Transcript Uniform Circular Motion

Chapter 5: Uniform Circular Motion
Chapter Goal: To learn
how to solve problems about
uniform circular motion.
Uniform circular motion:
the motion of an object
traveling at a constant speed
on a circular path.
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Uniform Circular Motion
•The velocity vector shows the
direction of motion at any point on the
circle.
•For uniform circular motion the
velocity vectors at each point are
tangent to the circle and are the same
length
•Period (T) – the time it takes to make
one revolution around the circle.
•T is in seconds, v is in m/s.
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Example Problem
Your roommate is working on his bicycle. The bicycle
wheel has a radius of 0.3 meters. As he spins the
bicycle wheel, you notice that a stone is stuck in the
tread. It goes by 3 times every second.
a. What is the period of the stone?
b. What is the speed of the stone?
Hint: the stone makes 3 revolutions per second (rps).
The period is how many seconds it takes to make
one revolution.
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Example Problem - Answer
Your roommate is working on his bicycle. The bicycle
wheel has a radius of 0.3 meters. As he spins the
bicycle wheel, you notice that a stone is stuck in the
tread. It goes by 3 times every second.
a. What is the period of the stone? T = 1/3 rps = .333s
b. What is the speed of the stone?
v = 2πr/T = 2π (.3m)/(.333s) = 5.7m/s
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Determining the direction of acceleration for
uniform circular motion using vector subtraction
Draw a velocity vector on
top of each dot,
tangent to the circle.
Starting with 0, take 2
adjacent velocity
vectors and do a
graphical vector
subtraction to find the
direction of ∆v and

therefore a .
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Acceleration of Uniform Circular motion
• For uniform circular motion, the
acceleration always points
towards the center of the circle.
• This is called centripetal (“centerseeking”) acceleration.
• Is centripetal acceleration equal to
∆v/ ∆t? Yes, but if the magnitude
of v remains constant, how do we
evaluate ∆v ?
• Using trig, it can be shown that
ac = ∆v/ ∆t = v2/r.
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Rank in order, from largest to smallest,
the centripetal accelerations (ar)a to (ar)e
of particles a to e.
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Rank in order, from largest to smallest,
the centripetal accelerations (ar)a to (ar)e
of particles a to e.
(ar)b > (ar)e > (ar)a = (ar)c > (ar)d
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Determining the direction of acceleration for
uniform circular motion using Newton’s 2nd Law
ΣF = ma
• tells us the net force vector and the acceleration
vector are in the same direction (since mass is a
scalar).
• ball on a string
• ball around a track
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FBD for Uniform Circular Motion Problems – The
radial/vertical (rz) coordinate system
• r axis – similar to a true horizontal
x axis, except positive is always
towards the center of the circle,
regardless of whether it is left or
right.
• z-axis – similar to a true vertical y
axis.
• velocity is tangent to the circle
(perpendicular to both r and z axes.
Note: if circular motion is vertical,
the r axis and z axis are the same!
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How to analyze motion problems for uniform circular motion,
using Newton’s 2nd Law
• ΣF = ma
• If the object is in uniform circular
motion then acceleration is a
centripetal acceleration, ac:
– |ac| = v2/r
– direction is towards the
center of the circle (positive
r axis).
– |ΣF| = mac = m v2/r
– ΣF must be in the positive r
direction.
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Car turning a “circular” corner at constant speed
Is there a net force on the car
as it negotiates the turn?
Where did it come from?
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Car turning a “circular” corner at constant speed
When the driver turns the wheel
the tires turn. To continue
along a straight line, the car
must overcome static friction
and slide. If the static friction
force is less than the
maximum, the tire cannot
slide and so has no choice but
to roll in the direction of the
turn.
fs-max = μs |n|
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Car turning a “circular” corner at constant speed –
example problem
What is the maximum speed at which a 1500 kg car can make a
turn around a curve of radius 50 m on a level road without
sliding out of the turn (skidding)?
Recall toward the center of the circle is positive r, even if it is to
the left in this example
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Car turning a “circular” corner at constant speed –
example problem
What is the maximum speed at which a 1500 kg car can make a
turn around a curve of radius 50 m on a level road without
sliding out of the turn (skidding)?
1. draw a free body diagram with appropriate axes and lists
knowns and “finds”
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Car turning a “circular” corner at constant speed –
example problem
2. Newton’s 2nd Law in
component form. Start
with z axis since there is no
acceleration:
ΣFz = maz = 0
n – FG = 0, therefore n = mg
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Car turning a “circular” corner at constant speed –
example problem
2. Newton’s 2nd Law in component form.
Now do forces in the r direction:
3. ΣFr = mac = mv2/r. The only force in the r
direction is the frictional force, therefore:
fs = mv2/r
vmax occurs at fs-max. Therefore:
μs |n| = mv2/r or μs mg = mv2/r
Copyright © 2008 Pearson Education, Inc., publishing as Pearson Addison-Wesley.
Car turning a “circular” corner at constant speed –
example problem
Note the mass term drops out, as all
forces depend on the mass.
Solve for vmax :
v = (us rg)-1/2 = 22 m/s, about 45
mph
Slow down!
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