MATLAB APPLICATION: BUNGEE JUMPER WITH CORD

Because these equations are not stiff, we can use ode45 to obtain the solutions and dis- play them on a plot: [t,y] = ode45bungee,[0 50],[0 0],[],30,0.25,68.1,40,8; plott,-y:,1,-,t,y:,2,: legendx m,v ms As in Fig. 23.11, we have reversed the sign of distance for the plot so that negative distance is in the downward direction. Notice how the simulation captures the jumper’s bouncing motion. 40 20 ⫺20 ⫺40 ⫺60 ⫺80 10 20 30 40 50 x m v ms FIGURE 23.11 Plot of distance and velocity of a bungee jumper.

23.5 CASE STUDY PLINY’S INTERMITTENT FOUNTAIN

Background. The Roman natural philosopher, Pliny the Elder, purportedly had an in- termittent fountain in his garden. As in Fig. 23.12, water enters a cylindrical tank at a con- stant flow rate Q in and fills until the water reaches y high . At this point, water siphons out of the tank through a circular discharge pipe, producing a fountain at the pipe’s exit. The foun- tain runs until the water level decreases to y low , whereupon the siphon fills with air and the fountain stops. The cycle then repeats as the tank fills until the water reaches y high , and the fountain flows again. When the siphon is running, the outflow Q out can be computed with the following formula based on Torricelli’s law: Q out = C 2gyπr 2 23.28 23.5 CASE STUDY 609

23.5 CASE STUDY continued

Neglecting the volume of water in the pipe, compute and plot the level of the water in the tank as a function of time over 100 seconds. Assume an initial condition of an empty tank y = 0, and employ the following parameters for your computation: R T = 0.05 m r = 0.007 m y low = 0.025 m y high = 0.1 m C = 0.6 g = 9.81 ms 2 Q in = 50 × 10 −6 m 3 s Solution. When the fountain is running, the rate of change in the tank’s volume V m 3 is determined by a simple balance of inflow minus the outflow: d V dt = Q in − Q out 23.29 where V = volume m 3 . Because the tank is cylindrical, V = π R 2 t y. Substituting this re- lationship along with Eq. 23.28 into Eq. 23.29 gives d y dt = Q in − C √ 2gyπr 2 π R 2 t 23.30 When the fountain is not running, the second term in the numerator goes to zero. We can incorporate this mechanism in the model by introducing a new dimensionless variable siphon that equals zero when the fountain is off and equals one when it is flowing: d y dt = Q in − siphon × C √ 2gyπr 2 π R 2 t 23.31 In the present context, siphon can be thought of as a switch that turns the fountain off and on. Such two-state variables are called Boolean or logical variables, where zero is equiva- lent to false and one is equivalent to true. Q in Q out R T y = y high y = y low y = 0 r FIGURE 23.12 An intermittent fountain.