03 Lecture Outline Part III

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


Uniform circular motion—Figure 3.27

• For uniform circular motion, the speed is constant

and the acceleration is perpendicular to the velocity.


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.



• 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



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


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.


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.


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


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


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.


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


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.


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


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

03 Lecture Outline Part III

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