Circular Motion Lab Results Part 1 Radius effect on velocity Mass centripetal force kept constant Part 2 Mass effect on velocity Radius centripetal force kept constant Part 3 Forces effect on velocity ID: 623488
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Slide1
Uniform Circular MotionSlide2
Circular Motion Lab Results
Part 1: Radius’ effect on velocity
Mass, centripetal force kept constant
Part 2: Mass’ effect on velocityRadius, centripetal force kept constantPart 3: Force’s effect on velocityRadius, mass kept constant
Slide3
Circular Motion Lab Results
If the sum of the forces on an object is pointed toward a common center of rotation, perpendicular to its velocity, it will travel in a circle
at a constant speed
.
m
∑F
V
rSlide4
Warm-up Question (1/10)
What is the average speed of the Moon in its orbit given the information below?
Answer:Slide5
Example Problem #2
What is the average speed of the Earth in its orbit around the Sun given the information below?
Answer:Slide6
Vector nature of circular motion
Draw a circle on your whiteboard.
Determine a direction of circular motion (clockwise or counterclockwise)
Draw 2 dots, labelling them A and B—B will be further along the motion than ADraw the instantaneous velocity vectors at each of the pointsDraw a resultant vector showing the change in velocity between A and BSlide7
Terminology reminders:
Centripetal Force
:
The net force, acting towards the center of rotation, the allows an object to travel in uniform circular motion.Centripetal Acceleration: The acceleration of the object, caused by the centripetal force, directed towards the center of the circular path.Slide8
Applying Newton’s 2nd Law
We know, from our lab, that:
Since Newton’s 2
nd
law tells us
And the centripetal force is a net force, then the centripetal acceleration must be:
Slide9
Question 7.1
Tetherball
a) toward the top of the pole
b) toward the ground
c) along the horizontal component of the tension force
d) along the vertical component of the tension force
e) tangential to the circle
In the game of tetherball, the struck ball whirls around a pole. In what direction does the
net force on the ball point?
W
TSlide10
The
vertical component of the tension
balances the
weight
. The
horizontal component of tension
provides the
centripetal force
that points toward the center of the circle.
W
T
W
T
Question 7.1
Tetherball
a) toward the top of the pole
b) toward the ground
c) along the horizontal component of the tension force
d) along the vertical component of the tension force
e) tangential to the circle
In the game of tetherball, the struck ball whirls around a pole. In what direction does the
net force
on the ball point?Slide11
You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you?
a) you are thrown to the right
b) you feel no particular change
c) you are thrown to the left
d) you are thrown to the ceiling
e) you are thrown to the floor
Question 7.2a
Around the Curve ISlide12
You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you?
Question 7.2a
Around the Curve I
The passenger has the tendency to continue moving in a straight line. From your perspective in the car, it feels like you are being thrown to the right, hitting the passenger door.
a) you are thrown to the right
b) you feel no particular change
c) you are thrown to the left
d) you are thrown to the ceiling
e) you are thrown to the floorSlide13
a) centrifugal force is pushing you into the door
b) the door is exerting a leftward force on you
c) both of the above
d) neither of the above
During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening?
Question 7.2b
Around the Curve IISlide14
a) centrifugal force is pushing you into the door
b) the door is exerting a leftward force on you
c) both of the above
d) neither of the above
During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening?
The passenger has the tendency to continue moving in a straight line. There is a centripetal force, provided by the door, that forces the passenger into a circular path.
Question 7.2b
Around the Curve IISlide15
a) car’s engine is not strong enough to keep the car from being pushed out
b) friction between tires and road is not strong enough to keep car in a circle
c) car is too heavy to make the turn
d) a deer caused you to skid
e) none of the above
You drive your dad’s car too fast around a curve and the car starts to skid. What is the correct description of this situation?
Question 7.2c
Around the Curve IIISlide16
The friction force between tires and road provides the centripetal force that keeps the car moving in a circle. If this force is too small, the car continues in a straight line!
a) car’s engine is not strong enough to keep the car from being pushed out
b) friction between tires and road is not strong enough to keep car in a circle
c) car is too heavy to make the turn
d) a deer caused you to skid
e) none of the above
You drive your dad’s car too fast around a curve and the car starts to skid. What is the correct description of this situation?
