Transcript Document

PHY
5200 Mechanical
Phenomena
Momentum
and Angular
Momentum
PHY 5200
Mechanical Phenomena
Newton’s Laws of Motion
Claude
A Pruneau
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Master
title style
Physics and Astronomy Department
Wayne State University
Dec
2005.
Claude
Click
A Pruneau
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Physics and Astronomy
Wayne State University
7/17/2015
Claude A Pruneau, PHY5200,
Chap 3
1
Principle of Conservation of Momentum
•
We consider the total momentum of a system of N particles.
– we label each particle with =1,…, N.
N
r
r
r
r
P  p1  p2 L  pN   p
 1
•
Provided the internal forces obey Newton’s third law, one can write
d r r
P  Fext
dt
•
This implies
Fext  0
P0
P  constant
If the net external force Fext acting on an N-particle system is zero, the
system’s total mechanical momentum P   m vr is constant.
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Claude A Pruneau, PHY5200, Chap 3
2
Example - Inelastic Collision of two bodies
Question:
Two bodies have masses m1 and m2, and velocities v1 and v2. The two bodies
collide and lock together - and subsequently move as a single unit. Note: This is
perfectly inelastic collision. Assuming external forces can be neglected during (and
shortly after) the collision, find the velocity of the combined masses v.
Before Collision
After Collision
2
v2
2
1
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v1
1
Claude A Pruneau, PHY5200, Chap 3
v
3
Example - Inelastic Collision of two bodies (cont’d)
Solution:
Use the fact there are no external forces acting on the two bodies. That implies the
total momentum of the system is conserved (I.e. constant).
r r
r
r
Pinitial  p1  p2  m1v1  m2v2
r
r
r
Pfinal  m1v  m2v  m1  m2 v
r
Pfinal  Pinitial
r
r
r
m1v1  m2v2  m1  m2 v
r
r
m1v1  m2 v2
v
Weighted average of the initial velocities.
m1  m2
Important Special Case: One of the two masses initially at rest:
r
m1v1
v
m1  m2
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Direction is the same, but the speed smaller
Claude A Pruneau, PHY5200, Chap 3
4
Motion of rockets
•
•
•
•
•
Quic kTime™ a nd a
TIFF (Un co mp res sed ) d eco mp re sso r
ar e n eed ed to see thi s p ictu re.
Rocket propulsion provides an excellent example of applicability of
“momentum conservation”.
Rocket motion is achieved from ejection of gas at high speed - no
eternal force is required to push or pull the rocket.
F=ma is not straightforwardly applicable given the mass is constantly
changing because of fuel consumption and ejection.
F=dp/dt is however applicable.
Consider
vex
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v
v
Speed of the rocket relative to the ground.
vex
Speed of the exhaust gas relative to the rocket.
Claude A Pruneau, PHY5200, Chap 3
5
Motion of rockets (cont’d)
•
Quic kTime™ a nd a
TIFF (Un co mp res sed ) d eco mp re sso r
ar e n eed ed to see thi s p ictu re.
At time t, the rocket has a momentum P(t).
P(t)  m(t)v
•
An instant dt later, i.e. at time t+dt, the mass of the rocket is reduced
m(t  t)  m(t)  dm
•
Its speed at that time is
v(t  t)  v(t)  dv
•
The momentum of the rocket is
•
•
The fuel ejected during the time interval dt is -dm
The velocity of the fuel relative to the ground is
P(t  t)  m(t)  dmv(t)  dv
v(t)  vext
•
The momentum of the fuel ejected during the time interval
v(t)
v(t)  vext
vext
dm v(t)  vext 
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Claude A Pruneau, PHY5200, Chap 3
6
Motion of rockets (cont’d)
•
In the presence of gravity one would have
FG 
•
•
P(t  t)  P(t)
t
For simplicity, assume no gravitational force, FG = 0.
Apply momentum conservation
0  P(t  t)  P(t)
0  P(t  t)  P(t)
m(t)v(t)  m(t)  dmv(t)  dv dm v  vext 
•
Distribute products, drop (t)
mv  mv  mdv  vdm  dmdv  vdm  vext dm
•
The above expression is evaluated in the limit t0, dmdv is thus negligible.
mv  mv  mdv  vdm  vdm  vext dm
•
Simplify terms in “mv” and “vdm”
0  mdv  vext dm
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Claude A Pruneau, PHY5200, Chap 3
7
Motion of rockets (cont’d)
•
One has
0  mdv  vext dm
•
Or
mdv  vext dm
•
Divide both sides by “dt”
dv
dm
m
 vext
