PHY5200 F07

Chapter 2: Projectiles and Particles

Reading

Taylor 2.3-2.4 (today) and 2.4-2.6 (Monday)

Recall

The trajectory of a particle subject to a linear drag force is

y(x) = [(vy0 + vter)/vx0]x + vterτ ln[1 - x/(vx0τ)]

Horizontal range

You might recall from introductory physics, that in the case of no air drag, the range of a projectile is

Rvac = 2vx0vy0 / g

The range when we take air resistance into account is the value of x that makes y=0, that is the value x=R that satisifies

0 = [(vy0 + vter)/vx0]R + vterτ ln[1 - R/(vx0τ)]

This is a transcendental equation; it doesn't have an analytical solution. It can be solved numerically, or approximations can be used. For instance, if the air drag is small, then vter and τ = vterg are very large. Then it is possible that the second term in the logarithm is small and we can replace the log with a Taylor series expansion:

0 ≈ [(vy0 + vter)/vx0]R + vterτ[R/(vx0τ) + (1/2)R²/(vx0τ)² + (1/3)R³/(vx0τ)³]

Now this expression can be simplified, dividing out a factor of R, and rewriting in the form:

R = (2vx0vy0)/g - 2R²/(3vx0τ)

This expression could be solved as a quadratic equation in R, but since we're already making approximations, a somewhat easier method is to substitute the value for the range in vacuum for R² on the right hand side. In vacuum, the range of a projectile is given by

Rvac = (2vx0vy0)/g

Substituting this on the right hand side yields

R ≈ Rvac - 2Rvac²/(3vx0τ)
R ≈ Rvac (1 - (4/3)(vy0/vter))  .

Writing the result this way emphasizes that, in the absence of air drag, the range is Rvac, and that the presence of air drag causes the range to be less than Rvac.

Quadratic Air Resistance

Now we can investigate what happens in the case that quadratic air resistance dominates. This is the situation that occurs most frequently with larger objects moving in air: baseballs, basketballs, parachutists, and projectiles. Linear air drag can be understood as coming from the impulse that must be given to the fluid in order to move it around the object passing through it. It is normally characterized by laminar flow around the object, that is, flow without turbulence. Quadratic air drag is normally characterized by turbulence behind the object as it moves through the fluid, and arises due to energy being dissipated in the turbulence (hence proportional to v² like kinetic energy).

The equation of motion for a projectile subject to a quadratic drag force is

m dtv = mg - c|v|v

This is a first order non-linear differential equation. The x and y component equations are coupled:

m dtvx = -c(vx²+vy²)1/2 vx

and

m dtvy = mg - c(vx²+vy²)1/2 vy

These equations are normally impossible to solve analytically and we must resort to numerical techniques.

In the special case of one-dimensional motion, these equations can be solved analytically. We will treat these cases in the following.

Horizontal motion with quadratic drag

For one-dimensional motion subject to a quadratic drag force, the equation of motion is

m(dv/dt) = -cv²

This is a first order, non-linear differential equation. It is relatively easy to solve if we treat the numerator and denominator of the derivative as separable quantities that can be multiplied and divided in the equation, to yield

mdv/v² = -cdt

In mathematics, these quantities should be treated as differentials, and the technique of moving all quantities dependent on v to one side and those dependent on t to the other is called separation of variables.

We now integrate both sides to solve for an equation in v and t

m∫ v0v (dv'/v'²) = -c∫ 0t dt'

where definite integrals are used so there are no arbitrary constants of integration. The integral is

m(1/v0 - 1/v) = -ct

which can be solved for v as a function of t to yield

v(t) = v0/(1+cv0t/m) = v0/(1+t/τquad)

where

τquad = m / cv0

is a time constant for the motion with quadratic air drag. Note that, except that it has the units of time, this constant is (1) different from the time constant determined for linear drag (the quadratic constant depends on the initial velocity, while the linear constant is independent of the initial velocity), and (2) that the behavior as a function of time differs, since this is a polynomial in time, and the other solution is exponential.

Integrate the velocity to find the position as a function of time

x(t) = x(t=0) + ∫0t v(t')dt'

which yields

x(t) = v0τquad ln(1+t/τquad)

Unlike the case of linear drag, this motion doesn't have a limit, x goes to infinity as t goes to infinity.


© 2007 Robert Harr