Free Fall:
•
Kinematics in two (or more) dimensions obeys the
same 1

D equations in each component
independently.
Physics 1710
Chapter 4: 2

D Motion
—
II
a
x
= 0
a
y
=

g
v
x
=constant
y = y
initial
–
½ gt
2
x = x
initial
+ v
x,initial
t
Kinematic Equations
Physics 1710
Chapter 4: 2

D Motion
—
II
REVIEW
x(t) = x
initial
+ v
initial
t + 1/2 a
o
t
2
v(t) = dx/dt = v
initial
+ a
o
t
a= dv/dt = a
o
Observe Air Track
1
′
Lecture:
•
The velocity and acceleration of a body in a moving
(and accelerating) frame of reference (FoR ) is equal to
that of a stationary FoR
minus
the velocity or acceleration
of the moving FoR.
v
′
=
v
–
=
v
frame of reference
a
′
=
a
–
=
a
frame of reference
•
Motion in a circle at a constant speed is due to an
acceleration toward the center of the circle, a
centripetal
acceleration of
a =

ω
2
r
and

a

= v
2
/
r
toward the center.
Physics 1710
Chapter 4: 2

D Motion
—
II
Frame of Reference
Physics 1710
Chapter 4: 2

D Motion
—
II
Physics 1710
Chapter 4: 2

D Motion
—
II
Fly
v
′
v
′
v
frame of reference
v
v = v
′ + v
frame of reference
and
v
′
= v

v
frame of reference
Relative Motion
and the Galilean Transformation:
r
′
=
㴠=
r
=
–
v
frame of reference
t
d
r
′
⽤/=㴠==
r
⽤/=
–
v
frame of reference
v
′
=
v
–
=
v
frame of reference
d
v
′
⽤/
=
=
㴠
=
搠
v
⽤/
=
–
=
搠
v
frame of reference
/dt
a
′
=
㴠
=
愠
–
=
a
frame of reference
Physics 1710
Chapter 4: 2

D Motion
—
II
Relative Motion
in a Free Falling Frame of Reference
Physics 1710
Chapter 4: 2

D Motion
—
II
In Lab
Frame
In Moving
Frame
a
′
=
㴠
=
愠
–
=
a
frame of reference
Everyday Physics:
•
When a bird is flying at the same
velocity(same speed and direction) as a car
what is its relative velocity
v
′
to=the=car?
=
•
When you brake an automobile, which
direction is the acceleration on the vehicle?
Which direction do the passengers sense as
the acceleration relative to the frame of
reference of the car?
Physics 1710
Chapter 4: 2

D Motion
—
II
Relative Motion
and the Galilean Transformation:
v
′
=
v
–
=
v
frame of reference
a
′
=
a
–
=
a
frame of reference
Physics 1710
Chapter 4: 2

D Motion
—
II
Uniform Circular Motion
θ
r
Physics 1710
Chapter 4: 2

D Motion
—
II
x = R sin
θ
y = R cos θ
R
2
= x
2
+ y
2
r
=
(x,y) = x
i
+ y
j
v
v
x
= R cos
θ d θ/dt
= Rω cos θ
v
y
=

R sin
θ d θ/dt
=

Rω sin θ
d θ/dt = ω = constant
Uniform Circular Motion
θ
r
Physics 1710
Chapter 4: 2

D Motion
—
II
v
v
x
= R cos
θ d θ/dt
= R ω cos θ
v
y
=

R sin
θ d θ/dt
=

R ω sin θ
a
x
=

R
ω
2
sin
θ
a
y
=

R
ω
2
cos
θ
a
=

ω
2
r

a

= v
2
/R, toward center
Uniform Circular Motion
—
in review
⃒
a
?ÿ
= a,
a constant value always pointing toward the center of the
circle:
Centripetal acceleration
.
a
= a
x
i
+ a
y
j =

ω
2
r
where
a
x
= a sin
θ =

(
v
2
/R) sin
θ =

ω
2
R
sin
θ
a
y
= a cos
θ =

(
v
2
/R) cos
θ =

ω
2
R
cos
θ
Physics 1710
Chapter 4: 2

D Motion
—
II
Uniform Circular Motion

a

= v
2
/R, toward center
The
Centripetal
acceleration, where v is the tangential
speed and R is the radius of the circle.
v =
ω R =
2
π
R / T ,
Where T is the “
period
” or time to make one revolution.

a

= 4
π
=
2
R/ T
2
=
ω
2
R
Physics 1710
Chapter 4: 2

D Motion
—
II
Uniform Circular Motion
Little Johnny on the Farm
—
part II
.
Physics 1710
Chapter 4: 2

D Motion
—
II

g
a
v

g
a
v
.
?
Uniform Circular Motion

g
a
FoR
Physics 1710
Chapter 4: 2

D Motion
—
II
In Frame of
Reference of Bucket
a
′ =

g
–
a
FoR
If
a
FoR
 ≽ g
In same (down)
direction,
a
′ is up!
Uniform Circular Motion
g
a
Physics 1710
Chapter 4: 2

D Motion
—
II
a = (2
π/T)
2
R,
T
≈
1. sec
R
≈
1. m
2
π
≈
6.
a
≈
(6./1.sec)
2
(1.m)
a
≈ 36. m/sec
2
a
> 3 g
Do you believe
this?
80/20 Summary:
•
In a moving or accelerating Frame of Reference
•
v
′
=
㴠
=
瘠
–
=
v
frame of reference
•
a
?¿?
=
a
–
=
a
frame of reference
•
The
Centripetal
acceleration is
?µ
a =

?ö
2
r
or

a

= v
2
/
r,
toward the center.
Physics 1710
Chapter 4: 2

D Motion
—
II
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