Transcript Motion with Drag

Modeling motion subject
to drag forces
PHYS 361
Spring, 2011

physics
• Goal is to predict the motion of an object
– position vs. time ... x(t)
– velocity vs. time ... v(t)
• Several forces may be acting on this object
• Connect motion to forces using Newton’s laws
– Obtain differential equation(s): “Equations of Motion”
dv Fnet

dt
m
Fnet  ma
Solution is trivial if Fnet is constant. Most interesting forces, such as
those involved in riding a bicycle, are not constant.
Fdrag  b
1v  b2v
2
Fped
P

v
deriving a useful Equation of Motion
We want a differential equation of the form
dx
 f (x,t)
dt
But Newton’s 2nd law does not, at first glance, have this form:
F  ma

Of course, this equation is interesting (i.e. worthy of a computational
colution) only if the force is not constant. It could be a function of time,
position, or even velocity.

Let’s consider a situation where force depends on velocity and time.
How could we rewrite Newton’s 2nd Law in the desired form?
F(v,t)  m
dv
dt
dv 
F(v,t)

dt
m
F  F(v,t)
Forces that depend on velocity
dv
F(v,t)  m
dt
A cyclist’s power output is more typically constant than applied force.
Power is defined
 as
dE d 1
dv
2
P
  2 mv  mv
 vF
dt dt
dt
Drag force: viscous and inertial

viscous drag:
F  b1v
Stokes Law. Valid for small v.
inertial drag:
F  b2v 2
pushing air out of the way
valid for larger v
Equation of motion for a cyclist
dv F(v,t)

dt
m

P
2
F   b2v
v
Assume inertial drag is much larger than viscous drag

Euler method
Our differential equation:
dv P 1 b2  2
     v
dt m v m 
remember: b2 is a constant

Euler method for obtaining a
finite difference equation:

b2  12 AC
dv x v i1  v i


dt t
t
Substituting into our equation, we can solve for vi+1
P  1

AC  2
v i1  v i  t   t
v i
m v i
 2m 