One Dimensional F(v)
But since dx/dt = v, we can write this as: F(v) = m dv(t)/dt .
Rearranging terms: dt = m dv / F(v) .
Now we integrate (if we can figure out how): toò tf dt = m voò 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 and 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 + toò 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: h = (F/A) / (dv/dy) where h 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 = rvD/h where r 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 h 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 = -6phrv (small object moving slowly, R < 10)
where h 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 = ½CDArv2 (300 < R < 300,000)
Where CD is a constant that depends on the shape, A is the cross-sectional area of the object, r 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),
P = F/A = (Dp/Dt)/A = D(mv)*Ds/(A*Ds*Dt) = mDv*v/DV = ½mv2/V = ½rv2,
F = ½Arv2
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 = ½CLArv2 , 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)CDAr.
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: 0òa (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 in text: 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 = DW/Dt where W = ò F dx.
Thus, P = D ò Fdx /Dt = F dx/dt = F v.
Therefore, F = P/v .
P/v = m dv/dt, or
dt = (mv/P) dv .
t = voò 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 .
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 = 1/2mv2 - 1/2mvo2 = DKE.
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 example: If you have some initial speed, no power applied, and some air resistance, will the object every completely stop?
a) Model #1: FAR = -bv, where b is a constant that depends on the shape of the object and the density of the air. From above, b = 6phr. We will assume that b holds constant during the motion.
-bv = m dv/dt
dt = -(m/b) (1/v) dv [recall that ò 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.
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: from above b = (1/2)CDAr. We will assume that b holds constant during the motion.
-bv2 = m dv/dt
dt = (-m/b)(1/v2)dv [recall that ò 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.
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.
c) Model #3: FAR = -beav where b and a are constants that depend on the shape of the object and the density of the air. We will assume that a and b hold constant during the motion. [Note that b has units of force and a has units of inverse speed (s/m).]
-beav = m dv/dt
dt = (-m/b) e-av dv [recall that òe-avdv = (-1/a)e-av but we must apply both lower and upper limits]
t = (m/ab)e-av - (m/ab)e-avo .
Inverting this, we get: v(t) = (-1/a)ln[(ab/m)t + e-avo] .
For v=0, we need the argument of the ln function to be 1, so the stopping time is:
tS = (m/ab)(1 - e-avo) . 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 a 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-avo) behaves this way or not. However, if we expand e-avo in a power series (for small avo), we get:
tS = (m/ab)(1 - [1 - avo + a2vo2/2 - ...]) = (m/ab)( avo - a2vo2/2 -...) = mvo/b - mavo2/2b +... .
From this, it is easy to see that as a 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/ab)(1 - e-avo) does indeed behave that way.