Transcript Uniform circular motion

11/18 do now – on a new sheet
• Consider the position-time plots below. Sketch
the shape of the corresponding velocity-time
graphs.
Hint:
1. Slope in p-t
graph is
velocity
2. The lines in
v-t graph are
straight, not
curved
• Due today
– worksheet 2.1.2B – Forces on Angles Packet
– Castle Learning Test – Due tonight
• Due tomorrow
– Homework: castle learning
• Due Friday
– Projectile test correction
– Project correction
– Worksheet 1.2.3 ground launched projectile packet
correction
• Homework is posted on google classroom code: nsdnlt
Circular Motion Chapter Outline
Lesson 1: Motion Characteristics for Circular Motion
Lesson 2: Applications of Circular Motion
Objective - Lesson 1: Motion
Characteristics for Circular Motion
1.
2.
3.
4.
Speed and Velocity
Acceleration
The Centripetal Force Requirement
Mathematics of Circular Motion
Uniform circular motion
• Uniform circular motion is the motion of an
object in a circle with a constant or uniform
speed. The velocity is changing because
the direction of motion is changing.
Speed is constant
v avg
d 2R
 
t
T
• R is radius of the circle
• 2πR is the circumference of the circle
• T is time to make one lap around the circle, also
known as period.
The average speed and the radius of the circle are
directly proportional.
Velocity is changing
The Direction of the Velocity Vector
The direction of the velocity
vector at every instant is in a
direction tangent to the
circle.
Acceleration
• An object moving in uniform circular motion is moving in a
circle with a uniform or constant speed. The velocity
vector is constant in magnitude but changing in direction,
which is tangent to the circle..
• Since the velocity is changing. The object is accelerating.
v v f v i
a

t
t
vi represents the initial velocity
vf represents the final velocity
t represents the time interval
Direction of the Acceleration Vector
• The velocity change is directed towards the center of the
circle.
• The acceleration is in the same direction as this velocity
change. The acceleration is directed towards the center of
the circle.
The Centripetal Force
• According to Newton's second law of motion, an
object which experiences an acceleration
requires a net force.
• The direction of the net force is in the same
direction as the acceleration. Since
acceleration is directed toward the center, the
net force must be toward the center. This net
force is referred to as the centripetal force. The
word centripetal means center seeking.
• In uniform circular motion, the Centripetal
Force is the NET force.
Inertia, Force and Acceleration
Centrifugal force is a fictitious force.
Centripetal force is the reason objects
moves in a circle
As a car makes
a turn in a
circle,
centripetal
force is
provided by
friction.
As a bucket of
water tied to a
string and spun in
a circle, centripetal
force is provided
by the tension
acting upon the
bucket.
As the moon
orbits the Earth
in a circle,
centripetal
force is
provided by
gravity acting
upon the moon.
Summary
• An object in uniform circular motion moves at
____________ speed. Its velocity is ___________
to the circle and its acceleration is directed toward
the ___________ of the circle. The object
experiences ____________ which is directed in
the same direction as the acceleration, toward
the _________ of the circle. This net force is
called ______________ force. The fact that the
centripetal force is directed perpendicular to the
tangential velocity means that the force can alter
the direction of the object's velocity vector without
altering its magnitude.
Check Your Understanding
• A tube is been placed upon the table and shaped into a
three-quarters circle. A golf ball is pushed into the tube
at one end at high speed. The ball rolls through the tube
and exits at the opposite end. Describe the path of the
golf ball as it exits the tube.
The ball will move along a path
which is tangent to the spiral at
the point where it exits the tube. At
that point, the ball will no longer
curve or spiral, but rather travel in
a straight line in the tangential
direction.
example
• The initial and final speed of a ball at two different points in
time is shown below. The direction of the ball is indicated by
the arrow. For each case, indicate if there is an acceleration.
Explain why or why not. Indicate the direction of the
acceleration.
No, no change
a.
in velocity
b.
yes, change in
velocity, right
c.
yes, change in
velocity, left
d.
yes, change in
velocity, left
example
• Identify the three controls on an
automobile which allow the car to be
accelerated.
The accelerator allows the car to increase
speed. The brake pedal allows the car to
decrease the speed. And the steering wheel
allows the car to change direction.
example
An object is moving in a clockwise direction around a
circle at constant speed.
1. Which vector below represents the direction of the
velocity vector when the object is located at point B on
the circle?
d
2. Which vector below represents the direction of the
acceleration vector when the object is located at point
C on the circle?
b
3. Which vector below represents the direction of the
force vector when the object is located at point A on
d
the circle?
Example
• The centripetal force acting on the space
shuttle as it orbits Earth is equal to the
shuttle’s
A.inertia
B.momentum
C.velocity
D.weight
Mathematics of Circular Motion
vavg
2   R

T
Fnet  m  a
m: mass
v: speed
R: radius
T: period
Fnet
v2
a
R
v2
 m
R
4  2  R
Fnet  m 
T2
Relationships between quantities
vavg
2   R

