Iterative prediction of motion
迭代

• Calculate all F acting on the system
• Apply momentum principle:



pfuture  pnow  Fnet t
•Find new position of system:



rfuture  rnow  vavg t

 p
v
m
The iterative method is an
example of a numerical method.
数值方法
It works for any situation, even if
the force is changing.
For some situations, we can use
algebra and calculus to find an
analytical solution.
代数
解析解
微积分
For most situations, there is no
analytical solution.
Three body
problem
No analytical
solution exists.
But it can be
solved with
numerical
methods.
Special case:
If the force is constant, there is always
an analytical solution.
Fan cart
The force
from the fan
is almost
constant.
Movies…
Your turn…
可略
Your turn…
Your turn…
Analytical solution:
constant force
Newton’s 2nd Law (y direction):
p f , y  pi , y  Fnet , y t
Divide by m (assuming v << c):
v f , y  vi , y 
Fnet , y
m
t
Find average velocity:
vavg , y 
vi , y  v f , y
2
(constant
force)
Analytical solution:
constant force
Newton’s 2nd Law (y direction):
p f , y  pi , y  Fnet , y t
Divide by m (assuming v << c):
v f , y  vi , y 
Fnet , y
m
t
Find average velocity:
vavg , y
1 Fnet , y
 vi , y 
t
2 m
(constant
force)
Analytical solution:
constant force
Position update:
y f  yi  v y ,avg t

1 Fnet , y 
 yi   vi , y 
t t
2 m


1 Fnet , y 2
y f  yi  vi , y t 
t
2 m
Note: Because Fnet is constant,
Δt does not have to be small.
Example:
You simultaneously drop two objects, a 5 kg
bowling ball and a 0.2 kg billiard ball, from the
top of the tower. If air resistance were
negligible, how long would it take each ball to
fall to the ground?
56 m
Leaning tower of Pisa
“Free-body diagram”
Four fundamental forces
Newton’s Universal Gravity
万有引力
The gravitational force on object 2 by
object 1 is given by

m1m2
Fon 2 by 1  G  2 rˆ
r
G = 6.67 x 10-11 m3/kg s2
Object 1

r
Object 2
Your turn…
Approximate gravitational force near
the Earth’s surface
Fobject,Earth  G
mobjectmEarth
2
REarth
 mEarth 
 mobject  G 2 
 REarth 
 mobject g
The acceleration g is approximately constant
(9.8 m/s), close to the Earth’s surface.
Your turn…
Example: compute the Earth’s
motion around the Sun.
Step 1:

 
 p f  pi  Fnet t

1  
 pi  p f 
 vavg 
2m
  
 rf  ri  vavg t
Your turn…
Step 2:

 
 p f  pi  Fnet t

1  
 pi  p f 
 vavg 
2m
  
 rf  ri  vavg t
Step 3:

 
 p f  pi  Fnet t

1  
 pi  p f 
 vavg 
2m
  
 rf  ri  vavg t
And so on…
Smaller time step
Quiz