Question 19.12: Inverse bungee jump A bungee cord with a 49-N/m spring const......
Inverse bungee jump
A bungee cord with a 49-N/m spring constant will be used for an inverse bungee jump by a 70-kg person. The person starts by hanging stationary at the end of a cord that stretches an unknown distance A (the unstretched cord is 32 m long). The person hanging at the end of the cord is then pulled down an additional distance A so the cord is now stretched a distance 2A. Finally, the person is released and vibrates up and down over a distance 2A. Determine the amplitude A of the vibration, the maximum speed of the jumper during each vibration, and the period of one vibration.
Sketch and translate We start by sketching the situation (see the sketch at the right). The person vibrates around an equilibrium position, the place where he hangs at the end of the cord when not vibrating. The gravitational force that Earth exerts on the person is constant during the entire vibration. However, the force that the cord exerts on the person continually changes—increasing as the cord stretches downward. We will use two systems for analysis: for the energy analysis the jumper, the cord, and Earth are the system. For the force analysis, the jumper will be the system.
Simplify and diagram Assume that the cord stretches linearly (that is, it does not exceed its elastic limit). Also, ignore air resistance and the mass of the bungee cord. The y -axis points upward with the origin at the equilibrium position where the jumper hangs at rest. Force diagrams for the jumper as a system when at the top of the vibration, while passing through equilibrium, and at the bottom of the vibration are shown at the right. Bar charts represent the energy of the jumpercord- Earth system when the person is at these three positions. Assume that the system is isolated and thus its energy is constant. When the person is at the highest point in the vibration (a distance A above the equilibrium position), the cord is not stretched and the system has only gravitational potential energy. When the person is passing through equilibrium, he has maximum speed and kinetic energy. Also, the system has elastic potential energy as the cord is now stretched a distance A. At the bottom of the vibration, the system has considerable elastic potential energy as the cord is stretched a distance 2A; there is also negative gravitational potential energy in the system as the person is below the origin of the vertical y-axis.
Represent mathematically We apply the vertical component form of Newton’s second law to the jumper to determine how far the cord is stretched when hanging at the equilibrium position (see the second force diagram in the figure above). As the positive y-direction is up, the component of the elastic force exerted on the jumper is positive and the component of the gravitational force exerted by Earth is negative:
kA + (-mg) = 0
where A equals the distance that the cord is stretched when the jumper of mass m hangs at rest (zero acceleration) at the equilibrium position. This distance A also equals the amplitude of the vibration—the cord will be relaxed at the top of the vibration and will be stretched 2A at the bottom of the vibration.
At the top of the vibration, the system’s energy will all be gravitational potential energy +mgA and at the bottom it will be elastic and gravitational \frac{1}{2} k(2 A)^2-m g A .
These energies are equal, and they equal the maximum kinetic energy plus elastic energy when the jumper passes through the equilibrium position \frac{1}{2} m v_{\max }^2+\frac{1}{2} k A^2 \text {. Thus, }
+m g A=\frac{1}{2} m v_{\max }^2+\frac{1}{2} k A^2=\frac{1}{2} k(2 A)^2-m g A
The period of the vibration will be
T=2 \pi \sqrt{\frac{m}{k}}
Solve and evaluate The cord will be stretched A when the jumper is at equilibrium:
A=\frac{m \text{g}}{k}=\frac{(70 \mathrm{~kg})(9.8 \mathrm{~N} / \mathrm{kg})}{(49 \mathrm{~N} / \mathrm{m})}=14 \mathrm{~m}
We can determine the jumper’s maximum speed using the left equality in the energy equation above:
v_{\max }=\left(\frac{2 m \text{g} A-k A^2}{m}\right)^{1 / 2}
=\left(\frac{2(70 \mathrm{~kg})\left(9.8 \mathrm{~m} / \mathrm{s}^2\right)(14 \mathrm{~m})-(49 \mathrm{~N} / \mathrm{m})(14 \mathrm{~m})^2}{70 \mathrm{~kg}}\right)^{1 / 2}
=11.7 \mathrm{~m} / \mathrm{s} \approx 12 \mathrm{~m} / \mathrm{s} \text { or } 26 \mathrm{mph}
The period for one down-up-down vibration will be
T=2 \pi\left(\frac{m}{k}\right)^{1 / 2}=2 \pi\left(\frac{70 \mathrm{~kg}}{49 \mathrm{~N} / \mathrm{m}}\right)^{1 / 2}=7.5 \mathrm{~s}
All of the numbers above have the correct units and reasonable
magnitudes.
Try it yourself: Suppose the bungee cord used for the inverse bungee jump had a spring constant of 40 N/m. Determine the amplitude of the ride, the maximum speed during a vibration, and the time interval for one vibration.
Answer: 17 m; 13 m/s; 8.3 s.