Transcript y 1

LCROSS crashes into the Moon
Image credit:
NASA
LCROSS studied plumes created by impact
trying to measure composition of the ejecta
and estimate amount of water ice.
Then, in 4 min, it also crashed.
Centaur rocket stage is released from the circular
orbit 86400 km above the lunar surface. Find the
speed with which the stage crashes into the
Moon. Moon’s mass is 7.3x1022 kg, radius 1740
km. Gravitational constant is 6.67x10-11 N m2 kg-2.
How much energy is released in the crash if the
rocket stage has mass of 2300 kg? Find TNT
equivalent (1 kg TNT = 4x106 J)
2.365 km/s; 6.3x10^9 J
Compare to most powerful explosions
Tsar Bomb 1961
50 MT
Castle Bravo 1954
15-25 MT
.
<http://eol.jsc.nasa.gov/debrief/Iss020/topFiles/ISS020-E-9048.htm> ( 10/15/2009 22:06:18)
Chapter 8
All forces are
CONSERVATIVE
or
NON-CONSERVATIVE
A force is conservative if:
The work
done
by the force in going


from r1 to r2 is independent of the path
the particle follows
or
The work done by the
 force when the
particle goes from r1 around a closed
path, back to r1 , is zero.
Non-conservative:
doesn’t
satisfy
above conditions
the
Theorem: if a force can be written as the gradient (slope)
of some scalar function, that force is conservative.
1D case:
dU
Fx  
dx
U(x) is called the  potential energy
function for the force F
If such a function exists, then the force is
conservative
x2
dU
Fx  
dx
W
W
con
x2
dU
  Fx dx   
dx
dx
x1
x1
 [U ( x2 )  U ( x1 )]
con
does NOT depend on path!
If Fx(x) is known, you can find the potential energy function as
U ( x)    Fx ( x) dx  C
Work-energy theorem:
K 2  K1  W
total
1 2
W
con
1 2
W
nc
1 2
K 2  K1  U 2  U1   W
nc
1 2
K 2  U 2  K1  U 1  W
nc
1 2
nc
1 2
If W
 0, K 2  U 2  K1  U1  const
Energy conservation law!
A strategy: write down the total energy E = K + U at the
initial and final positions of a particle;
Then use
K 2  U 2  K1  U1 , if W
nc
1 2
or
K 2  U 2  K1  U 1  W
nc
1 2
0
Examples
y
U ( y )  mgy  C
Force of gravity
Fy   mg
Fx  k ( x  x0 )
Spring force
x
k ( x  x0 )
U ( x) 
C
2
2
x0
A gun shoots a bullet at angle θ with the x axis
with a velocity of magnitude Vm.
What is magnitude of the velocity when the
bullet returns to the ground?
Hits the target at height H above the ground?
How high it will go?
Note: motion is 2D, but U(y) = mgy is still a function
of only one coordinate y.
Potential Energy Diagrams
• For Conservative
forces can draw
energy diagrams
• Equilibrium
points
– If placed in the
equilibrium point
with no velocity, will
just stay (no force)
Fx >0
Fx  
dU
dx
a) Spring initially compressed (or stretched) by A and released;
b) A block is placed at equilibrium and given initial velocity V0
0
Stable vs. Unstable Equilibrium
Points
The force is zero at both maxima and minima
but…
– If I put a ball with no velocity there would it stay?
– What if it had a little bit of velocity?
A block of mass m is (not) attached to a vertical spring,
spring constant k.
A
If the spring is compressed an amount A and the block
released from rest, how high will it go from its initial
position?
A particle is moving in one direction x and its potential
energy is given by U(x) = ax2 – bx4 .
Determine the force acting on a particle.
Find the equilibrium points where a particle can be at rest.
Determine whether these points correspond to a stable or
unstable equilibrium.
Block of mass m has a massless spring connected to
the bottom. You release it from a given height H and
want to know how close the block will get to the floor.
The spring has spring constant k and natural length
L.
L
H
y=0
Water Slide
Who hits the bottom with a faster speed?
The curve of fastest descent
Cycloid
Inverted cycloid:
Brachistochrone
http://curvebank.calstatela.edu/brach/brach.htm
H
Roller Coaster
You are in a roller coaster car of mass M that
starts at the top, height H, with an initial
speed V0=0. Assume no friction.
a) What is the speed at the bottom?
b) How high will it go again?
c) Would it go as high if there were friction?
H
Roller Coaster with Friction
A roller coaster of mass m starts at rest at height
y1 and falls down the path with friction, then
back up until it hits height y2 (y1 > y2).
Assuming we don’t know anything about the
friction or the path, how much work is done by
friction on this path?
Several dimensions: U(x,y,z)
U ( x, y, z )
U ( x, y, z )
U ( x, y, z )
Fx  
; Fy  
; Fz  
x
y
z
Partial derivative is taken assuming all other arguments fixed
Compact notation using vector del, or nabla:

     
F  U ,   i 
j k
x
y
z

dU
Another notation: F   
dr
Geometric meaning of the gradient U
Direction of the steepest ascent;
Magnitude U : the slope in that direction

F  U :
Direction of the steepest descent
Magnitude F : the slope in that direction
http://reynolds.asu.edu/topo_gallery/topo_gallery.htm
:
If
or

dU
F  
dr
U ( x, y, z )
U ( x, y, z )
U ( x, y, z )
Fx  
; Fy  
; Fz  
x
y
z
then

U ( r2 )
 


dU 
W   F  dr      dr    dU  U (r2 )  U (r1 )
L

dr
U ( r1 )
W
con


 [U (r2 )  U (r1 )]
Work-energy theorem:
K 2  K1  W
total
1 2
W
con
1 2
W
nc
1 2
K 2  K1  U 2  U1   W
nc
1 2
K 2  U 2  K1  U 1  W
nc
1 2
nc
1 2
If W
 0, K 2  U 2  K1  U1  const
Energy conservation law!
Checking if the force is conservative
Fx  a; Fy  b
Fx  ax; Fy  by
Fx  ay; Fy  bx
Fx  f ( x); Fy  g ( y )
Fx  axy; Fy  b
Checking if U(x,y) is a potential energy for any force
 2U ( x, y)  2U ( x, y)

xy
yx
Find the velocity of the block at points 0,B, and D
Repeat if there I friction on the table up to x = B
Power
Power is a rate at which a force does work
If work does not depend on time
(or for average power):
Otherwise:
W
P
t
 
W Fdr  
P

 Fv
dt
dt
Even if instantaneous power depends on time, one can
talk about the average power
Power could also define the rate at which any form
of energy is spent, not only mechanical
How many joules of energy does 100 watt light bulb use
per hour? How fast would a 70-kg person have to run to
have that amount of energy?