Transcript hp1f2013_class07_integration_of_eq_of_motion

Honors Physics 1
Class 07 Fall 2013
Center of mass – multivariable integration
Impulse – Integration of force
Integrating equations of motion
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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
This approach is used for calculating the average value of
a quantity for any distribution.
q
 qdV

q 
  dV
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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,  
= 2 we can pull it out of the integral.
A a
R

xiˆ  yjˆ  dxdy


M
Rx 


 x  dxdy 

M
M 
0
a /2
a 2 x
a /2
  ax 2
x3 

 
2  
M  2
3 
M
0
(which equals 2/3 of a/2.)
x

0
dydx 

M
a /2
 x  a  2 x  dx
0
 3a3 2a3  4  a3  a


  2   
24  a  24  6
 24
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Computing Center of mass for a
Shaped Object
1) Use symmetry to reduce the amount of direct calculation.
2) Set up the multivariable integral relation for calculating the
average value along the axis of interest.
3) Do the integration in one of the variables (e.g.- y), putting the
limitations on the shape (in terms of the other variables) into the
limits of the integral.
4) Repeat for as many directions as needed until you get to the last
one, where the upper and lower bounds are then the global max
and min of the shape.
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A Math Interlude: Differentials
When a function f of some variable x is "continuous and smooth"*
then we can use the first term in the Taylor series to
evaluate the change in f for a small change in x:
df
df 
dx
dx
df is called a differential and specifically uses the linear
approximation.
This idea is very useful for converting between variables.
*continuous= no jumps**; smooth=no jumps in the derivative.
** no discontinuities.
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Changing the variable of integration
b
x2
Consider the integral:  xe dx
a
A useful substitution is: x  u 2
dx
1
so x  u1/2 and then dx 
du  u 1/2 du
du
2
The integral then can be rewritten:
b
ub
ub
ua
ua
1
x
1/2 u 1 1/2
xe
dx

u
e
u
du



2
2
a
2
u
e
 du 

1 ub
e  e ua
2

with ua  a 2
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Impulse
The relation between force and momentum for a system of
particles is: F 
dP
.
dt
t
So Fdt  dP and therefore  Fdt ' 
0
P t 

dP  P (t )  P (0)
P (0)
t
The integral J   Fdt ' is called the impulse and is equal to the
0
change in momentum over the time the force is applied.
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Impulse example 1 (KK131)
0.2 kg ball bounces from floor in time 10-3
s. 8 m/s. Find average force.
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Propulsion example
Is it possible to build a jetpack using downward
firing machine guns? (Yes!)
We need to know:
–
–
–
–
mass of a machine gun, (AK 47 = 4.8 kg)
mass of a bullet, (0.008 kg)
rate at which bullets can be fired. (10/s)
speed of exiting bullet wrt gun (715 m/s)
http://what-if.xkcd.com/21/
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Propulsion example: AK47 jet pack
F t  p
one bullet: F t  mb vb  0
Nb
kg
m
b
 mb vb R  0.008  715  10  57 N  thrust
t
b
s
s
Weight of AK47=47N
Average Thrust-Weight=10N
Average child weighs 200N, so 20 guns firing simultaneously
should lift her up.
Randall Munroe’s drawing
My drawing
F  F t
Only 3 seconds of lift
We need more ammunition
and ammunition has mass.
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But what if the mass changes with time?
When bullets leave the gun, they leave at velocity u relative to the gun.
In time t we eject mass m.
Let's take the remaining mass of the gun at any time to be M.
P(t+t)=Mv+mu
dP
dv
dm
dv
dM
M
u
M
u
dt
dt
dt
dt
dt
If there is no external force on the gun, then
dv 1 dM
 u
dt M dt
To find the gun velocity:
t
dv
 dt ' dt ' 
0
vf
 dv  v f
 vi
vi
t
Mf
0
Mi
1 dM
 M u dt ' dt ' 

Mf 
1
udM  u ln 

M
M
 i 
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Filling in some details on rocket discussion
View system from lab frame.
Mass at time t is M+m
v(t)=v
v(t+t)=v+v
Initial momentum: P (t )   M  m  v
Final momentum: P (t  t )  M (v  v )  m  v  v  u 
P  M v  mu
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