dt
dt
•
Or
•
Where
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r&
&
mv  vext m
vext m&
has units of a force and is called “thrust”
Claude A Pruneau, PHY5200, Chap 3
8
Rocket Motion - Solution 1
•
The rocket motion is governed by
&
m&
v  vext m
•
Or equivalently
dv
dm
m
 vext
dt
dt
•
Eliminate “dt”
mdv  vext dm
•
Separate the variables
dm
dv  vex
m
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Claude A Pruneau, PHY5200, Chap 3
9
Rocket Motion - Solution 2
•
We need to integrate
dv  vex
•
•
•
•
•
dm
m
The “initial” velocity is “vo”.
The “final” velocity is “v”.
The “initial” rocket mass will be labeled “mo”.
The “final” is written “m”.
The integral can thus be written
v
m
dm
v dv  vex m m
o
o
•
•
Where it is understood that both “v” and “m” are functions of time “t”.
Integration yields:
v  vo  vex ln m mo
m
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Claude A Pruneau, PHY5200, Chap 3
10
Rocket Motion - Solution 3
•
Simplification yields
v  vo  vex ln m  ln mo 
•
Or equivalently
•
Or more simply
v  vo  vex ln m / mo 
v  vo  vex ln mo / m
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Claude A Pruneau, PHY5200, Chap 3
11
Rocket Motion - Notes
•
The solution is
v  vo  vex ln mo / m
Notes:
•
•
•
•
The mass mo, the initial mass of the rocket, includes fuel and payload.
This result places a severe constraint on the maximum speed of a rocket.
The ratio mo/m is largest when all the fuel is burned and “m” corresponds to the
rocket+payload mass.
If the initial mass is 90% fuel, then the ratio is “10”, since ln(10)=2.3, the gain in
speed only amounts to 2.3 the exhaust velocity.
– Rocket builders try to maximize the exhaust velocity, and/or design rockets
with multiple stages.
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Claude A Pruneau, PHY5200, Chap 3
12
Saturn V Rockets
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
July 16th, 1969 launch oF Saturn V rocket carrying
the crew of Apollo 11 to the Moon.
The Saturn V rocket was the largest rocket ever used
by NASA, and the only one able to lift the large
masses needed to land astronauts on the moon and
returning them safely. Saturn V rockets launched all
of the Apollo moon missions, and several to Earth
orbit as well.
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Claude A Pruneau, PHY5200, Chap 3
13
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Size
Height
Diameter
Mass
Stages
111 m (364 ft)
10 m (33 ft)
2,900,000 kg (6,500,000 lb)
3 (2 for Skylab launch)
Capacity
Payload to LEO
75,000 kg (2-stage)
Payload to the Moon
118,000 kg (3-stage)
47,000 kg
First Stage - S-IC
Quic kT ime™ and a
T IFF (Uncompres sed) decompres sor
are needed to s ee this picture.
Engines
Thrust
Burn time
Fuel
5 F-1 engines
33.4 MN (7,500,000 lbf)
150 s
RP-1 and liquid oxygen
Second Stage - S-II
Engines
Thrust
Burn time
Fuel
5 J-2 engines
5 MN (1,000,000 lbf)
360 s
Liquid hydrogen and liquid oxygen
Third Stage - S-IVB
Engines
Thrust
Burn time
Fuel
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Claude A Pruneau, PHY5200, Chap 3
1 J-2 engine
1 MN (225,000 lbf)
165 + 335 s (2 burns)
Liquid hydrogen and liquid oxygen
14
Center of Mass
•
Many ideas seen so far can be rephrased in terms of the notion of center of
mass.
•
Consider a group of N particles, labeled =1, …, N, with masses m and
positions r measured relative to some arbitrary origin O.
1
2
r1
r2
r3
rN
3
N
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Claude A Pruneau, PHY5200, Chap 3
15
C. M. of a continuous system
•
•
•
The notion of C. M. was 1st defined for a system of N particles.
In practice, one often deals with extended and continuous objects.
So we must consider the notion (extension) of CM for continuous objects.
•
The discrete sum over all particles, must be replaced by a continuous sum over
all infinitesimal elements of mass, dm.
N
m  
i
i 1
•
•
•
•
dm
volume
Where the integral must be carried over the entire volume I.e. to account for all
mass elements.
Here we face an integral in “dm” taken over a “volume”.
It is therefore more practical to transform the integral into a volume integral.
Introduce the mass density or density
m

