at the bottom of the valley?
How will his Normal Force (apparent weight) change as his speed
increases?
 The Normal Force (apparent weight) increases as the speed increases.
NB
W
Slow
a
Medium
Fast
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
7
FBD - Vertically Swinging Ball
 Which way is the Tension in the string acting on the ball when it is at the top of its





rotation (TT)?
Which way is the weight acting on the ball at the top?
Which way is the acceleration acting on the ball at the top of the rotation?
Which way is the Tension in the string acting on the ball when it is at the bottom
of its rotation (TB)?
Which way is the weight acting on the ball at the bottom?
Which way is the acceleration acting on the ball at the bottom of the rotation?
TB
a
TT
a
W
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
W
8
Tension and Apparent Weight
 In the same manner that the Normal Force and the apparent weight are the same
thing in the Ferris Wheel and Roller Coaster examples, the Tension and the
apparent weight are the same thing in the vertically swinging ball example.
 At which point in the rotation of the ball, top or bottom, will the speed of the ball
be the greatest?
 If you are spinning a ball in a vertical circle, then at which position, top or bottom,
would you expect the Tension (apparent weight) to be the greatest? Explain.
 Since the speed is higher at the bottom, the Tension will be higher at the bottom.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
9
Circular Motion and Centripetal Acceleration












Let us take a look at a Medieval Flail.
Why did the flail continue to rotate before the chain broke?
What does the chain do to the rotating spiked ball?
The chain pulls inward on the spiked ball.
A pull is an example of a …?
A force.
What two things can a force do to a body?
Accelerate (change speed or change direction) it and/or
deform it.
The force on the chain causes an inward acceleration that
makes the spiked ball want to constantly change direction.
This inward acceleration is known as radial or centripetal
acceleration.
Once the chain breaks, the force and its accompanying
acceleration no longer exist.
As a result, the flail no longer “wants” to turn and proceeds
in a tangential direction from the point where the chain
broke.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
2
v
a
r
10
Centripetal Force - Swinging Ball Examples
 Any body rotating about a fixed point will experience a centripetal (center







seeking) acceleration.
This acceleration is always directed inwards towards the center of the circle.
The velocity is always perpendicular to this acceleration.
The acceleration’s magnitude is found using the following equation where v is its
tangential velocity and r is the radius of the circle.
If asked to find the centripetal force, then simply multiply the acceleration by the
mass.
When viewing “Swinging Ball” problems, note that the body has a force (tension,
normal,…) acting on it that is pulling the body towards the center of the circle at
all times.
In such instances, the Tension (or Normal Force)
is known as the “apparent weight.”
a
In this problem that force is the tension in the string.
v2
ac 
r
v2
F  m
r
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
v
11
Centripetal Force - Swinging Ball Examples #27
 On this slide we will derive the equations relating the tensions in the top and bottom of a
vertically swinging ball to the velocities of the ball at the top and bottom of its swing.
 Notice that the tension is pulling the ball towards the center of rotation at all times.
 We start with Newton’s second law and proceed as below (top in left column bottom in
right column).
 Think about the results.
2
 Which tension would you expect to be the largest: top or bottom? Why?
 Do our equations agree with our speculations? Explain?
c
v
a 
r
F  W  TT  mac
a
TT
W
TB
vT2
F  W  TT  m
r
vT2
F  W  TT  m
r
2
v
TT  m T W
r
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
F  W  TB  mac
a
W
vB2
F  W  TB  m
r
vB2
F  W  TB  m
r
2
v
TB  m B  W
r
12
Minimum Speed
 In order to maintain its circular orbit, the ball must spin on the string fast enough




to complete its circular orbit at the top of the swing.
The Tension (apparent weight) at the top would be less than that at the bottom but
greater than zero.
When the ball is traveling at the minimum speed, the Tension (apparent weight) in
the string will equal zero at the top of the swing.
What would happen to the ball’s orbit if it was moving at a speed less than the
minimum speed?
The orbit would decay and no longer be circular.
Maintain
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
Minimum
Decay
13
Centripetal Force - Swinging Ball Examples #28
 Often you will be asked to calculate the minimum
speed that the ball must maintain so that the ball
a
continues moving in a circular fashion in the vertical
plane.
TT
 What do you think the value of the tension would be at
the top of the circle if the ball was traveling at its
minimum speed?
W
 TT = 0.
 As a result, our equation would become as follows.
vT2
TT  m W
 Notice that this equation has no final dependence on
r
mass.
 This means that any mass (a boulder, a marble, a
2
v
bowling ball,…) traveling in a circle of radius r must
0  m T W
have at least the speed given by this equation in order
r
to continue to travel in a vertical circle.
vT 
Wr
mgr

 gr
m
m
vT  gr
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
14
Vertically Swinging Ball Problem?




