One Dimensional F(v)
Basics
From Newton’s 2nd law: F(v) = F(dx/dt) = ma = m d2x(t)/dt2
But since dx/dt = v, we can write this as: F(v) = m dv(t)/dt .
Rearranging terms (called separation of variables): dt = m dv / F(v) .
Now we integrate (if we can figure out how):
tf
to
dt = m vovf [1/F(v)] dv .
If we can integrate the right side, then we get: t = t(v), that is, t as a function of v.
We now have to try to invert t(v) to get v(t). This is not always easy!
If we can get v(t), we use v = dx/dt to get:
dx = v(t) dt, and after integrating, we have:
x(t) = xo + tot v(t) dt .
Example of F(v): Drag
A fluid that flows over a surface (or through a cylinder) will have a force of friction due
to the stationary surface (or cylinder). Viscosity is a measure of how well the fluid flows,
and is defined to be:  = (F/A) / (dv/dy), where  is the symbol for viscosity, F is the
drag (frictional) force over the area, A, that the fluid flows over, and (dv/dy) is how the
velocity varies with the distance from the surface. This is based on laminar flow (where
one layer flows over the adjacent layer). If the flow is too fast, then the flow becomes
turbulent and eddies form.
The Reynolds number is a dimensionless number defined as: R = vD/
where  is the mass density of the medium that is flowing (or through which the object is
moving), v is the speed of the fluid (or the object through the fluid), D is the diameter of
the cylinder through which the fluid is flowing (or the diameter of the object), and  is
the viscosity of the fluid. We have generally found that when the Reynolds number is
below 2,000, we have laminar flow, and above 2,000 we have turbulent flow.
For small objects moving at low velocities through a medium, where the Reynolds
number is less than 10, the frictional drag force is approximately:
Fdrag = -6rv
(small object moving slowly, R < 10)
where  is the viscosity of the medium, r is the radius of the object, and v is the speed of
the object. This situation applies for the Milikan oil drop experiment.
For larger objects moving faster, where the Reynolds number is between 300 and
300,000, the frictional drag force is approximately:
Fdrag = ½CDAv2
Where CD is a constant that depends on the shape, A is the cross-sectional area of the
object,  is the density of the fluid, and v is the speed of the object through the fluid. CD
for a smooth sphere is 0.5, for a baseball it is 0.3, and for a cylinder it is 1.2.
For Reynolds numbers between 10 and 300, the drag is transitioning from the linear to
the quadratic dependence on v. For Reynolds numbers higher than 300,000, the CD
quantity is no longer constant.
A quick way to see the quadratic dependence of the drag force on the velocity is to start
with the definition of pressure (P=F/A), use Newton’s Second Law: F = p/t, use the
definition of velocity: v = s/t, and the definition of Volume: V = A*s:
P = F/A = (p/t)/A = (mv)*s/(A*s*t) = mv*v/V = ½mv2/V
= ½v2,
or
F = ½Av2
which gives the result above for CD = 1. A more aerodynamic shape may cause less air
resistance, and so the CD factor is introduced.
In the case of lift, Flift = ½CLAv2 , but the CL depends on the shape and if spinning the
spin/v. See the article by Rod Cross in The Physics Teacher, Feb. 2012 issue, page 81.
Density of air at STP is 1.3 kg/m3. Viscosity of air at 18oC is 1.8x10-5 N*s/m2 .
Density of water is 1,000 kg/m3. Viscosity of water at 20oC is 1.0x10-3 N*s/m2 .
Homework Problem: Problem #6: Assume a plane of mass m initially at rest is
accelerated by a jet engine that develops a constant thrust of Fo and the jet
encounters an air resistance of FAR(v) = –bv2. Here the b includes all of the
factors other than velocity: b = (1/2)CDA.
a) Find v(t) for the plane.
b) Is there a terminal velocity for the plane under these circumstances? If so,
what is it (in terms of the parameters given)?
HINT: 0a (du / [1-u²]) = 1/2 ln[1+a] - 1/2 ln[1-a]. Extra 10% if you can derive the
solution for the above integral.
In-class example: 2-14: Assume the engines of a propeller-driven airplane of mass "m"
deliver a constant power, P, at full throttle. Find the force, F(v). Neglecting friction find
the velocity and position of the airplane as it accelerates down the runway, starting from
rest at t=0. Check your answer by using Conservation of Energy.
Power = work done per time = W/t where W = Fdx.
Thus, P =  Fdx /t = F dx/dt = F v.
Therefore, F = P/v .
Now using Newton's Second Law: F = ma becomes:
P/v = m dv/dt, or
dt = (mv/P) dv .
t = vov (m/P)v dv .
In our case, since Power is constant, we get: t = (m/P) [v2/2 - vo2/2] , or reversing:
v(t) = [vo2 + (2P/m)t]1/2 .
Units: P has units of energy/time = (kg*m2/s2)/s = kg*m2/s3 ; m has units of kg; t has
units of s; so Pt/m has units of (kg*m2/s3)(s)/(kg) = m2/s2 which are the required units.
Basic behavior: As t goes up, v goes up (as expected).
As P goes up, v goes up (as expected). As m goes up, v goes down (as expected).
