Chapter 4 Two–Dimensional Kinematics

conjunctionfrictionMechanics

Nov 13, 2013 (3 years and 8 months ago)

120 views

Chapter 4 Two–Dimensional Kinematics
Section 4.1
Constant
velocity
motion

A
body
moving
with
constant
velocity
in
two
dimensions
travels
equal
distances
for
equal
time
intervals
(not
necessarily
the
same
distance
along
each
dimension,
i.e.
∆x
is
not
generally
the
same
as
∆y).
On a graph of x versus time, constant velocity still appears as a straight line where: positive
velocity
motion
has
a
positive
slope
(points
up),
zero
velocity
motion
is
a
horizontal
line,
and
negative
velocity
motion
has
a
negative
slope
(points
down).
The
slope
of
a
straight
line
on
the
graph
of
x
versus
time
is
equal
to
the
x
component
of
velocity.
Replacing the "x" in the previous paragraph with "y" would hold for the y component of
velocity as well.
Constant
acceleration
motion

A
body
moving
with
constant
vector
acceleration
travels
different
distances
for
equal
time
intervals.
In
two
dimensions,
there
are
three
possibilities:
1)
only
an
x
component
of
acceleration,
2)
only
a
y
component
of
acceleration,
or
3)
acceleration
components
along
both
x
and
y
directions.
(The
case
where
acceleration
is
zero
in
both
x
and
y
directions
is
just
the
same
as
the
constant
velocity
case
above)
Graphs
can
be
drawn
of
velocity
versus
time
for
the
two
dimensional
case,
but
separate
graphs
would
be
needed
for
v
x
and
v
y

versus
time
in
order
to
calculate
the
x
and
y
displacements
(∆x
and
∆y)
from
the
area
under
the
graph
line.
Kinematic equations in two dimensions
Motion in two dimensions (x and y) under
constant
acceleration
is
described
using
the
following
two
sets
of
equations
(Note
that
the
set
of
x
components
is
used
to
describe
one-dimensional
motion.):
x components
y components
v
x

=
v
0
x

+
a
x

t
v
y

=
v
0
y

+
a
y

t
x

x
0

=
1
2

(
v
0x

+
v
x
)
t
y

y
0

=
1
2

(
v
0y

+
v
y
)
t
x

x
0

=
v
0
x

t
+
1
2

a
x
t
2
y

y
0

=
v
0
y

t
+
1
2

a
y
t
2
v
x
2

=
v
0
x
2

+
2
a
x

(x

x
0
)
v
y
2

=
v
0
y
2

+
2
a
y

(y

y
0
)
Meaning
of
symbols
above
(for
y
components,
replace
x
below
with
y):
x – x
0
= displacement along x-axis (from origin)
a
x

=
x-component
of
acceleration
v
x

=
final
x-component
of
velocity
t
=
time
that
has
passed
(from
t
=
0)
v
0
x
= initial x-component of velocity
Displacement
and
velocity
can
also
be
represented
using
the
unit
vector
notation,
e.g.:

r



=






v
0
x
 t  +  
1
2
 a
x
t
2

x
^


+






v
0
y
 t  +  
1
2
 a
y
t
2

y
^


v



=
(
)
v
0
x
 + a
x
 t

x
^


+
(
)
v
0
y
 + a
y
 t

y
^


Variables in equations above for specific situations:
Two-dimensional projectile motion(vertical versus horizontal):
Either
v
0
x

or
v
0y

may
be
non-
zero.
v
x

will
not
change
over
time,
but
v
y


v
0
y
.
For
accelerations,
a
x
= 0 and a
y
= –g =
–9.81 m/s
2
.
Section 4.3
Case
1
(Zero
Launch
Angle):
v
0y
= 0,
v
0x


0.
The
path
traced
out
by
the
object
will
be
half
a
downward
parabola
(Fig
4-5).
v
x
=
v
0x

over
the
course
of
the
path
(a
x
= 0) while v
y

=
–gt
and
becomes
increasing
larger
in
the
negative
y
direction.
Section 4.4,5
Case
2
(General
Launch
Angle,
where
∆y
=
0):
v
0y


0,
v
0x


0.
The
path
is
a
symmetric
parabola
opening
down
and
centered
around
the
maximum
height
(Fig
4-7,8,
11).
If
given
the initial velocity magnitude and launch angle, the time of flight, maximum height, range,
and final velocity can be calculated. In the absence of air friction, the maximum horizontal
range
occurs
when
the
launch
angle
is
45°.
The
velocity
vector
has
the
same
magnitude
going
up
through
a
value
of
y
as
it
does
coming
back
down
through
the
same
value
of
y
(Fig
4-10).
The
velocity
takes
on
its
minimum
magnitude
at
the
maximum
height
(=
v
0x
).
Case
3
(General
Launch
Angle,
where
∆y


0):
v
0y


0,
v
0x


0.
The
path
is
no
longer
a
symmetric
parabola,
but
is
still
parabolic
in
nature
(Fig
4-20).
Problems
of
this
sort
that
ask
for
time
will
generate
a
quadratic
equation
in
time
that
must
be
solved.
Some
additional
hints
for
problem
solving
(some
of
these
are
repeated
from
Chapter
2)
:

Make
sure
that
you
understand
whether
the
motion
is
in
two
dimensions
or
in
one
dimension. If it is in two dimensions, you will need to consider both x and y
component
equations.
Make
sure
that
you
correctly
identify
what
each
of
the
physical
quantities (s
x
,
v
f
y
, a
x
,
etc..)
are
in
the
problem
you
are
working

this
includes
understanding
which
values
you
are
given
and
which
you
need
to
solve
for
in
the
problem.

Make
sure
that
you
understand
exactly
what
is
being
asked
for
BEFORE
you
start
looking
for
an
equation
to
apply.
For
example:
Are
initial
values
given
or
being
asked
to
solve
for?
Are
there
values
implied
but
not
stated?
(
i.e.
starting
from
rest
=
initial
velocity
of
zero,
or
object
in
free
fall
means
that
y-acceleration
is
–9.81
m/s
2
).

Make
sure
you
understand
whether
the
motion
is
free
fall
or
whether
the
object
experiences
some
other
acceleration
(in
either
x
or
y
direction
or
both).

The
component
equations
above
have
in
common
the
variable
of
time
(=
t).
Thus,
the
same
time
(t)
must
pass
for
simultaneous
x
and
y
motion.

If
acceleration
takes
on
different
values
over
different
parts
of
an
object's
motion,
the
problem must be divided into individual parts for each of the different accelerations.