Transcript hp1f2013_class06_momentum

Honors Physics 1
Class 06 Fall 2013
Newton’s Second Law as he wrote it
Momentum and force
Impulse
Center of mass motion
Center of mass
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Newton’s Second Law Revisited
d (ma )
to allow for the possibility
dt
that the mass could change (e.g.- be ejected). It turns out that the product ma
Newton actually wrote his second law as F 
is very useful in itself. It is called the momentum.
We have previously only treated the motion of a point particle or masses with
simple interactions. Momentum allows us to more easily calculate the behavior
of multi-particle systems.
Consider the motion of a pair of particles that only interact with each other (no
external force on the "system".)
dp1
dp
; f2  2
dt
dt
Since Newton's third law states that the force particle 1 exerts on particle 2 is
equal and opposite to the force exerted by 2 on 1, we can write:
d
f1  f 2   p1  p2   0. This says that the momentum of the system is
dt
a constant of the motion, unless an external force acts on it.
f1 
2
External force and momentum
If there are both external and internal forces on a system of particles, Newtons 2nd law
is: Fext   fint
Fext
dpi

. But the sum of all internal forces is zero, so
i dt
dP

where P   pi .
dt
i
This equation looks just like our old single particle one.
Extending the similarity: Fext  MR, where M is the total mass
and R can be determined by comparison the use of P.
mi ri





dP d
MR 
   pi     mi ri  which is true if R  i
.
 

dt dt  i
M
  i

R is called the Center of Mass of the system.
Note that the equations we have deduced only apply to the
translational motion of the center of mass (and not rotation.)
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Momentum
Because it is conserved in the absence of external forces momentum has great use in
the analysis of collisions. Consider two bodies once again interacting in isolation.
p1 (t1 )  p2 (t1 )  p1 (t2 )  p2 (t2 )
Example:
A student sits on a stationary low-friction cart. The mass of the cart and student
together is M=50 kg. The student throws an iron ball of mass 7 kg at a horizontal
speed of 10 m/s in the +x direction. What is the motion of the student and cart
after the throw? (Ignore friction.)
Initial momentum of ball, cart, student = 0=psc  pb therefore psc   pb .
pb  mb vb  70 (kg m/s)iˆ
70
vsc  iˆ
 iˆ1.4 m/s.
50
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Discussion
Two pucks collide elastically*. Using only
conservation of momentum, what can
you tell about the final velocities?
*You know m1, m2, v1i, v2i.
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Elastic and inelastic collisions
Momentum is still conserved, independent of the type of collision.
Elastic collision: Just use conservation of momentum.
For a completely specified initial condition (m1, m2, v1, v2) ,
we have one equation, two unknowns.
From what me know so far, we can deduce the relationship between
the velocities, but not their magnitude.
Completely Inelastic: The objects move off together after colliding.
This means that we have one equation and one unknown. Easy peasy.
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Different reference frames
Look at the previous example from a reference frame
that is initially moving wrt the cart.
Momentum still conserved, but it is not the same.
Change in speed is still the same.
 
Initial: vsci  vsc  v fr  v fr ; psci   msc v fr
Initial: vbi  v fr ; pbi   mb v fr
Final: vbf  10  v fr ;
Pi  Pf ;


