Mechanics
Lecture
4,
Slide
1
Classical Mechanics
Relative and Circular Motion
Today’s
Concepts:
a)
Rela ve
mo on
b)
Centripetal
accelera on
plug in audio
Mechanics
Lecture
4,
Slide
2
Stuff you asked about:
➢
I
dislike
Circular
moXon
and
I
bet
you
cant
make
me
like
it
:p
➢
I
studied
this
in
high
school,
but
the
expression
of
angular
velocity
is
new
to
me,
and
I
haven't
learned
about
centripetal
acceleraXon
in
terms
of
derivaXves
before
➢
We
never
covered
angular
velocity.
➢
The
most
confusing
concept
was
quesXon
number
2
which
involved
the
moving
walkway
and
the
dog.
RelaXve
moXon.
➢
It
was
stated
we
will
not
use
accelerated
frames
of
moXon,
but
a
single
example
to
demonstrate
the
difficulty
of
dealing
in
those
frames
would
be
interesXng
to
see,
if
only
once
for
my
first
semester
in
college.
➢
Why
the
acceleraXon
direcXon
and
the
velocity
direcXon
are
always
perpendicular?
➢
please
talk
about
how
you
get
the
formula
for
centripetal
acceleraXon.
➢
Further
dis[c]ussing
angular
velocity
➢
All
seems
fair
to
me
and
I
enjoy
whatever
my
prof
teach[es]
me
in
class
➢
We have now.
That’s the spirit!
Coriolus effect
Mechanics
Lecture
4,
Slide
2
Stuff you asked about:
➢
I
feel
like
the
pace,
at
which
we
are
covering
the
material
is
too
fast
for
myself
and
some
of
the
students
I
talked
to.
The
lectures
on
smartPhysics
are
very
brief
and
are
designed
mostly
to
give
students
an
idea
of
what
will
be
covered
in
the
lecture.
In
reality,
however,
our
lecture
consists
of
a
review
of
the
smartPhysics
lecture,
where
even
the
examples
(pictures)
are
the
same.
In
physics
it
is
someXmes
necessary
to
listen
to
two
different
explanaXons
in
order
to
fully
understand
the
concept.
Also,
coming
back
to
the
comment
about
the
speed
at
which
we
are
covering
the
material,
I
feel
like
a
weekly
problem
sheet,
containing
not
many,
but,
say
10
problems
is
in
order.
The
homework
on
smartPhysics
makes
use
use
a
parXcular
formula
once,
which
doesn't
lead
to
memorizing
it.
Since
our
knowledge
will
be
tested
in
a
wri en
form,
it
would
be
nice
to
have
some
pracXce.
Thank
you.
you’re welcome
Mechanics
Lecture
4,
Slide
3
How
familiar
are
you
with
the
concepts
of
relaXve
moXon
and
centripetal
acceleraXon
from
your
high
school
course.
A)
I
already
know
this
stuff
B)
It
seems
familiar,
but
I
need
a
review
C)
We
didn't
learn
this
in
high
school
Mechanics
Lecture
4,
Slide
3
How
familiar
are
you
with
the
concepts
of
relaXve
moXon
and
centripetal
acceleraXon
from
your
high
school
course.
A)
I
already
know
this
stuff
B)
It
seems
familiar,
but
I
need
a
review
C)
We
didn't
learn
this
in
high
school
StudentLogForQuestion
Mechanics
Lecture
2,
Slide
16
A
flatbed
railroad
car
is
moving
along
a
track
at
constant
velocity.
A
passenger
at
the
center
of
the
car
throws
a
ball
straight
up.
Neglec ng
air
resistance,
where
will
the
ball
land?
A)
Forward
of
the
center
of
the
car
B)
At
the
center
of
the
car
C)
Backward
of
the
center
of
the
car
correct
Demo
-‐
train
Ball
and
car
start
with
same
x
posi on
and
x
velocity,
Since
a = 0
they
always
have
same
x
posi on.
