Transcript Drag Forces - Humble ISD

AP Physics C
Here is the situation!
Suppose you drop a marble into a graduated cylinder full of
glycerin. Obviously there will be some drag force acting on
the marble. Initially the marble will have some
acceleration but after a short time the marble will be
moving downward at constant speed.
Physicists have a name for this “constant speed” the marble
reaches…it is “Terminal Velocity”. The acceleration at this
“Terminal Velocity” is zero.
Think about this: As the marble moves downward through
the glycerin, the acceleration of the marble decreases from
some downward value to zero at terminal velocity.
Force Diagram
Let the drag force provided by the glycerin be defined by:
kv
Fd  kv
Such that the drag force increases
proportionally with the velocity.
Now write Newton’s Second Law
equation for the marble.
mg
 F  mg  kv  ma
Finding Terminal Velocity
 F  mg  kv  ma
kv
Remember that the key to “terminal velocity” is
a=0
so…
mg  kv  0
mg
Now solve for velocity…
in this case terminal velocity
mg
vT 
k
Finding Instantaneous Velocity
step 1
We will now develop an expression for velocity as a function of time,
v(t), to find the instantaneous velocity at any given point in time, t
Let’s begin with our Newton’s
second law equation
k
ag v
m
Then, solve for “a”
Remembering that
mg  kv  ma
a  dv dt
dv
k
g v
dt
m
this equation becomes…
The equation in this form is called a
“differential equation”
Finding Instantaneous Velocity, v(t)
step 2
Starting with the differential equation…
dv
k
g v
dt
m
We will need to find an anti-derivative,
but first we need to get the equation in
the right form…
In other words, all of the terms with the
variable (v) on one side of the equation
and all others on the other.
k 

dv   g  v dt
m 

Time for a little algebra…
dv
 dt
g  kmv
Finding Instantaneous Velocity, v(t)
step 3
Working from the equation …
dv
1

dv  dt
g  kmv g  kmv
It is time to find the anti-derivative and set the limits

v
0
t
1
dv   dt
0
g  kmv
Finding Instantaneous Velocity, v(t)
step 4

v
0
t
1
dv   dt
0
g  kmv
If we evaluate the right side of the equation it becomes…
t
t
0
0
 dt  t
t
Finding Instantaneous Velocity, v(t)
step 5
To evaluate the left side of the equation

v
0
t
1
dv   dt
0
g  kmv
We want to use a familiar “anti-derivative” form to finish so we will use

1
du  ln u
u
Because it most closely matches the form of our equation.
Finding Instantaneous Velocity, v(t)
step 6
We need to use a method of
integration (finding the antiderivative) called “u”
substitution…here is how it works.
Let
u  g  kmv
so, then

v
0
1
dv
g  kmv
du   k m dv
We need to rearrange the “du” equation so that we can substitute for “dv”
dv   m k du
next we substitute these
expressions into our anti-derivative
Finding Instantaneous Velocity, v(t)
step 7
Substituting the
expressions for “u”
and “du” into this
equation causes it to
become…
We can factor the “k/m” outside of the
integral (antiderivative) because it
is a constant.
1
dv
0 g  k m v
1 m
(

du
)
k
u
1
 m k  du
u
v
Note: We have
temporarily
stopped using
the limits for
the integral
during the
substitution
process.
Finding Instantaneous Velocity, v(t)
step 8
So now our equation with the substitutions is in the
general form shown earlier can be evaluated as follows:
1
 m k  du   m k ln u
u
Now we need to substitute our expression
for “u” back into the equation, so…

m
k
ln g 
k
m
v
v
0
Note:
We re-established
the limit
Finding Instantaneous Velocity, v(t)
step 9
So we have this...
 m k ln g  k m v
v
0
Now it is time to evaluate the limits, so…
 m k ln g  k m v  ln g  k m 0  
k v
g

m
m
 k ln
g
Finding Instantaneous Velocity, v(t)
step 10
Okay, let’s put the right and left sides back together to get:
k v
g

m
 m k ln
t
g
Remember that our goal is to solve for velocity, v, as a function of time.
More algebra…move the “–k/m” to the other side and then get rid of the “ln”
g  kmv
ln
 kmt
g
Next step: get rid of the “ln”
Remember… ln x
So…
e
x
Finding Instantaneous Velocity, v(t)
step 11
From the previous page…
A little algebra (magic)…
g  k mv
ln
g
 k mt
e
e
g  kmv
 k mt
e
g
More algebra…
g  k m v  ge
solving for velocity as a
function of time we get…
v(t ) 
 k mt
1  e 
 k mt
mg
k