Question 7.2c
Around the Curve III
Follow-up
: What could be done to the road or car to prevent skidding?Slide17
Question 7.3
Missing Link
A Ping-Pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the Ping-Pong ball leaves the track,
which path will it follow
?
a
b
c
d
eSlide18
Question 7.3
Missing Link
Once the ball leaves the tube, there is no longer a force to keep it going in a circle. Therefore, it simply continues in a straight line, as Newton’s First Law requires!
A Ping-Pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the Ping-Pong ball leaves the track,
which path will it follow
?
Follow-up
: What physical force provides the centripetal force?
e
d
a
b
cSlide19
Question 7.4
Ball and String
a)
T
2
= ¼
T
1
b)
T2 = ½T1c) T
2 = T
1
d)
T
2
= 2
T
1
e)
T
2
= 4
T
1
Two equal-mass rocks tied to strings are whirled in horizontal circles. The
radius
of circle 2 is
twice
that of circle 1. If the
period
of motion is the
same
for both rocks,
what is the tension in cord 2 compared to cord 1
?
1
2Slide20
The centripetal force in this case is given by the tension, so
T
=
mv
2
/
r
. For the same period, we find that
v
2 = 2v1
(and this term is squared). However, for the denominator, we see that r2 = 2r1 which gives us the relation T2 = 2
T1.
Question 7.4
Ball and String
Two equal-mass rocks tied to strings are whirled in horizontal circles. The
radius
of circle 2 is
twice
that of circle 1. If the
period
of motion is the
same
for both rocks,
what is the tension in cord 2 compared to cord 1
?
1
2
a)
T
2
= ¼
T
1
b)
T
2
= ½
T
1
c)
T
2
=
T
1
d)
T
2
= 2
T
1
e)
T
2
= 4
T
1Slide21
Question 7.5
Barrel of Fun
A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?
a
b
c
d
eSlide22
The
normal force
of the wall on the rider provides the
centripetal force
needed to keep her going around in a circle. The
downward force of gravity is balanced by the upward frictional force
on her, so she does not slip vertically.
Question 7.5
Barrel of Fun
A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?
Follow-up
: What happens if the rotation of the ride slows down?
a
b
c
d
eSlide23
Practice Problem #3
An astronaut rotates at the end of a test machine whose arm has a length of 10.0 m. If the maximum acceleration she reaches must not exceed 5
g
, what is the maximum number of revolutions per minute of the arm?Slide24
Practice Problem #4
A neutron star has a radius of 50.0 km and completes 1 rotation every 25.0
ms.
What is the centripetal acceleration experienced at the equator?Hypothetically, what is the centripetal force acting on a 50.0 kg object at the surface of the neutron star? How many times its weight is this?Slide25
Question 7.6a
Going in Circles I
a)
N
remains equal to
mg
b)
N
is smaller than
mgc) N is larger than mg
d) none of the above
You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the
normal force
N
exerted by your seat is equal to your
weight
mg
. How does
N
change at the top of the Ferris wheel when you are in motion?Slide26
Question 7.6a
Going in Circles I
a)
N
remains equal to
mg
b)
N
is smaller than
mg
c)
N
is larger than
mg
d) none of the above
You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the
normal force
N
exerted by your seat is equal to your
weight
mg
. How does
N
change at the top of the Ferris wheel when you are in motion?
You are in circular motion, so there has to be a centripetal force pointing
inward
. At the top, the only two forces are
mg
(down)
and
N
(up)
, so
N
must be smaller than
mg
.
Follow-up
: Where is
N
larger than
mg
?Slide27
R
v
a)
F
c
=
N
+
mg
b)
F
c
=
mg
–
N
c)
F
c
=
T
+
N
–
mg
d)
F
c
=
N
e)
F
c
=
mg
A skier goes over a small round hill with radius
R
. Because she is in circular motion, there has to be a
centripetal force.
At the top of the hill, what is
F
c
of the skier equal to?
