Transcript Lecture7_Torque_Newtons3rdLaw

The distribution of material within an
object can be as important in determining
how it will respond to efforts to move it…
…as is the direction of the applied force
F
with respect to its center of mass.
F
R
d
r
Work is performed
by applying a force
(down on this end)
over the distance d.
h
Which changes the
potential energy of
the load on this end.
By how much?
PE  mgh
W  Fd
Conservation of energy requires:
Fd  mgh
mgh h
F
 mg
d
d
h =?
d
d
R
d
A.
B.
C.
h
r
r
D. r
E. h
R
r
r
F  mg
R
d
h
R
r
Raising the handlebars d raises the load h
Again: Fd = mgh
h should be in the same proportion as
d
1. R/r
2. r/R
3. R/d
4. r/h
h
r
So F  mg  mg
d
R
When an object is on a ramp its weight
pulls it into the supporting surface
and down along its surface.
This component represents the unbalanced
force that will accelerate the car downhill.
Weight
This component
of the weight is
compensated by
the normal force.
Gravity’s pull straight DOWN is both
pulling both into and along the ramp
at the same time!
At which position does the pedalist get
the greatest acceleration by bearing
down with his full weight?
In both position
A and D
the force is
exerted straight
along a line
toward the
center of the axle
which can do
nothing in
cranking the
sprocket.
r
r
F
F
When F and r are
perpendicular we
define the torque as
r
F
T  F r
Notice the units on torque are Newton-meters.
CENTER
OF MASS
This “off-center” push
consists of a component
of force directed along
the line to the center
(which will launch the
book into an trajectory)
and a component
perpendicular to the line
to center (the torque which
will set it spinning as well).
Together these two torques work to
rotate this refrigerator out from its corner.
R
F1
R1
R2
F2
F2 is perpendicular to the line back to center.
so it contributes a torque = F2R2
F1 is not perpendicular to R1, the line to center.
Instead we use the “lever arm” R, the distance
F1 misses the center by. This torque = F2R2
The light turns green and you’re in a hurry!
Will the car accelerate faster if you floor
the pedal and “burn rubber” or if instead
you accelerate so as to just avoid skidding
your wheels?
A. Skid your tires and burn rubber.
B. Just barely avoid skidding the tires.
Beginner skiers
learn to ski-plow
to slow down or
stop.
The skis dig into the snow, and
do work in plowing the snow aside.
This work comes at a cost: it depletes the
skier’s kinetic energy…slowing her to a stop.
But the snow also pushes back on the skier!
Perhaps more
tangible if the
encounter was
with a boulder
rather than a
pile of soft snow.
How do you turn right
when skiing downhill?
You push LEFT against the snow!
When snow-plowing you actually push out
with your LEFT skit o turn right! In more
advanced parallel skiing you lean right to
dig the inside (right) edge of your skis into
the snow…pushing LEFT against the hill!
The net pushes against
the tennis ball (see how
it has been deformed?)
slowing it to a stop,
then sending it back.
The tennis ball
pushes into the
racquet’s netting,
stretching its
strings.
Contact always produces pairs of forces.
The place-kicker’s toe
pushes into the football.
Note how it has been
deformed! This force
will send it flying.
The kicker’s
toe will feel
the football.
This force
will ever-so-slightly
slow his foot down.
The cue ball traveling with speed v
strikes a stationary billiard ball head-on.
A. The cue ball rebounds backward, while
its target is sent moving forward.
B. The cue ball stops while its target
continues forward with the speed v.
C. The cue ball and target ball roll
forward together with a speed <v.
D. The cue ball comes to rest in place
next to the target ball.
Consider this rear end collision with a parked car.
If left in neutral without
the parking brake set, the
force of impact will send
this car rolling forward.
The force of impact
stops this car.
If this involves speeds over 5-10 mph, both
vehicles will sustain damage. Which one
“feels” or experiences the force of impact?
