PHYD38. PROBLEM SET #1. SOLUTIONS .

Problems 1.1a & 1.2b

Solve Strogatz: 2.4.2, and 2.4.4.

SOLUTIONS

Will be discussed in tutorial.

Problem 1.2. Vertical fall with air drag

This is strictly speaking a 2-dimensional system, although in reality it works like two separate 1-d problems: the vertical motion under constant gravitational acceleration -g, in the presence of quadratic drag force proportional to v². The function to be found is v(t). The second variable is the vertical distance traveled z. It can be found from dz = v dt, as an afterthought, or by the method specified later in this problem. In any case, z is not involved in finding v(t). Downward fall in air is discussed in the textbook (exercise 2.2.13) but you should solve it here in terms of the variables defined below. The air drag force $F_d$ on balls, jumpers, cars, airplanes (but not necessarily very small particles like dust) is expressed as $$F_d = C_d A \frac{\rho \mbox{v}^2}{2},$$ where $\rho$ = air density, $C_d$ is a nondim. drag coefficient of order one, and $A$ the cross sectional area of the body. Vertical axis points upward, thus the drag force has positive sign when the object moves down (but only then). Gravitational acceleration is negative and equal -g. When acceleration of drag $F_d$/M (where M = mass of the body) reaches the same absolute value as gravity's acceleration

g = 9.81 m/s²,

then the body reaches terminal velocity v=V and is falling through air without accelerating or decelerating. So all those detailed properties of the object and air are conveniently contained in one single number, V (terminal velocity). In practice it ranges from ~10 m/s for small bodies that tend to have large A/M ratios, to 100 m/s or more for slender bodies with smaller A/M. You will just assume that in a given problem V is known, and use a nondimensional velocity defined as

u = v/V

Derive the equation of motion using u, g, and V. Show that in both upward and downward motion

du/dt = -(g/V) (1 + u |u|).

The free-fall starts at time t=0 with u=0 (initial conditions). Find the exact solution for u(t). Sketch it in a diagram. Use the nondimensional time on the horizontal axis: gt/V. You may use the calculator and compute a few points, to get the shape right.

Is the convergence to the terminal speed at large times exponential or algebraic? What is the time scale? Next, find the vertical distance traveled. When u < 0, 1+u|u| = 1-u². Instead of integrating v(t) = V u(t) over time to find z(t), use the following shortcut method:

dz = v dt, while
dt = -(V/g) du/(1-u²) by separating variables in the equation of motion, so
dz = -(V²/g) u du/(1-u²).

You can integrate the last ODE much more easily than dz = v(t) dt. True, you'll get z(u) not z(t), but u = u(t) is already known, so if you ever need z(t), you can get it as z(t) = z(u(t)). But you don't need z(t) in this problem. You may assume z=0 at u=t=0, for simplicity. Distance z(u) will then be negative. (If needed, the initial height can easily be redefined to be > 0).

SOLUTIONS

Problem 1.3. Upward vertical throw in air and combined up-and-down motion

Now consider a twin problem of vertical throw of the same body upward, until it stops rising. It is NOT goverened by exactly the same diff. equation as the fall. Namely, in an upward motion, u > 0, 1 + u |u| = 1+u², gravity and air drag work hand in hand and brake the motion quickly. They do not oppose each other like in the previous problem, and so never end up in approximate balance of forces.

Formulate and integrate the equation of motion to obtain u(t), with initial condition u(t0) = u0 > 0, at some negative time t0, and final velocity equal v=0 at time t=0. These final conditions match the initial conditions of the fall (prev. problem). Sketch u(t) or, more precisely, u(gt/V).

Formulate the ODE for the distance traveled, and solve it to obtain z_max = z(u=0)=z(t=0) in terms of V, g, and u0.

Show that the maximum height of a vertical throw in air is always smaller than in vacuum (to obtain the latter, simplify your solution in the appropriate limit of u0, corresponding to very small air density and drag force). For example, how much smaller is z_max in air than in vacuum if u0 = 1 in both cases (upward initial velocity = terminal velocity)?

What comes up must come down and, in our problem, cover the same vertical distance |z| = z_max. Given some u0, what is the final nondimensional speed u, with which the body strikes the ground? Check that the final speed u satisfies |u| < u0, i.e. |v| < v0 (give a physical reason for this) and that during the descent |u| < 1 (why?).

Does the fraction u/u0 depend on anything (u0, g, V?). Show that u/u0 is the same for all bodies thrown on all planets in the universe, and only depends on u0= v0/V.

SOLUTIONS

Contains part of prob. 2 solution (distance traveled down)

 

Problem 1.4

Analyze fully the 1-D system $$\dot{x} = 1- e^{-x} -(1+d)\,x^2$$ where $d$ is the last digit of your student number (state it in the solution!).
Find the fixed points with accuracy of 6 digits, by any method that works (can be numerical), and characterize stability analytically or graphically.

SOLUTIONS

One (unstable) root is x=0 (for any d), and one is stable but impossible to compute analytically. Graphing the f(x) function, one can roughly estimate that the root is not far from $x=1/(1+d)$, so that can be the starting value.

Method 1. ODE can be evolved toward the stable root by some simple time-integration method like Euler method, with timestep dt. It is interesting that dt does not need to be very small. The final point is an attractor and, somewhat curiously, will be achieved independent of step, even for dt=1! That way one does need to make thousands of steps, which for any other ODE would typically be required, given that Euler method is only 1st order accuracy. With timestep dt=1, and naming the right-hand side of ODE $f(x)$, the method consists of iteration: $x_{next} = x + f(x)$. It converges very quickly, in 10-14 steps one gets double precision accuracy of 14 digits. Try dt > 1 and see what happens, though.

Method 2. A very fast method of finding zeros is Newton's method. Read about it on wiki. It will work in case of unstable fixed points (method 1 won't). We watched a clip from the movie "21" where Kevin Spacey (playing me :-) discussed the Newton-Raphson method. [BTW, it isn't necessarily true that 'Newton stole it from Raphson' (cf. movie). Newton had a formula practically of the same form as he much later published (later than Raphson, that much is true).] The method is an iteration: $x_{next} = x - f(x)/f'(x)$. It is quadratic for well behaved zeros like our, i.e. the number of accurate digits doubles(!) in every iteration. Four to five iterations suffice for 14 digit accuracy. That's really hard to beat, though there are algorithms with even faster convergence!

Program 1.3.py solves this problem numerically, using both methods Here is it's output.