PHY5200 F06

Chapter 2: Projectiles and Particles

Reading

Taylor 2.4-2.5 (today) and 3.1-3.2 (Monday)

Recall

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

y(x) = [(v_y0+v_ter)/v_x0]x + v_terτ ln[1 - x/(v_x0τ)]

Horizontal range

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

R_vac = 2v_x0v_y0 / g

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

0 = [(v_y0+v_ter)/v_x0]R + v_terτ ln[1 - R/(v_x0τ)]

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 v_ter and τ=v_terg 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 ≈ [(v_y0+v_ter)/v_x0]R + v_terτ[R/(v_x0τ) + (1/2)R²/(v_x0τ)² + (1/3)R³/(v_x0τ)³]

Now this expression can be simplified to give:

R = (2v_x0v_y0)/g - 2R²/(3v_x0τ)

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 rhs. In vacuum, the range of a projectile is given by

R_vac = (2v_x0v_y0)/g

This yields

R ≈ R_vac - 2R_vac²/(3v_x0τ)

R ≈ R_vac (1 - (4/3)(v_y0/v_ter)) .

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

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

mv[dot] = mg - c|v|v

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

mv_x[dot] = -c(v_x²+v_y²)^1/2 v_x

and

mv_y[dot] = mg - c(v_x²+v_y²)^1/2 v_y

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∫_v₀^vdv'/v'² = -c∫₀^tdt'

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/τ_quad)

where

τ_quad = 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 (this depends on the initial velocity, but is independent of g), and (2) that the behavior as a function of time differs, since this is a polynomial in t, and the other solution is exponential.

Integrate to find the position as a function of time

x(t) = x(t=0) + ∫₀^t v(t')dt'

which yields

x(t) = v0τ_quad ln(1+t/τ_quad)

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

Vertical motion with quadratic drag

For one-dimensional motion subject to both gravity and quadratic drag, the equation of motion is

m(dv/dt) = mg - cv²

where I've chose gravity in the positive direction (y axis pointing downward). When the drag force equals the weight, the net force on the object is zero and it ceases to accelerate or decelerate. This occurs at the terminal velocity

v_ter = (mg/c)^½.

This is the speed that a falling object will eventually obtain, independent of its initial velocity.

Using the terminal velocity, we can rewrite the equation of motion as

v[dot] = g(1 - [v²/v_ter²])

and use the technique of separation of variables to obtain

dv / (1-v²/v_ter²) = gdt .

This can be integrated to yield

(v_ter/g) arctanh(v/v_ter) = t

or, solving for v as a function of t

v(t) = v_ter tanh(gt/v_ter)

where tanh is the hyperbolic tangent function

tanh(z) = (e^z - e^-z) / (e^z + e^-z)

and arctanh is the inverse hyperbolic tangent function.

The hyperbolic functions are a set of functions connected to the standard trigonometric fucntions: sin, cos, tan, and their inverses. The hyperbolic tangent appears in a number of situations. It has the properties of tanh(0)=0, d(tanh(z))/dz=1 for z=1, tanh(∞)=+1, tanh(-∞)=-1. The hyperbolic sine and cosine functions are defined as

sinh(z) = ½(e^z - e^-z) cosh(z) = ½(e^z + e^-z)

They have the following properties: sinh(0) = 0, d(sinh(z))/dz = 1 for z=0, sinh(∞) = ∞, sinh(-∞) = -∞, cosh(0) = 1, d(cosh(z)/dz = 0 for z=0, cosh(∞) = cosh(-∞) = ∞. The hyperbolic tangent has the same relation to the hyperbolic sine and cosine functions as tangent has with sine and cosine, that is, tanh(z) = sinh(z)/cosh(z).