T
v2
ac 
R
v2
Fc  m 
R
vavg and R are direct: R double, v double
vavg and T are inverse: T double, v halves
ac and v are direct squared: v double, ac quadruple.
Fc and v are direct squared: v double, Fc quadruple.
ac and R are inverse: R double, ac halves.
Fc and R are inverse: R double, Fc halves.
Fc and m are direct: m double, Fc doubles.
ac and m are not related: m double, ac unchanged.
example
•
1.
2.
3.
4.
A car going around a curve is acted upon
by a centripetal force, F. If the speed of
the car were twice as great, the centripetal
force necessary to keep it moving in the
same path would be
F
F = m v2 / r
2F
F in directly proportional to v2
F/2
4F
2
F in increased by 2
example
• Anna Litical is practicing a centripetal force
demonstration at home. She fills a bucket with water, ties
it to a strong rope, and spins it in a circle. Anna spins the
bucket when it is half-full of water and when it is quarterfull of water. In which case is more force required to spin
the bucket in a circle?
It will require more force to accelerate a half-full bucket
of water compared to a quarter-full bucket. According to
the equation Fnet = m•v2 / R, force and mass are
directly proportional. So the greater the mass, the
greater the force.
Example
•
1.
2.
3.
4.
The diagram shows a 5.0-kilogram cart
traveling clockwise in a horizontal circle of
radius 2.0 meters at a constant speed of 4.0
meters per second. If the mass of the cart was
doubled, the magnitude of the centripetal
acceleration of the cart would be
doubled
halved
unchanged
quadrupled
ac = v2 / R
example
•
1.
2.
3.
4.
A 60.-kilogram adult and 30.-kilogram child are passengers on
a rotor ride at an amusement park. When the rotating hollow
cylinder reaches a certain constant speed, v, the floor moves
downward. Both passengers stay "pinned" against the wall of
the rotor, as shown in the diagram. The magnitude of the
frictional force between the adult and the wall of the spinning
rotor is F. What is the magnitude of the frictional force between
the child and the wall of the spinning rotor?
F
Ff = µFNorm
2F
½F
FNorm = Fnet = mv2/R
¼F
The child is half as massive as the adult, therefore FNorm of
the child is half of that of the adult and its Ff is half of that
of friction of the adult also.
example
•
1.
2.
3.
4.
Two masses, A and B, move in circular paths as
shown in the diagram. The centripetal acceleration of
mass A, compared to that of mass B, is
the same
twice as great
one-half as great
four times as great
F = m v2 / r
11/19 do now
• Identify the resultant in
the following vector
addition diagram.
Finally, indicate which
two vectors were
added to achieve this
resultant (express as
an equation such as X
+ Y = Z).
• Homework: castle learning
• Due Friday
– Projectile test correction
– Project correction
– Worksheet 1.2.3 ground launched projectile packet
correction
• Due Monday
– Force and Vector Application Packet correction
– Castle learning test correction
– Circular and satellite motion packet
• Homework is posted on google classroom - code:
nsdnlt
Solving problems using
Equations
•
•
•
•
Identify the given and unknown.
Choose an equation
Substitute number with units
Answer with units
vavg
2   R

T
v2
ac 
R
v2
Fc  m 
R
Example
• A 900-kg car moving at 10 m/s takes a turn
around a circle with a radius of 25.0 m.
Determine the acceleration and the net force
acting upon the car.
Given: = 900 kg; v = 10.0 m/s R = 25.0 m
Find: a = ? Fnet = ?
a = v2 / R
a = (10.0 m/s)2 / (25.0 m)
a = (100 m2/s2) / (25.0 m)
a = 4 m/s2
Fnet = m • a
Fnet = (900 kg) • (4 m/s2)
Fnet = 3600 N
example
• A 95-kg halfback makes a turn on the football field. The
halfback sweeps out a path which is a portion of a circle with
a radius of 12-meters. The halfback makes a quarter of a
turn around the circle in 2.1 seconds. Determine the speed,
acceleration and net force acting upon the halfback.
Given: m = 95.0 kg; R = 12.0 m; Traveled 1/4 of the circumference in 2.1 s
Find: v = ?
a = ? Fnet = ?
a = v2 / R
a = (8.97 m/s)2/(12.0 m)
v=d/t
v = (1/4•2•π•12.0m) /(2.1s)
a = 6.71 m/s2
v = 8.97 m/s
Fnet = m*a
Fnet = (95.0 kg)*(6.71 m/s2)
Fnet = 637 N
example
• Determine the centripetal force acting upon a 40-kg child
who makes 10 revolutions around the Cliffhanger in 29.3
seconds. The radius of the barrel is 2.90 meters.
Given: m = 40 kg; R = 2.90 m; T = 2.93 s (since 10 cycles takes 29.3 s).
Find: Fnet
Step 1: find speed: v = (2 • π • R) / T = 6.22 m/s.
Step 2: find the acceleration: a = v2 / R = (6.22 m/s)2/(2.90 m) = 13.3 m/s2
Step 3: find net force: Fnet = m • a = (40 kg)(13.3 m/s/s) = 532 N.
example
• A go-cart travels around a flat, horizontal, circular track
with a radius of 25 meters. The mass of the go-cart with
the rider is 200. kilograms. The magnitude of the
maximum centripetal force exerted by the track on the
go-cart is 1200. newtons. Which change would increase
the maximum speed at which the go-cart could travel
without sliding off this track?
A. Decrease the coefficient of friction between the go-cart
and the track.
2
v2 a  v
Fc  m 
c
B. Decrease the radius of the track.
R
R
C. Increase the radius of the track.
2   R
v