Train Demo Clicker Question
v
train car
Mechanics
Lecture
4,
Slide
4
Relative Motion
What
you
just
did
Mechanics
Lecture
4,
Slide
5
PreLecture 4, Question 1
vgr = vgL + vLr = vgL b vrL
Mechanics
Lecture
4,
Slide
5
PreLecture 4, Question 1
so
velocity of green with respect to red
=
velocity of green with respect to Lab + velocity of Lab with respect to red.
=
velocity of green with respect to Lab – velocity of red with respect to Lab
+ 6 m/s – ( –2 m/s) = +8 m/s
Recall that the
velocity of a with respect to c
=
velocity of a with respect to b + velocity of b with respect to c
Mechanics
Lecture
4,
Slide
6
v
belt,ground
= 2 m/s
v
dog,belt
= 8 m/s
A)
6 m/s
B)
8 m/s
C)
10 m/s
CheckPoint
A
girl
stands
on
a
moving
sidewalk
that
moves
to
the
right
at
2 m/s
relaXve
to
the
ground.
A
dog
runs
toward
the
girl
in
the
opposite
direcXon
along
the
sidewalk
at
a
speed
of
8 m/s
relaXve
to
the
sidewalk.
What
is
the
speed
of
the
dog
relaXve
to
the
ground?
Mechanics
Lecture
4,
Slide
7
v
dog, ground
=
v
dog, belt
+
v
belt, ground
= (−8 m/s) + (2 m/s) = −6 m/s
+
x
v
belt,ground
= 2 m/s
v
dog,belt
= 8 m/s
What is the speed of the dog relative to the ground?
A)
6 m/s
B)
8 m/s
C)
10 m/s
Mechanics
Lecture
4,
Slide
8
About
65%
of
you
got
this
right
–
lets
try
it
again.
CheckPoint
A
girl
stands
on
a
moving
sidewalk
that
moves
to
the
right
at
2
m/s
relaXve
to
the
ground.
A
dog
runs
toward
the
girl
in
the
opposite
direcXon
along
the
sidewalk
at
a
speed
of
8
m/s
relaXve
to
the
sidewalk.
What
is
the
speed
of
the
dog
relaXve
to
the
girl?
v
belt,ground
= 2 m/s
v
dog,belt
= 8 m/s
A)
6 m/s
B)
8 m/s
C)
10 m/s
Mechanics
Lecture
4,
Slide
9
A)
Because
the
girl
is
actually
moving
and
the
two
vectors
are
opposite,
so
together
they
make
6
m/s
B)
Because
the
girl
is
not
moving
rela ve
to
the
belt,
and
the
dog
is
going
8
m/s
rela ve
to
the
belt,
the
dog
is
also
moving
8
m/s
rela ve
to
the
girl..
C)
The
dog
and
girl
are
running
towards
each
other
so
when
you
add
the
two
veloci es
together
it
would
be
8+2.
What is the speed of the dog relative to the girl?
v
belt,ground
= 2 m/s
v
dog,belt
= 8 m/s
A)
6 m/s
B)
8 m/s
C)
10 m/s
A)
Because
the
girl
is
actually
moving
and
the
two
vectors
are
opposite,
so
together
they
make
6
m/s
B)
Because
the
girl
is
not
moving
rela ve
to
the
belt,
and
the
dog
is
going
8
m/s
rela ve
to
the
belt,
the
dog
is
also
moving
8
m/s
rela ve
to
the
girl..
Mechanics
Lecture
4,
Slide
10
B)
Because
the
girl
is
not
moving
rela ve
to
the
belt,
and
the
dog
is
going
8
m/s
rela ve
to
the
belt,
the
dog
is
also
moving
8
m/s
rela ve
to
the
girl.
Using
the
velocity
formula:
v
dog, girl
=
v
dog, belt
+
v
belt, girl
= −8 m/s + 0 m/s
= −8 m/s
What is the speed of the dog relative to the girl?
v
belt,ground
= 2 m/s
v
dog,belt
= 8 m/s
A)
6 m/s
B)
8 m/s
C)
10 m/s
Mechanics
Lecture
4,
Slide
11
Tractor Demo (moving cardboard)
Which
direc on
should
I
point
the
toy
bulldozer
to
get
it
across
the
cardboard
fastest?
