Module 3
Solving Newton 2 Law and Velocity Dependent Forces
nd
We will consider a direct approach to solving Newton’s 2nd Law for some
simple but important types of forces.
I.
Net Force Is A Function Only Of Time
In this case, Newton’s 2nd Law is of the form

dv 
m  Ft  .
dt
The solution method is to separate all terms containing linear
momentum to one side and place all terms with containing time on the
other side as follows:

v
t

vo
0
 
m  dv   Ft dt



m v  m v o   Ft dt
t
0



m v  m v o   Ft dt
t
0
1 t
 
v  v o   Ft dt
m0
The integral on the right hand side can’t be solved without knowing
the exact nature of the time dependence of the net external force. The
integral represents the area under the force-time graph or when
including the 1/m term with the integral, it is the area under the
acceleration-time graph. If the force in the problem is constant then
we obtain the kinematic equations.
To find the position vector, we replace the velocity with its definition
and integrate again with respect to time.

dx 
1 t
 v o   Ft dt
dt
m0

x
t
1 tt


 dx   v o dt  m  Ft dt dt

xo
0
00
1 tt
 

x  x o  v o t    Ft dt dt
m00
1 tt



x  x o  v o t    Ft dt dt
m00
II.
Net Force Is A Function Of Velocity
A.
General Case
For a 1-dimensional problem, if the net force is a function of only
velocity then we have
m
dv
 Fv 
dt
Separating the variables gives us
1
m
t
v
0
vo
dv
 dt   F(v)
t v dv

m vo F(v)
The right-hand-side of the equation depends on the nature of the
velocity dependent force (ex. The drag force). For sliding friction and
rolling friction, the force’s only dependence on velocity is that the
force points in the opposite direction. In these cases, we can use the
kinematic equations and avoid this treatment altogether. For slipping
between lubricated surfaces, the velocity dependence is usually to
complicated and tabular data from experiments is used to solve the
problem. Thus, it is possible to solve this equation directly only in a
few special cases as in the material in chapter 2 and the end of the
chapter problems. One fact that can be stated is that the general
solution for the velocity is a function that depends on the initial
velocity of the object and the ratio of time to mass.
t 

v   v o ; 
m

If this function  is found then the position of the object can be found
by
dx
t 

   vo ; 
dt
m

x
t
xo
0

t 
 dx     v o ; m  dt
t
t 

x  x o     v o ;  dt
m
0 
t
t 

x  x o     v o ;  dt
m
0 
B.
Special Case 1: F = -bv where b is a positive constant
Consider the case where a boat with its motor turned off is initially
gliding in the water in the +x direction at a speed vo. If the drag force
of the water upon the boat depends on the first power of the boat’s
velocity then Newton’s 2nd Law in the horizontal direction gives us
dv
 bv
dt
m
Using separation of variables, we have
v
b t
dv
dt   

m0
v
vo

b
t  ln(v)  ln(v o )
m

b
v
t  ln( )
m
vo
e

b
t
m

v
vo
v  vo e

b
t
m
Thus, we see that the velocity of the boat exponentially decreases
toward zero. Although mathematically, the boat would appear to
never stop, its motion will eventually become insignificance
compared to other motion (like the waves passing the boat). For a
given minimum velocity the result can be rearranged in order to
determine the time that it takes for the boat to reach this minimum
velocity. The example in your textbook for a falling object is similar
except that gravity attempts to accelerate the object. In this case the
initial velocity exponentially fades out and a new velocity (terminal
velocity) exponentially fades in.
To find the position of the boat as a function of time, we do the
following
b
 t
dx
 vo e m
dt
x
t
 dx  v o  e
xo

b
t
m
dt
0
x  x o   vo
m
e
b

b t
t
m
0
b
 t
v m
x  x o  o (1 e m )
b
Thus, the boat will eventually undergo a displacement equal to
Vom/b. If the boat is more massive or has a greater initial velocity
then it will travel farther. The greater the drag force (i.e. larger b) then
the shorter the distance traveled by the boat.
Let us examine our results for small time periods using a Taylor series
expansion for the exponentials. For velocity, we get
   b t    b t  2   b t 3







v  v o 1 

 
  ..... 
  m (1!)   m (2!)   m (3!) 

For small amounts of time t, each term in the expansion is much
smaller than the proceeding term so our result to 1st order in t is
approximately
v  vo 
vo b
m
t
But, F(t  0)  - v o b so 
vo b
 a(t  0) and our result is just the
m
kinematic result with the acceleration determined by the initial
velocity of the object. This makes sense as the velocity doesn’t change
much for small time intervals so the drag force and corresponding
acceleration is approximately constant.
v  vo  a o t
By the same expansion technique, we find that for small times
1 bv 
x  x o  vo t    o  t 2
2 m 
You should work this out!! The expansion technique not only checks
our algebra for physical sense but is also used in a powerful method
for solving differential equations called perturbation theory which I
will cover shortly.
C.
Special Case 2: F = ±kv where k is a positive constant
For larger moving objects far from their terminal velocities, the drag
force may be dependent on a higher power of velocity. If the power
dependence is even, you will have to break the problem into parts
based upon whether the object is moving in the + or – x direction and
apply the necessary signs to the drag force so that it opposes the
direction of motion.
For the example below, I will consider our boat again moving in the
+x direction but with a drag force that is proportional to the square of
the velocity of the boat. For this situation, we have from Newton’s 2nd
Law that
m
dv
  k v2
dt
Using separation of variables, we have
v
k t
dv
dt    2

m0
vo v
v
k
1
t
m
v vo
k
1 1
t 
m
v vo
1 1 k

 t
v vo m
1 m  k vo t

v
vo m
v
vo m
m  k vo t



1 
v  vo 
k vo t 
1


m 
We can define a time constant for the problem as
τ
m
k vo
Thus, our result can be written as


