nd/or z components
Circular motion
Curved paths and projectile motion
Frames of reference
Radial and tangential acceration
Physics 207: Lecture 5, Pg 2
Chapter 4: Motion in 2 (and 3) dimensions
Chapter 4: Motion in 2 (and 3) dimensions
3
3
-
-
D Kinematics
D Kinematics
The position, velocity, and acceleration of a particle in
3-dimensions can be expressed as:
r
r = x
i
i + y
j
j + z
k
k
v
v = v
x
i
i + v
y
j
j + v
z
k
k
(
i
i ,
j
j ,
k
k unit vectors )
a
a = a
x
i
i + a
y
j
j + a
z
k
k
See text: 4-1
Which can be combined into the vector equations:
r
r =
r
r(t)
v
v = d
r
r / dt
a
a = d
2
r
r / dt
2
a
d x
dt
x
=
2
2
a
d y
dt
y
=
2
2
a
d z
dt
z
=
2
2
v
dx
dt
x
=
v
dy
dt
y
=
v
dz
dt
z
=
x
x(t
=
)
y
y t
=
( )
z
z t
=
( )
Physics 207: Lecture 5, Pg 3
Instantaneous Velocity
Instantaneous Velocity
The instantaneous velocity is the limit of the average
velocity as t approaches zero
The direction of the instantaneous velocity is along a line
that is tangent to the path of the particles direction of motion.
v
The magnitude of the
instantaneous velocity
vector is the speed.
(The speed is a scalar
quantity)
Physics 207: Lecture 5, Pg 4
Average Acceleration
Average Acceleration
The average acceleration of a particle as it moves is
defined as the change in the instantaneous velocity
vector divided by the time interval during which that
change occurs.

a
The average
acceleration is a
vector quantity
directed along v
Physics 207: Lecture 5, Pg 5
Instantaneous Acceleration
Instantaneous Acceleration
The instantaneous acceleration is the limit of the average
acceleration as v/ t approaches zero
The instantaneous acceleration is a vector with components
parallel (tangential) and/or perpendicular (radial) to the
tangent of the path
Changes in a particles path may produce an acceleration
The magnitude of the velocity vector may change
The direction of the velocity vector may change
(Even if the magnitude remains constant)
Both may change simultaneously (depends: path vs time)
Physics 207: Lecture 5, Pg 6
Motion along a path
Motion along a path
( displacement, velocity, acceleration )
( displacement, velocity, acceleration )
3-D Kinematics : vector equations:
r
r =
r
r(t)
v
v = d
r
r / dt
a
a = d
2
r
r / dt
2
Velocity :
v = d
r
r / dt
v
av
= r
r / t
Acceleration :
a = d
v
v / dt
a
av
= v
v / t
v
v
2 v
v
-
-
v
v
1
v
v
2
y
x
path
v
v
1 Page 2
Physics 207 Lecture 5
Physics 207: Lecture 5, Pg 7
General 3
General 3
-
-
D motion with non
D motion with non
-
-
zero acceleration:
zero acceleration:
Uniform Circular Motion is one specific case:
a
a
v
path
and time
t
a
a
a
a
a
a = 0
Two possible options:
Change in the magnitude of
v
Change in the direction of
v
a
a
= 0
a
a
= 0
a
a
a
a
a
a =
+
Animation
Physics 207: Lecture 5, Pg 8
Uniform Circular Motion
Uniform Circular Motion
What does it mean ?
How do we describe it ?
What can we learn about it ?
See text: 4-4
Physics 207: Lecture 5, Pg 9
Average acceleration in
Average acceleration in
UCM
UCM
:
:
Even though the
speed
speed is constant,
velocity
velocity is
not
not constant
since the direction is changing: must be some acceleration !
Consider average acceleration in time t
v
v
2
R
R
v
v
1
See text: 4-4 v
v
-
-
v
v
1
v
v
2
a
av
= v
v / t
seems like v
v (hence v
v/ t )
points toward the origin !
Physics 207: Lecture 5, Pg 10
Instantaneous acceleration in
Instantaneous acceleration in
UCM
UCM
:
:
Again: Even though the
speed
speed is constant,
velocity
velocity is
not
not
constant since the direction is changing.
As t goes to zero in v
v / t
d
v
v / dt =
a
a
a
a = d
v
v / dt
Now
a
a points
in the -
R
R direction.
R
R
See text: 4-4
Physics 207: Lecture 5, Pg 11
Acceleration in
Acceleration in
UCM
UCM
:
:
This is called
This is called Centripetal Acceleration.
Calculating the magnitude:
|v
1
| = |v
2
| = v
v
v
1
v
v
2 v
v
v
v
2
v
v
1
R
R