Watch the animation below as the ball rolls down the hill.
What “type” of problem is one?
What is the chief defining characteristic of a swinging ball problem?
In this example, the Normal Force behaves in the same way as the Tension in a
vertically swinging ball problem.
 Therefore, this problem would be classified as a “Swinging Ball” problem.
TNTT
a
W
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
15
Circular Motion and Force #29
 Synchronized jet fighters perform a perfect
vertical circular loop maneuver.
 The jets are going 1622 km/h at the bottom of
the loop.
 What is the minimum diameter of the loop
needed in order to prevent the pilot from
experiencing no more than 5.5 g (5.5 times the
normal value of Earth’s gravity)?
 What is the apparent weight of the pilot at the
top and bottom of the loop?
 Before answering this question, identify the
“type” of problem that this example is.
 This problem is similar to which of the circular
types discussed so far?
 In which direction is the normal force acting on
the pilot at the top of the loop? Bottom?
 This problem is a swinging ball problem
because the normal force always acts towards
the center of the loop.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
NB  m(5.5g)
16
Centripetal Force – Ferris Wheel Examples
 As the man riding the Ferris wheel is moving in a
circle, he experiences a centripetal force (red) at
all points as shown.
 The seat is also exerting an upward normal force
on him (blue), and his net force (green) at any
point is the vectorial sum of these forces.
 At the top, the centripetal force counters the
normal force; therefore, the net force acts up and
is small.
 At the bottom, the centripetal force acts in the
same direction as the normal force; therefore, the
net force acts up and is large.
 Where would Dr. Physics’ apparent weight be
the highest: top or bottom?
 When viewing “Ferris Wheel” problems, note
that the force (tension, normal,…) acting on the
body changes directions (towards or away from
the center) as the body rotates.
 Notice that the normal force acts away from the
center at the top of the Ferris wheel and towards
the center at the bottom of the Ferris wheel.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
17
Centripetal Force – Ferris Wheel Examples #30
 On this slide we will derive the equations relating the Normal Force in the top and
bottom of a vertically spinning Ferris Wheel to the velocities of a body at the top
and bottom of its rotation.
 We start with Newton’s second law and proceed as below (top in left column
bottom in right column).
v2
ac 
 Think about the results.
r
 Which normal force would you expect to be the largest: top or bottom? Why?
 Do our equations agree with our speculations? Explain?
NT
vT2
F  W  NT  m
r
a
W
NB
F  W  NT  mac
vT2
F  W  NT  m
r
2
v
NT  m T  W
r
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
F  W  NB  mac
vB2
F  W  NB  m
r
a
W
vB2
F  W  NB  m
r
2
v
NB  m B  W
r
18
Maximum Safe Speed
 Previously were observed that an unsecured rider would eventually be ejected from a
roller coaster (Ferris Wheel) as the roller coaster increased speed.
 We also observed that the magnitude of the Normal Force (apparent weight) reduced
as the speed at the top increased.
 If the car continues to speed up as it goes over the top of the hill, then what value
will the Normal Force (apparent weight) eventually reach?
 The speed, where the Normal Force becomes zero, is the maximum safe speed of the
coaster for an unsecured rider.
 If the speed of the coaster increases beyond this point, then the rider will be ejected.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
19
Centripetal Force – Ferris Wheel Examples #31
 Often you will be asked to calculate the maximum speed that
a body may have in order to remain in the Ferris Wheel
without being buckled in.
 What do you think the value of the normal force would be at
the top of the circle if the body was traveling at this
maximum speed?
 N = 0.
 As a result, our equation would become as follows.
 Notice that this equation has no final dependence on mass.
 This means that any mass (a boulder, a marble, a bowling
ball,…) sitting in a Ferris wheel traveling in a circle or
radius r must have a speed not greater than that given by this
equation in order to keep from being thrown from the Ferris
Wheel.
 If a man (65.0 kg) and a girl (17.0 kg) were both in a Ferris
wheel and not strapped in, then both would be thrown from
the ride.
NT
a
N  m
2
T
v
W
r
W
vT2
0  m  W
r
vT 
Wr
mgr