v
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Claude A Pruneau, PHY5200, Chap 3
16
C. M. of a continuous system
•
•
The mass density may not be uniform across an object.
– E.G. in the human body, the bones, muscles, and other internal organs vary
in density.
The notion of density must therefore be defined for infinitesimal mass or volume
elements.
dm
r
 (r ) 
•
•
Note that the density is here noted as a function of the position.
With this expression in hand, it is now possible to write the expression for the
CM of continuous system.
rCM 
•
•
dV
1
M
1
r
rdm


M
volume
r dm
r
dV

dV
volume
Where the last expression indeed corresponds to a volume integral.
Insert the definition of density in the above expression.
rCM
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1

M
r r
r (r )dV

volume
Claude A Pruneau, PHY5200, Chap 3
17
C. M. of a continuous system (cont’d)
•
•
The expression for the center of mass is a vector equation.
It can be decomposed into three Cartesian components.
rCM
•
•

1
r
X

x

(
r
)dV
 cm

M volume


1
1
r r
r

r

(
r
)dV

Y

y

(
r
)dV

cm


M volume
M
volume


1
r
Z

z

(
r
)dV
 cm

M volume

Also note the volume element dV may be written as a product dV=dxdydz.
The above integrals are thus of the form
Xcm
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1

M

r
x(r )dxdydz
volume
Claude A Pruneau, PHY5200, Chap 3
18
C.M. Continuous System - Example 1
•
•
Calculate the position of the C.M. of a rectangular box of sides a, b, and c and
uniform density .
Let’s perform the calculation in a reference frame oriented along the sides of the
box as follows:
z
c
a
x
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y
b
Claude A Pruneau, PHY5200, Chap 3
19
Example 1 - cont’d
•
•
•
The calculation proceeds similarly for all three directions x, y, and z.
We show the calculation along “x” only.
The definition of c.m. along x is
Xcm
•
1

M
volume

M

xdxdydz
volume
The volume integral is calculated on a rectangular volume and may thus be
written:
Xcm 
•
xdxdydz
The density is constant and can therefore be factorized out of the integration.
Xcm 
•


c
b
a
0
0
0
dz  dy  xdx

M
This yields
Xcm
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
a
 x2 
 a2
 z 0 y 0   
bc
M
 2 0 M 2
c
b
Claude A Pruneau, PHY5200, Chap 3
20
Example 1 - cont’d
•
We have
Xcm 
•
 a2
M 2
bc
The mass can also be written as the product of the density and the volume of
the box.
M  V  abc
•
The C.M. position (along x) is thus
Xcm 
•
 a2
abc 2
bc 
a
2
By symmetry, the answer is similar along the y and z axes. The C.M. position of
a rectangular box of sides a, b, and c is thus:
 a b c
rcm   , , 
 2 2 2
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Claude A Pruneau, PHY5200, Chap 3
21
Example 2 C.M. of a solid cone
•
Let’s calculate the C.M. position of the solid cone illustrated below.
z
• By symmetry, the
position of the c.m.
along x and y is (0,0).
• We need only calculate
the position of the c.m.
along “z”.
R
h
R=Rz/h
y
O
x
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Claude A Pruneau, PHY5200, Chap 3
22
Example 2 - Cont’d
•
The position of the c.m. along z is given by
Zcm 
•
M

zdxdydz
volume
Given the cylindrical symmetry, it is more convenient to carry the integration in
cylindrical coordinates (,,z).
Z cm 
•
•


M

zd ddz
volume
The density is assumed constant and can thus be factored out.
The integral may be written
Zcm 

h
Rz /h
0
0
zdz 

M
2
d   d
R
h
R=Rz/h
y
0
x
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z
Claude A Pruneau, PHY5200, Chap 3
O
23
Example - cont’d
•
•
The integral is
Zcm 

h
Rz /h
0
0
zdz 

M
2
d   d
0
Integration over  yields a factor 2.
Zcm  2
Z cm  2
Zcm  2

h
Rz /h
0
0
zdz 

M

h

h
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Rz /h
 
zdz
2 
M 0
 0
M

2
2
Rz / h 

z
dz
2
0
 R 
3
 
z
dz

 h M 0
2
Zcm
d 
h
4
R2 h2
 R  h
 

 h M 4
4M
2
Zcm
Claude A Pruneau, PHY5200, Chap 3
24
•
You can verify that the volume of the cone is
1 2
V  R h
3
•
z
The mass can be evaluated, and the CM position is thus
Zcm 
 R h
2
4M
2