R
R
v
v
R
R
=
Similar triangles:
But R = v t for small t
So:
v
v
v t
R
=
R
v
t
v
2
=
a
v
R
=
2
Physics 207: Lecture 5, Pg 12
Period and Frequency
Period and Frequency
Recall that 1 revolution = 2 radians
Period (T) = seconds / revolution = distance / speed = 2 R / v
Frequency (f) = revolutions / second = 1/T (a)
Angular velocity ( ) = radians / second
(b)
By combining (a) and (b) = 2 f
Realize that:
Period (T) = seconds / revolution
So T = 1 / f = 2 / R
R
v
v
s
= 2 / T = 2 f
v
v
R
s Page 3
Physics 207 Lecture 5
Physics 207: Lecture 5, Pg 13
Recap: Centripetal Acceleration
Recap: Centripetal Acceleration
UCM results in acceleration:
Magnitude:
a = v
2
/ R = 2
2
2
2
R
Direction:
-
r
r (toward center of circle)
R
a
a See text: 4-4
Physics 207: Lecture 5, Pg 14
Lecture 5,
Lecture 5,
Exercise 1
Exercise 1
Uniform Circular Motion
Uniform Circular Motion
A fighter pilot flying in a circular turn will pass out if the
centripetal acceleration he experiences is more than about
9 times the acceleration of gravity g. If his F18 is moving
with a speed of 300 m/s, what is the approximate radius of
the tightest turn this pilot can make and survive to tell
about it ? (Let g = 10 m/s
2
)
(a) 10 m
(b) 100 m
(c) 1000 m
(d) 10,000 m
UCM (recall)
Magnitude: a =
v
2
/ R
Direction:
(toward center of
circle)
r Physics 207: Lecture 5, Pg 15
Lecture 5,
Lecture 5,
Exercise 1
Exercise 1
Solution
Solution
a
v
R
g
=
=
2
9
R
m
m
= 10000
9 8
1000
.
Answer (c)
Physics 207: Lecture 5, Pg 16
Example: Newton & the Moon
Example: Newton & the Moon
What is the acceleration of the Moon due to its motion
around the earth?
T = 27.3 days = 2.36 x 10
6
s
(period ~ 1 month)
R = 3.84 x 10
8
m
(distance to moon)
R
E
= 6.35 x 10
6
m
(radius of earth)
R
R
E
Physics 207: Lecture 5, Pg 17
Moon...
Moon...
So = 2.66 x 10
-6
rad s
-1
.
Now calculate the acceleration.
a = 2
R = 0.00272 m/s
2
= .000278 g
direction of
a
a is toward center of earth (-
R
R
).
1
-
6
s

rad

10
x
66
.
2
rot
rad
2
x
s
day
86400
1
x
day
rot

3
.
27
1 = Calculate angular frequency:
Physics 207: Lecture 5, Pg 18
Radial and Tangential Quantities
Radial and Tangential Quantities
a
v
For uniform circular motion Page 4
Physics 207 Lecture 5
Physics 207: Lecture 5, Pg 19
Radial and Tangential Quantities
Radial and Tangential Quantities
a
v
What about non-uniform circular motion ?
a is along the direction of motion
a
r
is perpendicular to the direction of motion
Physics 207: Lecture 5, Pg 20
Lecture 5,
Lecture 5,
Exercise 2
Exercise 2
The Pendulum
The Pendulum
1m

= 30°
A) v
r
= 0 a
r
= 0
v
0 a
0
B) V
r
= 0 a
r 0
v
0 a = 0
C) v
r
= 0 a
r 0
v = 0 a
0
Which statement best describes
the motion of the pendulum bob
at the instant of time drawn ?
i.
the bob is at the top of its swing.
ii.
which quantities are non-zero ?
Physics 207: Lecture 5, Pg 21
Lecture 5,
Lecture 5,
Exercise 2
Exercise 2
The Pendulum
The Pendulum
Solution
Solution
a NOT uniform circular motion :
is circular motion so must be a
r
not zero,
Speed is increasing so a not zero
1m = 30°
a
O
At the top of the swing, the bob
temporarily stops, so v = 0.
C) v
r
= 0 a
r 0
v = 0 a
g
In the next lecture we will learn about forces and
how to calculate just what a is.
a
r
Physics 207: Lecture 5, Pg 22
Relative motion and frames of reference
Relative motion and frames of reference
Reference frame S is stationary
Reference frame S is moving at v
o
This also means that S moves at v
o
relative to S Define time t = 0 as that time
when the origins coincide
Physics 207: Lecture 5, Pg 23
Relative Velocity
Relative Velocity
Two observers moving relative to each other generally
do not agree on the outcome of an experiment
For example, observers A and B below see different
paths for the ball
Physics 207: Lecture 5, Pg 24
Relative Velocity, equations
Relative Velocity, equations
The positions as seen from the two reference frames
are related through the velocity
r

= r v
o
t
The derivative of the position equation will give the
velocity equation
v

= v v
o
These are called the Galilean transformation
equations Page 5
Physics 207 Lecture 5
Physics 207: Lecture 5, Pg 25
Central concept for problem solving:
Central concept for problem solving:
x
x
and
and
y
y


components of motion treated independently.
components of motion treated independently.
Again: man on the cart tosses a ball straight up in the air.
You can view the trajectory from two reference frames:
Reference frame
on the ground.
Reference frame
on the moving train.
y(t) motion governed by
1) a = -g y
2) v
y
= v
0y
g t
3) y = y
0
+ v
0y
g t
2
/2
x motion: x = v
x
t
Net motion
:
R = x(t) i + y(t) j (vector)
Physics 207: Lecture 5, Pg 26
Acceleration in Different Frames of Reference
Acceleration in Different Frames of Reference
The derivative of the velocity equation will give the
acceleration equation
v

= v v
o
a

= a
The acceleration of the particle measured by an
observer in one frame of reference is the same as
that measured by any other observer moving at a
constant velocity relative to the first frame.
Physics 207: Lecture 5, Pg 27
Lecture 5,
Lecture 5,
Exercise 3
Exercise 3
Relative Motion
Relative Motion
You are swimming across a 50 m wide river in which the
current moves at 1 m/s with respect to the shore. Your
swimming speed is 2 m/s with respect to the water.
You swim across in such a way that your path is a straight
perpendicular line across the river.
How many seconds does it take you to get across?
2m/s
1m/s
50m
s

25
2
50
=
a)
s

50
1
50
=
b)
s

29
3
50
=
c)
s

35
2
50
=
d)
Physics 207: Lecture 5, Pg 28
Lecture 5,
Lecture 5,
Exercise 3
Ex