 gr
m
m
vT  gr
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
20
Circular Motion and Force #32
 A test dummy (m = 62.5 kg) rides freely in a high-speed roller coaster with a radius of
curvature of r = 52.00 m.
 What type of a problem is this one?
 A Ferris wheel problem.
 If the coaster was moving at vT = 18.00 m/s
at the top of the coaster, then what is the
apparent weight of the test dummy at this
point?
 What is the maximum safe speed of the
roller coaster?
RUN
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
21
Circular Motion and Force #33
 An airplane, whose engine failed, was gliding to the ground.
 Once the engine restarted, the pilot “pulled up” in order to keep the plane from
crashing into the ground.
 The radius of curvature of the pull up is 175.0 m, and the plane’s speed at the
lowest point of this curve is 78.0 m/s.
 What is the apparent weight of the 58.2 kg pilot at this lowest point in flight?
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
22
Friction Types for Rolling and Skidding Tires
 What type of friction acts between the road and a tire that is rolling without
skidding?
 What type of friction acts between the road and a tire that is skidding?
 Watch the red dot on the tire below as it rolls and then slides.
 When the tire is rolling, the red spot is stationary above the position on the
road that it is in contact with; therefore, stronger static friction acts on the
tire.
 When the tire is skidding, the red spot slides relative to road’s surface;
therefore, the weaker kinetic friction acts on the tire.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
23
Skidding and Curves
 When an ATV safely negotiates a curve, the
centripetal acceleration produced by the
inward acting static friction is able to
accelerate (change the direction of) the ATV.
 The tracks left behind by this ATV would
look like the ones to the right.
 When an ATV goes around the curve at too
high a speed, the tires begin to skid.
 The centripetal acceleration produced by the
inward acting kinetic friction is not able to
accelerate (change the direction of) the ATV
quickly enough causing the ATV to go out of
control.
 The tracks left behind by this ATV would
look like the ones to the right.
 If you are a race car fan, then you have likely
seen similar marks left behind by many cars
on the surface of the race track.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
24
Negotiating Curves
 Watch the animation to the right and
explain what has happened.
 What force acted on the ATV that
prevented it from slipping off the road
during the first lap and in which direction
was it acting?
 The static friction directed inwards towards
the center of curvature.
 What forces were acting on the ATV when
it slipped off the road?
 Excess speed caused the tires to skid and to
slide changing the static friction to a much
smaller kinetic friction that was not able to
make the ATV change direction.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
N
ac
FsFk
W
25
Negotiating Unbanked Curves #36
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
N
 v2 
Fx  FFS  m  
r
ac
Fy  N W  0
Fs
FFS  s N
FFS  s N
W
N  W  mg
Force of Friction v. Time
1
0
-1 0
FFS  s mg
Friction (N)
 When a car negotiates an unbanked
curve, the only force causing the
centripetal acceleration is the static
friction acting inwards along the
radius of curvature.
 This static friction accelerates the car
and causes it to turn.
 Use Newton’s laws to derive an
expression for the maximum speed
the ATV can have to safely negotiate
a curve of radius r when the
coefficient of static friction between
the tires and the road is s.
 The equation for static friction is as
follows.
 Since we are considering the
maximum static friction, the static
friction equation becomes as follows.
 Is the size of the body of any
importance in this equation? Why or
why not?
2
4
6
8
10
-2
-3
-4
-5
-6
v 
s mg  m  
r
2
-7
Time (s)
v  s gr
26
Circular Motion and Force #37
 A semi tractor approaches a curve at 20.0 m/s.
 The coefficient of static friction between the truck’s tires
and the road is 0.68.
 The radius of curvature of the road is 45.0 m.
 Will the truck skid on this road?
 What is the radius of curvature if the maximum safe
speed around this curve is 35.0 m/s?
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
27
Curves and Merry-Go-Rounds
 A motorcycle runs on a curved (circular) track next to a boy riding a Merry-GoRound.
 What kind of friction acts between the tires and the track? Boy’s shoes and MGR?
 What would happen if the motorcycle went to fast? MGR spins too fast?
 As a result, a MGR can be thought of as a curve problem.
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
28
Circular Motion and Force #38
 Two boys play on a merry-go-round as shown.
 The white boy (m = 38.0 kg) is a distance of 1.35 m from the center traveling at a
speed v = 3.33 m/s,
 The black boy (m = 31.0 kg) is 1.20 m from the center raveling at a speed v = 2.96
m/s.
 The coefficient of static frictions s = 0.68.
 Which boy will slide off the merry-go-round? Why?
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
29
Cresting Hills
 A truck carrying a crate negotiates a hill as shown below.
 What would happen to the crate if the truck went too fast?
 What type of problem could you classify the truck cresting a hill as being? Why?
© 2001-2005 Shannon W. Helzer. All Rights Reserved.
30
Circular Motion and Force #20 and #21
 A young child (m = 33.00 kg) rides in the bed of a pickup truck as the truck
approaches a hill with a radius of curvature of 25.00 m. If the truck is traveling at a
speed of 15.00 m/s, then what will be the apparent weight of the boy at the very top
of the hill?
 On the return trip, the driver of the vehicle is traveling at 16.00 m/s. Will the child
be safe riding in the back of the vehicle? Hint: find the maximum safe speed and
properly interpret your findings.
 How would this maximum speed vary if it was a full grown adult riding in the back
of the truck instead of a child.
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