Physics 374
Fall 2004
Prof. E. F. Redish

HW4, Problem 4

a.    There are two forces on the ball: gravity and the drag force
                                          (F^net) = -mg -bv
      We should make sure this makes sense.  "-mg" is always negative, which is "down", which is good.  If the ball is traveling up (v > 0), then the drag force (-bv) is negative, which is "down", which makes sense.  If the ball is traveling down (v > 0), then the drag force 
(-bv) is positive, which is "up", which makes sense.  So from Newton's Second Law:
 					m(dv/dt) = -mg -bv
 					(dv/dt) = -g - (b/m)v
 					
 b.    There are many ways of using Mathematica to solve this equation and then graph the solution.  Some are certainly more efficient than others.  I will try to be as efficient as possible.  I'll define a function called "drag" to which you can pass a value of b and a value of m.  Make sure you understand why the code below translates into English as "Let the function drag[b,m] return a plot of the value of v[t] evaluated according to the rules of numerically solving the equation (dv/dt) = -g - (b/m)v subject to the initial condition v(0)=10 for values of t from 0 to 20sec.  Keep the vertical range of the graph between -20m/s and 20m/s."

drag[b_, m_] := 
      Plot[Evaluate[
          v[t] /. NDSolve[{v'[t] == -9.81 - ((b*v[t])/m), 
                v[0] == 10}, v, {t, 0, 20}]], {t, 0, 20},  
        PlotRange -> {-20, 20}]

drag[1, 1]

drag[5, 1]

drag[ .5, 1]

drag[ .5,  .5]

As we'd physically expect, we see all the velocity graphs eventually leveling off at some terminal velocity (which is of course always negative...i.e. downwards).  Setting (dv/dt) = 0 in the equation from part "a" tells us to expect that terminal velocity = -(gm/b) .  Indeed, all the graphs put the terminal velocity there.  We see that terminal speed should increase with increasing mass (which makes physical sense...think of a golf ball and a ping-pong ball falling through the same air) and we see terminal speed should decrease as the medium becomes more viscous (i.e. b goes up).  That makes sense too...think of a golf ball falling in air vs. a golf ball falling in water.  
       We also see that for the same drag coefficient, a larger mass takes longer to reach its terminal velocity than a smaller mass takes...see the last two graphs above.  That matches what we'd physically expect.  Also, looking at the first two graphs, the same mass will reach its terminal velocity more quickly in a more viscous medium, again matching our physical intuitions.

(Solution by T .  Bing)