As t goes to infinity, v goes to infinity (if there is no resistance, this makes sense).
By considering t = (m/P) [v2/2 - vo2/2], we can see that P*t = ½mv2 - ½mvo2 = KE.
Since Power is Energy per time, energy should be power times time!
Problem with numerical method here! Since F = P/v, the force becomes infinite if v=0.
Additional In-class problem: If you have some initial speed, no power applied, and
some air resistance, will the object ever completely stop?
a) Model #1: FAR = -bv, where b is a constant that depends on the shape of the object
and the viscosity of the air. From above, b = 6r. We will assume that b holds
constant during the motion.
-bv = m dv/dt , or dt = -(m/b) (1/v) dv [recall that dv/v = ln(v), but we must apply
both lower and upper limits]
t = -(m/b) ln(v/vo) .
Inverting this, we get: v = voe-(b/m)t . v goes to zero only as t goes to infinity. Thus in
this model, the object will never quite stop.
Units: since F = -bv, b has units of Nt/(m/s) = (kg*m/s2)/(m/s) = kg/s. (b/m) has units of
(kg/s)/(kg) = (1/s) as required.
Behaviors: as b goes up, v slows down faster; as m goes up, v slows down more slowly.
For small time, e-(b/m)t can be approximated (by power series) as 1 - (b/m)t, and so we get
v(small time) = vo[1 - (b/m)t] = vo - (bvo/m)t . We recognize this as the equation for
constant acceleration where the acceleration is bvo/m = Fo/m since Fo = bvo in this model.
To find x(t), we use dx/dt = v = voe-(b/m)t , or dx = voe-(b/m)t dt . Integrating gives x(t) =
xo + (mvo/b)(1 - e-(b/m)t ). Note that as t goes to infinity, x goes to xo + (mvo/b) so there is
a finite stopping distance even though there is no finite stopping time!
b) Model #2: FAR = -bv2 where b includes all of the factors other than velocity: b = (1/2)CDA.
We will assume that b holds constant during the motion.
-bv2 = m dv/dt
dt = (-m/b)(1/v2)dv [recall that v-2dv = -v-1, but we must apply both lower and upper
limits]
t = (m/bv) - (m/bvo).
Inverting this, we get: v = vo / [1 + (bvo/m)t] . v again goes to zero only as t approaches
infinity. Thus in this model also, the object will never quite stop.
Units: since F = -bv2, b has units of (Nt)/(m/s)2 = (kg*m/s2)/(m2/s2) = kg/m. (bvo/m) has
units of (kg/m)*(m/s)/kg = (1/s) as required.
Behaviors: as b goes up, v slows down faster. As m goes up, v slows down more slowly.
For times close to zero, we can use the approximation (by power series): 1/(1+x) = 1 - x
so that for small times we have:
v(small time) = vo[1 - (bvo/m)t] = vo - (bvo2/m)t. We recognize this as the equation for
constant acceleration where the acceleration is bvo2/m = Fo/m since Fo = bvo2 in this
model.
To find x(t), we use dx/dt = v = vo / [1 + (bvo/m)t] , or dx = {vo / [1 + (bvo/m)t] }dt .
Integrating gives x(t) = xo + (m/b)*ln[b*vo*t/m].
Note that as t goes to infinity, x goes to infinity (although fairly slowly) so there is NO
finite stopping distance and NO finite stopping time!
c) Model #3: FAR = -bev where b and  are constants that depend on the shape of the
object and the density of the air. We will assume that  and b hold constant during the
motion. [Note that b has units of force and  has units of inverse speed (s/m).]
-bev = m dv/dt
dt = (-m/b) e-v dv [recall that e-vdv = (-1/)e-v but we must apply both lower and
upper limits]
t = (m/b)e-v - (m/b)e-vo .
Inverting this, we get: v(t) = (-1/)ln[(b/m)t + e-vo] .
Units: since F = -bev , b has units of Nt, and  has units of (s/m). Therefore (1/) has
units of m/s as required, and (b/m) has units of (s/m)*(kg*m/s2)/kg = (1/s) as required.
Behaviors: For v=0, we need the argument of the ln function to be 1, so the stopping
time is:
tS = (m/b)(1 - e-vo) . Thus, in this model there is a stopping time and hence a stopping
distance. To find the stopping distance, we would have to solve for x(t) and then
substitute tS for t.
As  goes to zero, the force should approach the constant b, and the velocity should then
approach the relation: v = vo - (b/m)t and the stopping time should approach tS = mvo/b .
At first glance, it is hard to see whether our expression tS = (m/ab)(1 - e-vo) behaves this
way or not. However, if we expand e-vo in a power series (for small vo), we get:
tS = (m/b)(1 - [1 - vo + 2vo2/2 - ...]) = (m/b)( vo - 2vo2/2 -...) = mvo/b - mvo2/2b
+... .
From this, it is easy to see that as  approaches zero, tS approaches mvo/b as it should.
It is easy to see that as b goes to zero, the air resistance will go to zero and so the time for
stopping should go to infinity. Our expression tS = (m/b)(1 - e-vo) does indeed behave
that way.