Transcript Lecture8_Momentum

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
F
d
We’ve noted that an unbalanced force
acting continuously over a distance d
delivers kinetic energy to the object
being pushed:
Fd  mv
1
2
work done
2
kinetic energy
Often the distance over which the
forces act in a collision becomes
difficult to measure directly.
d
Particularly for sudden,
jarring “impulses”
where the contact forces act
for only brief instances.
Impulse is a physics term describing
how sudden the application of force
during such collisions is.
Analogous to our definition of work,
consider:
Force  time over which it acts
Ft  (ma)t
Recall:
 m(at ) v = v0 + at
 m( v  v0 )
 mv  mv0
producing a change in “momentum”
Momentum is inertia of motion
While inertia depends on mass
Easy to start
Hard to start
Momentum
depends on both mass and velocity
Easy to stop
Hard to stop
m
m
v
Easy to stop
Hard to stop
m
v
m
momentum = mass  velocity
“Quantity of motion”
v
v
To change velocity
 Force
To change momentum  Impulse
Ft (mv)
short “twang”
small momentum
t
F  (mv)
F  (mv)
t
long “twang”
larger momentum
Ft (mv)
Small forcemay not break! Short time large force