Question 7.6b
Going in Circles IISlide28
R
v
F
c
points toward the center of the circle (
i.e
., downward in this case).
The
weight vector
points
down
and the
normal force
(exerted by the hill) points
up
. The magnitude of the net force, therefore, is
F
c
=
mg
–
N.
mg
N
A skier goes over a small round hill with radius
R
. Because she is in circular motion, there has to be a
centripetal force.
At the top of the hill, what is
F
c
of the skier equal to?
Question 7.6b
Going in Circles II
Follow-up
: What happens when the skier goes into a small dip?
a)
F
c
=
N
+
mg
b)
F
c
=
mg
–
N
c)
F
c
=
T
+
N
–
mg
d)
F
c
=
N
e)
F
c
=
mgSlide29
R
v
top
a)
F
c
=
T
–
mg
b)
F
c
=
T
+
N
–
mg
c)
F
c
=
T
+
mg
d)
F
c
=
T
e)
F
c
=
mg
You swing a ball at the end of string in a vertical circle. Because the ball is in circular motion there has to be a
centripetal force.
At the top of the ball’s path, what is
F
c
equal to?
Question 7.6c
Going in Circles IIISlide30
R
v
T
mg
F
c
points toward the center of the circle (
i.e
., downward in this case).
The
weight vector
points
down
and the
tension
(exerted by the string) also points
down
. The magnitude of the
net force, therefore, is
F
c
=
T
+
mg.
Question 7.6c
Going in Circles III
Follow-up
: What is
F
c
at the bottom of the ball’s path?
a)
F
c
=
T
–
mg
b)
F
c
=
T
+
N
–
mg
c)
F
c
=
T
+
mg
d)
F
c
=
T
e)
F
c
=
mg
You swing a ball at the end of string in a vertical circle. Because the ball is in circular motion there has to be a
centripetal force.
At the top of the ball’s path, what is
F
c
equal to?Slide31
Vertical Uniform Circular Motion
Uniform Circular Motion (UCM) implies that the tangential speed of a object moving in a circular path around a center
will be constant
.Therefore, there is a constant net force (the Centripetal Force) acting toward the center of the circular path.Vertically, we must now always consider the weight of the objectIn order to keep the net force constant so that the speed remains constant, what must happen to the tension (or normal, or whatever additional force is acting to give UCM)Slide32
At the very TOP of a circular path
Imagine a stopper at the end of a string that is being whirled in a vertical circle.
What forces are acting on the stopper at the very top of its path?
Draw a force diagram to illustrate the forces acting on the stopperIn a 2nd color, draw in the vector of the net forceWrite a vector equation that can be used to determine the magnitude of the tension in the string.Slide33
At the very BOTTOM of a circular path
Imagine a stopper at the end of a string that is being whirled in a vertical circle.
What forces are acting on the stopper at the very bottom of its path?
Draw a force diagram to illustrate the forces acting on the stopperIn a 2nd color, draw in the vector of the net forceWrite a vector equation that can be used to determine the magnitude of the tension in the string.Slide34
A stunt pilot in an air show performs a loop-the-loop in a vertical circle of radius
3.21 x
10
3
m. During this performance the pilot whose weight is
806
N, maintains a constant speed of
2.25 x
102
m/s.
What is the magnitude of the normal force acting on the pilot at the top of his path?
What is the magnitude of the normal force acting on the pilot at the bottom of the path?Slide35
Practice problem #2
Keys, m= 150.0 g, are attached to a string and swung around in a vertical circle with radius of 35.0 cm.
What is the minimum speed at which the keys must be moving in order to remain in uniform circular motion?
What is the tension in the string when the keys, at this speed, are at the bottom of their circular path?Slide36
Practice Problem #3
A student is to swing a bucket of water in a vertical circle without spilling
any
If the distance from his shoulder to the center of mass of the bucket of water is 1.2 m, what is the minimum speed required to keep the water from coming out of the bucket at the top of the swing
?
If the combined mass of the water and bucket is 7.5 kg, what is the force the student must supply at the bottom of the swing in order to keep the system in uniform circular motion?