PHY5200 F06

Chapter 2: Projectiles and Particles

Reading

Taylor 2.1-2.3 (today) and 2.4-2.5 (Wednesday)

Recall

The frictional force can be approximated with the empirical relation

f(v) = bv + cv²

The frictional force is directed opposite the motion of the projectile

f = -f(v) vhat

Linear Air Resistance

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 vhat = -bv.

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

mv[dot] = mg - bv

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

mv_x[dot] = -bv_x

and

mv_y[dot] = mg - bv_y .

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

f = -cv v = -c sqrt(v_x² + v_y²)v

and this yields equations that involve both v_x and v_y, as shown in the text. We'll see that when the x and y components separate, the resulting differential equations are intrinsically more simple to solve (in closed form). In cases where they don't separate, a judicious change of variables can sometimes produce equations where the new variables are. When the components don't separate, the equations are coupled, meaning that v_x and v_y may appear in both equations such that both equations must be solved simultaneously.

For the linear drag force the equations decouple and can be solved separately. This is what we will do next. The solution to each component is also the solution to an independent one-dimentionsal problem on its own.

Horizontal motion with linear drag

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:

v_x[dot] = -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:

v_x(t) = A e^-kt

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^vdv'/v' = -k∫₀^tdt'

where I've used primes to avoid confusion with the limits of integration, and I've set v=v₀ at t=0. This integral yields

ln(v) - ln(v₀) = -kt

and after rearranging terms and exponentiating becomes

v = v₀e^-kt