  msc  mb  v f  (10 - v fr )mb  vscf  v f msc
Same as problem on previous page from here.
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Motion in the center of mass frame
The laboratory is actually an arbitrary reference frame. We can choose any other
inertial frame as well. Sometimes the center of mass is useful.
CoM frame is the one in which the initial momentum of the system is zero.
Example: One dimensional collision
m1: Moves at velocity v1i in lab frame.
m2 : Moves at velocity v2i in lab frame.
Primed coordinates are the cm frame.
v '1i  v1i  vcm ; v '2i  v2i  vcm
P 'i  m1v '1i  m2 v '2i  m1  v1i  vcm   m2  v2i  vcm   m1v1i  m2 v2i  vcm  m1  m2 
The center of mass system is the one where P '  0  m1v1i  m2 v2i  vcm  m1  m2 
vcm 
m1v1i  m2 v2i
 m1  m2 
The final momenta in the cm system are easy to deduce.
1) Elastic collision: Particles bounce back out with their momenta reversed.
2) Completely Inelastic: Particle stick together and have zero motion.
3) Explosive: Particles leave with equal and opposite momenta from each other.
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Calculating the Center of Mass
The center of mass is found by integration
over the volume.
1
1
1
xcom   xdM x ; ycom   ydM y ; zcom   zdM z
m
m
m
You can frequently compute the
center of gravity/mass using simple symmetries.
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Example: Calculating the COM
Let's try the calculation for a T-square that is 40 cm long
with a T top that is 30 cm across, and where all of the
parts are 3 cm wide. Take the thickness to be 2 mm.
1) Make a sketch.
2) Choose a set of axes.
3) Set up the integrals by breaking them up into simple sections.
4) Compute.
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Calculating the COM
y
The T-square is symmetric about
x x=0, so the xcom is at x=0.
It is also symmetric about z=0. ( zcom  0)
We have to actually do the y-integral.
ydzdxdy
1
ycom     ydzdxdy    
V
   dzdxdy
The thickness is constant over the whole area,
so the z integral is easy.
0
1.5 cm
1mm
3
15 cm
1mm
0
15 cm 1mm
z Vycom   y (  (  dz )dx)dy   y (  (  dz )dx)dy
37
top view
Vycom
1.5 cm 1mm
0
1.5 cm
3
15 cm

 2mm   y (  dx)dy   y (  dx)dy 
0
15 cm
 37 1.5cm

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T-square COM continued
Vycom
Vycom
3
0

 0.2cm  3cm  ydy  30cm  ydy 
0
37


2 3
2 0

y 
y 
 0.2cm  3cm    30cm   

 2  0 
 2  37



Vycom  0.3 0  37 2   3  32  0   0.3  1369   27  410.7  27  383.7
3 15
0 1.5

V     dydxdz     dydxdz  0.2    dydx    dydx 
0 15
0 15 0.1
37 1.5 0.1

 37 1.5
3
0

V  0.2  3  dy  30  dy   .6(37)  6(3)  22.2  18  40.2
0

 37
Vy
383.7
 9.5cm
ycom  com 
40.2
V
0 1.5 0.1
3 15 0.1
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T-square COM by serial logic
Identify pieces of the object for which you
can guess the COG. (Rectangles,
circles…)
Replace the piece with that mass at the
COG.
Find the balance point of the masses.
COG of top piece is at (0,1.5,0)
Mass(top)=Density*Vtop=D*18
COG of bottom piece is at (0,-18.5,0)
Mass(bottom)=D*Vbottom= D*22.2
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T-square COM by serial logic
Find the balance point:
 y  y  DV   y  y
 DV
 y  y V   y  y
V
1.5  y 18   y  18.5 22.2
cogtop
cog
cogtop
cog
cog
top
top
cog
cog
cogbottom
cogbottom
bottom
bottom
cog
27  18 ycog  22.2 ycog  410.7
 22.2  18 ycog  40.2 ycog  383.7
ycog  9.5cm
ycog
ycogtop
ycogbottom
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Center of Mass Integration
KK p 145 note 3.1
KK p119 nonuniform rod
KK p120 triangular sheet
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Center of mass integration:
Nonuniform rod
Take the mass per unit length to be:   0
dM   ( x)dx
L
L
x
.
L
2L
L

 x
x
M   dM    dx   0 dx  0  xdx  0
L
L
L 2
0
1
R
M

0

1
xiˆ  yjˆ  zkˆ dM 
M
0


0

0 L2
L 2
1
xiˆ  0 ˆj  0kˆ  dx 
M
L

x
0
0 L
2
0 x
L
dx
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Center of Mass Integration
We calculate the center of mass for a uniform triangular plate with edges at
x  0, y  0, and y  a  2 x
1
1
1
R
rdm


(
x
,
y
)
r
(
x
,
y
)
dxdy

 ( x, y )r ( x, y ) dxdy



M
M
M
M 4M
Since the density is constant,  
=
we can pull it out of the integral.
2
A a
R

xiˆ  yjˆ  dxdy


M
Rx 


a /2 a  2 x
xiˆ  dxdy 


M
M  
0
a /2
  ax 2
x3 

2 

M 2
3 
0
0
xdxdy 

M
a /2
 x  a  2 x  dx
0
  3a3
2a 3  4  a 3  a



  2   
M  24
24  a  24  6
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