The cue ball loses its energy
in this head-on collision.
The force of the impact
pushes back in stopping it.
The force of impact pushes the target forward.
The cue ball decelerates from v to 0 in the same
fraction of a second (the time both balls in
contact) that its target accelerates from 0 to v.
A boxer’s
right hook
delivers a
knock-out
punch!
The force this punch delivers to his
opponent’s face/head/neck is
A. greater than
B. exactly equal to
C. less than
the force the boxer’s hand experiences
from the blow.
A fly is struck against by windshield of
a car traveling 65mph down the highway.
The force experienced by the fly on impact
A. is greater than
B. is equal to
C. is less than
the force of impact experienced by the
windshield due to the impact.
Nothing’s moving, but not from lack of trying!
1. Stranded motorist pushes on car.
2. Car pushes back on her. How do we know?
3. Because it is mired in sand, the car’s tires
have a mound of sand to push up against.
4. Sand pushes back on car. How do we know?
5. With feet dug in, she pushes
back into the sand.
6. The sand pushes back on her.
This is what balances 2.
What needs to be changed to get out?
How do you walk? What are the forces
involved that allow you to walk?
As bracing yourself to push a car showed, you
push back against the ground below you
to propel yourself forward.
Imagine
trying to
walk
across a
surface
without
friction!
Micro-polished glass
Smooth plastic surface
500 m
A smoothly varnished surface.
50 m
Polished carbon steel surfaces
Since even the smoothest of surfaces
are microscopically rough, friction
results from the sliding up and over
of craggy surfaces, and even the
chipping and breaking of jagged peaks.
There are TWO TYPES of friction.
Static Friction
Acts to prevent objects from starting to slide
Forces can range from zero to an upper limit
Sliding Friction
Acts to stop objects that are already sliding
Forces of sliding friction have a fixed value
that depends on the particular surfaces
involved.
Frictional forces increase when you:
force the sliding surfaces together more
tightly (increase an object’s weight).
The peak static force is always greater
than sliding force
Surface features interpenetrate more
deeply when stationary objects settle.
Friction force drops when sliding begins
Cold welds are broken and moving objects
ride across the craggy surfaces higher.
f
W
The force of friction, f, is directly
proportional to the total force (usually
W for objects sliding horizontally) that
presses the sliding surfaces together:
f W
We write: f = W
where  is known as the
“coefficient of friction”
Typical coefficients of friction
maximum
Material
Rubber on dry concrete
Steel against steel
Glass across glass
Wood on wood
Wood on leather
Copper on steel
Rubber on wet concrete
Steel on ice
Waxed skis on snow
Steel across teflon
Synovial joints (hip, elbow)
static sliding
0.90
0.74
0.94
0.58
0.50
0.53
0.30
0.10
0.10
0.04
0.01
0.80
0.57
0.40
0.40
0.40
0.36
0.25
0.06
0.05
0.04
0.01
What happens when objects slide to rest?
Where does the lost kinetic energy go?
It generates heat,
an additional form of energy.
Rotation
Velocity
Wheels can circumvent friction by using
the fact that objects can roll without sliding
If friction prevents slipping at this point,
the foot planted at bottom stays stationary
as the entire assembly tips forward,
rotating about its axis.
Notice while the
planted foot stays
put, the axle
moves forward
at half the speed
that the top edge
of our wheel does!
Remember:pathlength
out a distance r from
the center of a rotation:
s=r
and the tangential
speed at that point:
v=r
2v
v
v=0
Each time this tethered ball comes
around, a wack of the paddle
gives it a boost of speed speed v .
Fd  m( v )
1
2
r
2
m
But this v is directly related to
an angular velocity,  (in radians/sec)
v = r
Fd  m( r )
1
2
2
Fd  mr (  )
1
2
2
2
For an individual mass m rotating
in an orbit of radius r
I  mr
2
Fd  I (  )
1
2
2
rotational
kinetic energy