Falling Body Problems
Consider a vertically falling body of mass m that is being influenced only by gravity g and an air resistance that is proportional to the velocity of the body. Assume that both gravity and mass remain constant and, for convenience, choose the downward direction as the positive direction.
Newton's second law of motion: The net force acting on a body is equal to the time rate of change of the momentum of the body; or, for constant mass, `F=m(dv)/(dt)` .
For the problem at hand, there are two forces acting on the body: the force due to gravity given by the weight w of the body, which equals mg, and the force due to air resistance given by —kv, where k > 0 is a constant of proportionality. The minus sign is required because this force opposes the velocity; that is, it acts in the upward, or negative, direction. The net force F on the body is, therefore, F = mg-kv.
Substituting this result we obtain `mg-kv=m(dv)/(dt)` or `(dv)/(dt)+k/m v=g` as the equation of motion for the body.
If air resistance is negligible or nonexistent, then k = 0 and equation simplifies to `(dv)/(dt)=g` .
When k > 0, the limiting velocity `v_l` is defined by `v_l=(mg)/k` .
Given differential equation is valid only if the given conditions are satisfied. This equation is not valid if, for example, air resistance is not proportional to velocity but to the velocity squared, or if the upward direction is taken to be the positive direction.
Example 1. A steel ball weighing 2 Ib is dropped from a height of 3000 ft with no velocity. As it falls, the ball encounters air resistance numerically equal to `v/8` (in pounds), where v denotes the velocity of the ball (in feet per second). Find the limiting velocity for the ball and the time required for the ball to hit the ground.
Locate the coordinate system with the ground now situtated at x=3000. Here, w=2 lb and `k=1/8` . Assuming gravity g is `32 ft/s^2` , we have from the formula w=mg that `m=w/(g) =2/32=1/16` slug. So, differential equation is `(dv)/(dt)+(1/8)/(1/16)v=32` or `(dv)/(dt)+2v=32` .
This is first-order linear differential equation. Its solution is `v(t)=ce^(-2t)+16` .
We are given that initially there is no velocity, so v(0)=0: `0=ce^(-2*0)+16` or `c=-16` .
And differential equation becomes `v(t)=-16e^(-2t)+16` .
From this equation we see that `v->16` when `t->oo` , so limiting velocity is 16 `ft/s^2` .
To find the time it takes for the ball to hit the ground (x = 3000), we need an expression for the position of the ball at any time t. Since `v=(dx)/(dt)` then `(dx)/(dt)=-16e^(-2t)+16` . Integrating both sides gives `x=8e^(-2t)+16t+c_1` . To find constant use the fact that x=0 when t=0: `0=8e^(-2*0)+16*0+c_1` or `c_1=-8` .
So, position of particle is given as `x(t)=8e^(-2t)+16t-8` .
We need to find such t that `x(t)=3000` : `3000=8e^(-2t)+16t-8` or `376=e^(-2t)+2t` .
This equation cannot be solved explicitly for t, we can approximate the solution by trial and error, substituting different values of t until we locate a solution to the degree of accuracy we need. Alternatively, we note that for any large value of t, the negative exponential term will be negligible. A good approximation is obtained by setting 2t = 376 or t = 188 sec. For this value of t, the exponential is essentially zero.
Example 2. A body is propelled straight up with an initial velocity of 500 ft/sec in a vacuum with no air resistance. How long will it take the body to return to the ground?
Since there is no air resistance then k=0, so `(dv)/(dt)=g=32` , so `v=32t+C`.
Since `v(0)=-500` (because we are moving up in negative direction) then `-500=32*0+C` or `C=-500` .
It is known that `(dx)/(dt)=v=32t-500` . Integrating this equation gives `x(t)=16t^2-500t+C_1` . Since `x(0)=0` then `0=16*0^2-500*0+C_1` or `C_1=0` .
So, `x(t)=16t^2-500t` . We need to find time when `x(t)=0` .
`16t^2-500t=0` or `t(16t-500)=0` which gives `t=500/16=31.25` seconds.