 R h
2
4
R
2
3
3

h
 R2 h 4
h
R=Rz/h
y
O
x
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Claude A Pruneau, PHY5200, Chap 3
25
Angular Momentum
•
The angular momentum is defined as
r r
l rp
•
angular momentum
•
Note that the vector position r is defined relative to a given origin O, the angular
momentum l, is thus also defined relative to that same origin, I.e. its value
depends on the choice of the origin.
One should therefore refer to l as the angular momentum relative to O.
•
Now consider the time rate of change of the angular momentum.
•
The derivative of a product is obtained with the product rule.
r
dl r& d r r
 l  r  p 
dt
dt
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r
r
dr r r dp r& r r r&
l 
 pr 
r  pr  p
dt
dt
Claude A Pruneau, PHY5200, Chap 3
26
•
The time rate of change if l is
r
p  mv
r& r r r&
l r  pr  p
•
Remember the expression of the force
r
r
F  p&
•
Also remember the momentum
is parallel to the velocity.
r r r
l  r& p  r  F

r r
l rp
O
0
•
We get an expression for the net torque about O.
r
r
l r F
•
Note many textbooks use the letter  to denote the torque, but here we will use
the same notation as Taylor.
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Claude A Pruneau, PHY5200, Chap 3
27
Torque
•
•
In many problems, one can choose the origin such that the net Torque is null.
In such cases, that implies the angular momentum about that origin is a constant
of motion.
Example of the motion of planets around the Sun.
•
Clearly,

r
O
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GmMrˆ
F
r2
r& r
l r F0
Angular momentum is constant
r and p remain in the same plane.
The motion of the planet (orbit) is
confined to a plane.
The motion is reduced to two
dimensions…
Claude A Pruneau, PHY5200, Chap 3
28
Kepler’s Second Law
•
•
One of the greatest achievements of Newton lies in that he was able to explain the Kepler’s 2nd law as a
simple consequence of conservation of angular momentum (Principia 1687).
We will discuss Kepler’s first and third laws later, here we focus on the 2nd law.
Kepler’s 2nd Law
As each planet moves around the sun, a line drawn from the planet to
the sun sweeps out equal areas in equal times.
Q

P

dA
O
dA

P’
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
Q’
r r
dr  vdt
Claude A Pruneau, PHY5200, Chap 3
29
Q

P

dA
•
Calculate the area of the triangles
O
1 r r
dA  r  vdt
2
•
dA

P’

Q’
r r
dr  vdt
Replace v by p/m, divide by dt.
r
dA 1 r p
1 r
 r

l
dt 2
m 2m
•
Since the planet angular momentum is conserved for a central force, then dA/dt
is constant, hence we demonstrated Kepler’s 2nd law.
7/17/2015
Claude A Pruneau, PHY5200, Chap 3
30
Angular Momentum of Several Particles
•
For a system of N particles, =1, 2, …, N, each with angular momentum
r r
l  r  p
•
•
All measured relative to the same origin O.
The total angular momentum is defined
N
r
r r
L   l   r  p
N
 1
•
 1
Let’s now consider the torque
r& N r
r
L   l   r  F
N
 1
•
 1
The rate of change of the angular momentum is simply the net torque.
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Claude A Pruneau, PHY5200, Chap 3
31
Net Torque
•
The net force acting on each particle can be written
r
r ext
F   F  F
 
•
Remember action=reaction
r
F   F
•
The angular momentum can thus be written
r
r ext
r
r
L    r  F   r  F
  
•

Focus on the 1st term, it can be re-written

r
r
r
r
r
  r  F    r  F  r  F
  
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  
Claude A Pruneau, PHY5200, Chap 3

32

r
r
r
r
r
  r  F    r  F  r  F
  
•
  

Again invoking action = reaction

r
r
r
r
r
  r  F    r  F  r  F
  
•
  

Or equivalently


r
r
r r
  r  F    r  r  F
  
•
  
Note that by construction, the above vector
product is null because the two vectors are
parallel to one another.
r
r  r
F
r
r
O
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Claude A Pruneau, PHY5200, Chap 3
33
•
So, we conclude that only the external forces have relevant contributions to the
net torque.
r
r ext
r
r
L    r  F   r  F
  

0
r ext
r
L  r  F  ext

The time rate of change of the angular
momentum is determined by the net
torque.
If the net external torque is null, the angular
momentum of the system is constant.
7/17/2015
Claude A Pruneau, PHY5200, Chap 3
34
Moment of Inertia
•
Calculations of the angular momentum may often be simplified and formulated in
terms of angular variables.
•
For example:
Angular velocity of rotation
Lz  I
Moment of Inertia
Uniform disk (mass M, radius R)
I
1
MR 2
2
Solid Sphere (mass M, radius R)
I
2
MR 2
5
N
In general
I   m 2
 1
7/17/2015
Claude A Pruneau, PHY5200, Chap 3
35
Net torque & C.M.
•
We found
r ext
r
L  r  F  ext

•
Where both L and G are measured relative to some origin in some reference
frame.
•
We will state without demonstration that this result also applies for non inertial
frames provided the origin is chosen to be the C.M..
7/17/2015
Claude A Pruneau, PHY5200, Chap 3
36