 1 
v  vo 
t
1 
 τ
We can find the location of the boat by



dx
1 
 vo 
t
dt
1 
 τ


 1 
 dx  v o   t  dt
xo
0 1 
 τ
x
t
1
x  x o  vo τ

1
t
τ
dη
η
where η 1
t
τ
t
x  x o  v o τ ln (1 )
τ
x  x o  vo
x  xo 
m
t
ln (1 )
k vo
τ
m
t
ln (1 )
k
τ
or
x  xo 
v kt
m
ln (1 o )
k
m
It should be noted that the velocity initially falls off faster in the case
of the drag force proportional to the square of the velocity, but the
long time dependence is only 1/t in this case while the velocity dies
out exponentially in the case of the linear velocity dependence. The
object can travel an infinite distance in the square velocity dependence
while it has a finite value in the linear case. Thus, the drag of a fluid
on a real world object must be more complicated than the square
velocity dependence with the drag force becoming linearly dependent
on velocity at large times. Thus, one possible drag force would be
F   k v2  b v
The first term dominates initially (small time values) while the second
term dominates as the object approaches terminal velocity or for large
values of time.
If the force is analytic then we could of course write any velocity
dependence as a Taylor series expansion in v (the possible solution
above is just the second and third terms of the expansion).
F  Fo  b v  k v 2  c v 3  d v 4  e v 5 ..........
D.
Finding The Distance An Object Must Travel To Obtain A Given
Velocity
In some problems, we either don’t know the time or want to express
velocity as a function of position. In these cases, it is useful to
remember that acceleration can be rewritten using calculus as follows:
dv dv dx

dt dx dt
dv
dv
v
dt
dx
dv 1

dt
2
 
d v2
Plugging this relationship into Newton’s 2nd Law gives us a different
type of differential equation to solve.
 
m d v2
 F(v2 )
2 dx
“Newton’s 2nd Law”
We have already seen this method when dealing with forces which
depend only on position, but the same method also works on velocity
dependent forces where we have rewritten the force in terms of speed
squared.
x
v2
 
m d v2
dx

 2 2 F(v2 )
xo
v
o
 
v2
m d v2
x  xo  
2 v 2 F(v2 )
o
v2
 
m d v2
x  xo 
2 v2 F(v2 )
o
III.
Expansion Solution Method (Perturbation Method) For Solving D.E.
A useful method for physicists and engineers is to initially model a
system in such a way as to determine the dominant response of the
system. For instance, we only include the dominant force like gravity
in a falling body problem and ignore smaller effects like air resistance.
The model is then improved iteratively by adding additional smaller
effects. Such solution methods are called perturbation methods (each
new effects slightly perturbs our original solution).
To illustrate this powerful technique for solving differential equations,
I am going to solve the problem of a ball dropped through air where
the drag force is dependent on the ball’s velocity to the first power. I
am also going to define upwards (away from the Earth) as the +x
direction. Thus, Newton’s 2nd Law for this problem would be
m
dv
  mg  b v
dt
Step 1: We make a zeroth order approximation to the problem by
ignoring air drag and solving for the velocity of the ball which we
designate as v(0) {zeroth approximation of velocity}.
dv0 
m
  mg
dt
dv0 
g
dt
We can solve this by separation of variables or by realizing that we
have already solved all constant acceleration problems (kinematic
equations).
v 0    g t
Step2: We now approximate the drag force using our zeroth order
approximation of the velocity. We then solve the resulting equation to
find our 1st order approximation of the velocity.
dv(1)
m
  mg  b v (0)
dt
dv(1)
bg
g  t
dt
m
v(1)
 dv
t
(1)
0
bg t
  g  dt 
 t dt
m
0
0
v (1)   g t 
1 bg 2
t
2m
Step3: We can do a better job approximating the drag force by using
our 1st order approximation of the velocity. We then solve the
resulting equation to find our 2nd order approximation of the velocity.
m
dv(2)
  mg  b v (1)
dt
dv(2)
b g b2 g 2
g  t t
dt
m 2 m2
v(2)
 dv
0
t
(2)
bg t
b2 g t 2
  g  dt 
 t dt  2 m 2  t dt
m
0
0
0
2
v
(2)
1 bg 2 1  b 
g t 
t    g t2
2 m
6 m
This process can be continued until the solution is obtained to
sufficient procession with each application yielding an additional term
to the expansion.
However, this process can often be stopped after just a few iterations
and a general solution in terms of a power series can be obtained by
inspecting the individual terms as in this case where
 b   t 
v  g  
i  0  m  i 1!

i
i 1
This method is obviously extremely useful for small time intervals
where the series can be truncated to just a few terms.
To find the displacement of the object, we can simply integrate our
velocity function with respect to time as in the past examples.