D. Increase the mass of the go-cart.
avg
T
Lesson 2: Applications of
Circular Motion
1. Newton's Second law - Revisited
2. Amusement Park Physics
Newton's Second Law - Revisited
Where Fnet is the sum (the resultant) of all forces acting
on the object.
Newton's second law is used in combination of circular
motion equations to analyze a variety of physical
situations.
Note: centripetal force is the net force!
Steps in solving problems involving forces
1. Drawing Free-Body Diagrams
2. Determining the Net Force from
Knowledge of Individual Force Values
3. Determining Acceleration from Net
Force Or Determining Individual Force
Values from Knowledge of the
Acceleration
Case 1: car driven in a circle
When a car is moving in a horizontal circle on a
level surface, there is a net force (centripetal
force) acting on it, the net force is FRICTION, which
is directed toward the center of the circle.
example
• A 945-kg car makes a 180-degree turn with a speed of 10.0 m/s.
The radius of the circle through which the car is turning is 25.0 m.
Determine the force of friction and the coefficient of friction acting
upon the car.
Given: m = 945 kg; v = 10.0 m/s; R = 25.0 m
Find: Ffrict = ? μ = ?
m*v2/R
Ff = Fnet =
Ff = (945 kg)*(10m/s)2/25m
Ff = 3780 N
FNorm
Ff
Ff = μFNorm; FNorm = mg;
Ff = μmg
3780 N = μ(9270 N);
μ = 0.41
Fgrav
example
• The coefficient of friction acting upon a 945-kg car is 0.850.
The car is making a 180-degree turn around a curve with a
radius of 35.0 m. Determine the maximum speed with which
the car can make the turn.
Given: m = 945 kg; μ = 0.85; R = 35.0 m
Find: v = ? (the minimum speed would be the speed achieved with
the given friction coefficient)
Ff = Fnet
FNorm
Ff = μFN = μFN ;
Ff = (0.85)(9270N) = 7880 N
Fnet = m*v2/R
7880 N = (945 kg)(v2) / (35.0 m);
Ff
Fgrav
v = 17.1 m/s
Case 2: a swing bucket
In a swing bucket, the net force (Fc) is the
combinations of gravity and tension.
At top:
Fnet = Ftens + Fg
Fnet = mv2/R
At bottom:
Fnet = Ftens - Fg
Fnet = mv2/R
Example
• A 1.50-kg bucket of water is tied by a rope and whirled in a circle with
a radius of 1.00 m. At the top of the circular loop, the speed of the
bucket is 4.00 m/s. Determine the acceleration, the net force and the
individual force values when the bucket is at the top of the circular
loop.
Fgrav = mg = 14.7 N
a = v2 / R = 16 m/s2
Fnet = ma = 24 N, down
Fnet = Fgrav + Ftens
24 N = 14.7 N + Ften
Ftens = 24 N - 14.7 N = 9.3 N
m = 1.5 kg
a = ________ m/s/s
Fnet = _________ N
Example
• A 1.50-kg bucket of water is tied by a rope and whirled in a
circle with a radius of 1.00 m. At the bottom of the circular
loop, the speed of the bucket is 6.00 m/s. Determine the
acceleration, the net force and the individual force values
when the bucket is at the bottom of the circular loop.
Fgrav = m • g = 14.7 N
a = v2 / R = 36 m/s2
Fnet = ma = 54 N, up
Fnet = Ften - Fgrav
54 N = Ften – 14.7 N
Ften = 54 N +14.7 N = 68.7 N
m = 1.5 kg
a = ________ m/s/s
Fnet = _________ N
Case 3: Roller Coasters and
Amusement Park Physics
In a roller coaster, the centripetal force is provided by the
combination of normal force and gravity.
example
• Anna Litical is riding on The Demon at Great America. Anna
experiences a downwards acceleration of 15.6 m/s2 at the top of the
loop and an upwards acceleration of 26.3 m/s2 at the bottom of the
loop. Use Newton's second law to determine the normal force acting
upon Anna's 864 kg roller coaster car at top and at bottom of the
loop.
Given:
m = 864 kg
atop = 15.6 m/s2 , down
abottom = 26.3 m/s2 , up
Fgrav = m•g = 8476 N.
Find:
Fnorm at top and bottom
top of Loop
• Fnet = m * a
Fnet = (864 kg)∙(15.6 m/s2 ) = 13 478 N, downward
Fnet = Fnorm + Fgrav
13478 N = Fnorm + 8475 N
Fnorm = 5002 N
Bottom of Loop
• Fnet = m * a
Fnet = (864 kg) * (26.3 m/s2 ) = 22 723 N, upward
Fnet = Fnorm + (-Fgrav)
22723 N = Fnorm + (-8476 N)
Fnorm = 31199 N