A)
To
the
left
B)
Straight
across
A
B
C
Mechanics
Lecture
4,
Slide
12
Clicker Question
Moving
sidewalk
2 m
/s
1 m/s
4 m
A
man
starts
to
walk
along
the
do ed
line
painted
on
a
moving
sidewalk
toward
a
fire
hydrant
that
is
directly
across
from
him.
The
width
of
the
walkway
is
4 m,
and
it
is
moving
at
2 m/s
relaXve
to
the
fire-‐
hydrant.
If
his
walking
speed
is
1 m/s,
how
far
away
will
he
be
from
the
hydrant
when
he
reaches
the
other
side?
A) 2 m
B) 4 m
C) 6 m
D) 8 m
Mechanics
Lecture
4,
Slide
13
Time
to
get
across:
Δ
t = distance / speed
= 4m / 1m/s
= 4 s
1 m/s
4 m
If the sidewalk wasn’t moving:
Mechanics
Lecture
4,
Slide
14
4 m
2 m
/s
Just the sidewalk:
Mechanics
Lecture
4,
Slide
15
1 m/s
4 m
2 m
/s
Combination of motions:
Mechanics
Lecture
4,
Slide
16
D = (speed of sidewalk) · (time to get across)
= (2 m/s)
·
(4 s)
= 8 m
1 m/s
4 m
2 m
/s
D
Mechanics
Lecture
4,
Slide
17
Three
swimmers
can
swim
equally
fast
rela ve
to
the
water.
They
have
a
race
to
see
who
can
swim
across
a
river
in
the
least
me.
Rela ve
to
the
water,
Beth
swims
perpendicular
to
the
flow,
Ann
swims
upstream
at
30
degrees,
and
Carly
swims
downstream
at
30
degrees.
Who
gets
across
the
river
first?
A)
Ann
B)
Beth
C)
Carly
x
y
Ann
Beth
Carly
Clicker Question
Mechanics
Lecture
4,
Slide
18
A
B
C
V
y,
Beth
= V
o
30o 30o
V
y,
Ann
= V
o
cos(30o)
V
y,
Carly
= V
o
cos(30o)
Time
to
get
across =
D /
V
y
D
Look at just water & swimmers
x
y
Mechanics
Lecture
4,
Slide
19
x
y
Clicker Question
Three
swimmers
can
swim
equally
fast
rela ve
to
the
water.
They
have
a
race
to
see
who
can
swim
across
a
river
in
the
least
me.
Rela ve
to
the
water,
Beth
swims
perpendicular
to
the
flow,
Ann
swims
upstream
at
30
degrees,
and
Carly
swims
downstream
at
30
degrees.
Who
gets
across
the
river
second?
A)
Ann
B)
Carly
C)
Both
same
Ann
Carly
Mechanics
Lecture
4,
Slide
20
A
girl
twirls
a
rock
on
the
end
of
a
string
around
in
a
horizontal
circle
above
her
head
as
shown
from
above
in
the
diagram.
If
the
string
brakes
at
the
instant
shown,
which
of
the
arrows
best
represents
the
resul ng
path
of
the
rock?
A
B
C
D
Top
view
looking
down
After
the
string
breaks,
the
rock
will
have
no
force
ac ng
on
it,
so
it
cannot
accelerate.
Therefore,
it
will
maintain
its
velocity
at
the
me
of
the
break
in
the
string,
which
is
directed
tangent
to
the
circle.
CheckPoint (& Demo)
Mechanics
Lecture
4,
Slide
21
Show Prelecture
Mechanics
Lecture
4,
Slide
22
Centripetal Acceleration
Mechanics
Lecture
4,
Slide
23
v = ω
R
Once
around:
ω = Δθ
/ Δ
t = 2π
/ Τ
v = Δ
x / Δ
t = 2π
R / T
ω is
the
rate
at
which
the
angle θ changes:
θ
Mechanics
Lecture
4,
Slide
24
dθ
v dt
R
=R dθ
Another
way
to
see
it:
v = ω
R
v = Rω
Mechanics
Lecture
4,
Slide
25
We can ignore this acceleration due to Earth's rotation since its small
1700 km/h
