Copyright © 2012 Pearson Education Inc.
PowerPoint® Lectures for
University Physics, Thirteenth Edition
– Hugh D. Young and Roger A. Freedman
Lectures by Wayne Anderson
Chapter 3
Motion in Two or
Three Dimensions
Part III
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Uniform circular motion—Figure 3.27
• For uniform circular motion, the speed is constant
and the acceleration is perpendicular to the velocity.
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Acceleration for uniform circular motion
• For uniform circular motion,
the instantaneous acceleration
always points toward the
center of the circle and is
called the centripetal
acceleration.
• The magnitude of the
acceleration is arad = v2/R.
• The period T is the time for
one revolution, and arad =
4π2R/T2.
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Acceleration for uniform circular motion
• From the two similar triangles:
Δs/R = Δv/v1 => Δv = v1(Δs/R)
arad = lim (Δv/ Δt) = lim v1(Δs/R)/Δt
Δt 0 Δt 0
When Δt tends to 0, Δs/Δt = v. Also,
v1 can be simply written as v
Then, arad = v2/R
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Acceleration for uniform circular motion
v can be also written using the
period T, which corresponds to
the time of one revolution as:
v = 2πR/T
By replacing v in the expression of
arad , we get:
arad = 4π2R/T2
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Nonuniform circular motion—Figure 3.30
• If the speed varies, the
motion is nonuniform
circular motion.
• The radial acceleration
component is still
arad = v2/R, but there
is also a tangential
acceleration
component atan that
is parallel to the
instantaneous velocity.
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Frame of reference
• A frame of reference is a coordinate system plus a time
scale.
• In this course, we assume that we are in an inertial, non
accelerating reference frame
• There is no way to distinguish between motion at rest
and motion at constant velocity in an inertial reference
frame.
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Motion is relative
• Imagine you are sitting in a chair watching a train
travel at 50 m/s
- From your reference frame, a cup of water you see
through the train window is moving at 50 m/s
- If you were on the train, however, the cup of water
would appear to be at rest
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Calculating Relative Velocities
• Consider two objects, A and B. Calculating the
velocity of A with respect to reference frame B (and
vice versa) is straight forward.
– Example: Walking on the train, speed of a person
with respect to the train frame
• Now, consider more objects A, B and C
vA/C = vA/B + vB/C
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Relative velocity in one dimension
• If point P is moving relative to
reference frame A, we denote the
velocity of P relative to frame A
as vP/A.
• If P is moving relative to frame
B and frame B is moving relative
to frame A, then the x-velocity
of P relative to frame A is vP/A-x
= vP/B-x + vB/A-x.
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Relative velocity in one dimension
• Question: A train (T) travels at 60 m/s to the east with
respect to the ground (G). A businessman (M) on the train
runs at 5 m/s to the west with respect to the train. Find the
velocity of the man with respect to the ground.
• Answer: First determine what information you are given.
Calling east the positive direction, you know the velocity of
the train with respect to the ground (vTG=60 m/s). You also
know the velocity of the man with respect to the train (vMT=-5
m/s). Putting these together, you can find the velocity of the
man with respect to the ground.
vM/G = vM/T + vT/G = - 5 + 60 = 55 m/s
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Relative velocity in two or three dimensions
• We extend relative velocity to two or three dimensions by using
vector addition to combine velocities.
• In Figure 3.34, a passenger’s motion is viewed in the frame of
the train and the cyclist.
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Relative velocity in two dimensions
• Question: The USA president’s airplane (P), Air Force One,
flies at 250 m/s to the east with respect to the air (A). The air
is moving at 35 m/s to the north with respect to the ground
(G). Find the velocity of Air Force One with respect to the
ground.
• Answer: In this case, it’s important to realize that both
vPA and vAG are two-dimensional vectors. You can find vPG by
vector addition.
vP/G = vP/A + vA/G
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Relative velocity in two dimensions
Looking at the diagram, you can easily solve for the magnitude of the velocity
of the plane with respect to the ground using the Pythagorean Theorem.
You can also find the angle of Air Force One using basic trig functions.
The velocity of Air Force One with respect to the ground is 252 m/s at an
angle of 8° north of east