Taylor 2.1-2.2 (today) and 2.3-2.5 (Wednesday)

In chapter 1 we reviewed Newton's laws and the mathematical objects used to express it, vectors. We learned what differential equations are and saw two ways to solve them:

- direct integration (not always possible), and
- guessing the solution (not always easy or reliable).

Here we revisit the motion of projectiles near the Earth's surface, this time adding air resistance to the problem, and we'll discuss the motion of charged particles in magnetic fields. This will bring us more differential equations to learn to handle, and some mathematical functions to become familiar with.

Frictional forces are complex, involving microscopic and mesoscopic processes.
Yet, we can treat many problems with the simple approximation for sliding friction f = μN, where N is the magnitude of the normal force to the surface, and possibly with a different coefficient of friction for static friction, μ_{s}, and kinetic friction, μ_{k}.

In general, air resistance is complex. But in a similar way, a simple approximation can be used to treat many problems. On what quantity do we expect air friction to depend, in analogy to the normal force for sliding? Obviously, the normal force plays no role for a projectile, so what should take its place? Velocity is the appropriate quantity, and we can approximate air friction with two terms, one proportional to velocity (the linear term) and one proportional to the square of velocity (the quadratic term)

f(v) = bv + cv²

The frictional force is directed opposite the motion of the projectile

Note that we are clearly neglecting certain cases, for instance the lift on an airplane wing. Nevertheless, this approximation is useful in many situations, and allows us to move from the simple problems of introductory physics to more realistic situations.

The two constants, b and c, must be determined, but there is a useful approximation for spherical objects. For spherical objects, the constants can be parameterized in terms of the diameter, D,

b = βD and c = γD²

where, for air at STP

β = 1.6×10^{-4} N⋅s/m² and γ = 0.25 N⋅s² / m^{4}

Now we need to solve the equation of motion when air resistance is included. The difficulty varies depending on whether one drag force dominates so that the other can be neglected, or if both must be included. The easiest case to solve is when the linear drag force dominates; the case where the quadratic force dominates is more difficult; and when both must be included, numerical calculations are required.

To determine if one force dominates, it is convenient to look at their ratio

Calculate the ratio of quadratic to linear drag force for:

- a baseball with D = 7 cm and v = 5 m/s
- a raindrop with D = 1 mm and v = 0.6 m/s
- an oil drop for a Millikan oil drop experiment with D = 1.5 μm and v = 5×10
^{-5}m/s

For the baseball

For the raindrop

For the oil drop

The quadratic force dominates for the baseball; the linear force dominates for the oil drop; and both forces are comparable for the raindrop.

We will begin with the easiest case: when the linear drag force dominates, as for the oil drop in a Millikan oil drop experiment.
Note that the linear force (vector) can be written as **f** = -*bv* **v**hat = -*b***v**.

The equation of motion of a projectile subject to gravity and a linear drag force is

a first-order, linear differential equation for **v**.
The feature that makes this easy to solve is that the component equations separate into x and y components of velocity naturally

and

Note that, for other force functions, the component equations don't necessarily separate into x and y velocity components. For example, the quadratic drag force can be written

and this yields equations that involve both *v _{x}* and

For the linear drag force the equations

The equation for the v_{x} component is also an equation for motion in one dimension of an object subject to linear drag.
The general form is:

d_{t}*v*_{x} = -*kv*_{x} ,

with k=b/m.
This differential equations says that the derivative of v_{x} is proportional to a (negative) constant times the original v_{x}.
A function that fits this description is the exponential function, so we choose a solution of the form:

where A is a constant of integration (depends on the initial conditions).

Notice that we can get this solution by direct integration if we treat the derivative like a fraction of infinitisimals:

dv/dt = -kv

becomes

dv/v = -kdt

which can be integrated as

∫_{v0}^{v}dv'/v' = -k∫_{0}^{t}dt'

where I've used primes to avoid confusion with the limits of integration, and I've set v=v_{0} at t=0.
This integral yields

ln(v) - ln(v_{0}) = -kt

and after rearranging terms and exponentiating becomes

v = v_{0}e^{-kt}

where k=m/b. The velocity tends to zero as time goes to infinity.

The horizontal position as a function of time is found by integrating the velocity

x(t) = x(0) + ∫_{0}^{t} v_{x}(t')dt'

x(t) = x_{∞}(1 - e^{-t/τ})

where I've set x=0 at t=0 and used x_{∞} = v_{x0}τ

The motion in the vertical direction is the same as that for an object dropped from a height and subject to a linear drag force,

m d_{t}v_{y} = mg - bv_{y} .

The solution of this equation is not much different from the case for horizontal motion and is

v_{y} = A e^{-kt} + mg/b

as can be checked by substitution, where k = b/m.
This motion differs from the previous case in that, instead of the velocity tending to zero as t tends to infinity, v_{y} tends to mg/b = g/k.
We call this the terminal velocity, since an object dropped from rest will achieve this velocity,

v_{ter} = mg/b = g/k

but won't exceeed it.

If the object is dropped from rest, then the velocity as a function of time can be written as

where τ = 1/k = v_{ter}g is the time constant.
Expressions like this arise for the problem of charging a capacitor through a resistor.
In this case, it says that the velocity will increase from zero and reach about 63% of its final value, v_{ter}, after one time constant, τ.
After two time constants the velocity reaches 86% of its final value, and 95% after 3 time constants.

If the object doesn't start from rest but with initial velocity v_{y0}, then the velocity as a function of time is

v_{y}(t) = v_{ter} + (v_{y0} - v_{ter}) e^{-t/τ} .

This is basically the result applied to the Millikan oil drop experiment. To determine the charge on a drop, one needs the mass. The above result shows that no matter how the drop forms it will quickly equalibrate and fall through the air with the terminal velocity. By measuring the terminal velocity in air, the mass can be found.

The position as a function of time is found by integrating the velocity

y(t) = ∫_{0}^{t} v_{y}(t') dt'

y(t) = v_{ter}t + (v_{y0} - v_{ter}) τ (1 - e^{-t/τ})

Now we have the solutions for the motion in both x and y, in particular

x(t) = v_{x0}τ (1 - e^{-t/τ})

y(t) = (v_{y0} + v_{ter}) τ (1 - e^{-t/τ}) - v_{ter}t

where the sign of vy = [(v_{y0} + v_{ter})/v_{x0}]x + v_{ter}τ ln[1 - x/(v_{x0}τ)]

One of your homework problems is to show that as the air drag is reduced to zero, the expression reduces to what we expect from introductory physics.

© 2007 Robert Harr