Design a "bungee jump" apparatus for adults. A bungee jumper falls from a high platform with two elastic cords tied to the ankles. The jumper falls freely for a while, with the cords slack. Then the jumper falls an additional distance with the cords increasingly tense. Assume that you have cords that are 8 m long, and that the cords stretch in the jump an additional 22 m for a jumper whose mass is 100 kg, the heaviest adult you will allow to use your bungee jump (heavier customers would hit the ground). (a) It will help you a great deal in your analysis to make a series of 5 simple diagrams, like a comic strip, showing the platform, the jumper, and the two cords at the following times in the fall and the rebound: 1 while cords are slack (shown here as an example to get you started) 2 when the two cords are just starting to stretch 3 when the two cords are half stretched 4 when the two cords are fully stretched 5 when the two cords are again half stretched, on the way up On each diagram, draw and label vectors representing the forces acting on the jumper, and the jumper's velocity. Make the relative lengths of the vectors reflect their relative magnitudes. (b) At what instant is there the greatest tension in the cords? (How do you know?) A At the bottom, when the person has fallen 30 m. O At the top, when the person has fallen 0 m. O When the person has fallen between 0 m and 8 m. O when the person has fallen 8 m O when the person has fallen between 8 m and the bottom. (c) What is the jumper's speed at this instant, when the tension is greatest in the cords? v= m/s (d) Is the jumper's momentum changing at this instant or not? (That is, is da/dt nonzero or zero?) Yes, the jumper's momentum is changing. O No, the jumper's momentum is not changing. (e) Which of the following statements is a valid basis for answering part (d) correctly? After a very short time the momentum will be upward (and nonzero). O Since the momentum is zero, the momentum isn't changing. Since the net force must be zero when the momentum is zero, and since dp,/dt is equal to the net force, doy/dt must be zero. If the momentum weren't changing, the momentum would remain zero forever. A very short time ago the momentum was downward (and nonzero). Check to make sure that the magnitudes of the velocity and force vectors shown in your diagram number 4 are consistent with your analysis of parts (c), (d), and (e). (f) Focus on this instant of greatest tension and, starting from a fundamental principle, determine the spring stiffness k, for each of the two cords. k₂-60.74 ✔N/m (g) What is the maximum tension that each one of the two cords must support without breaking? (This tells you what kind of cords you need to buy.) Fy-1336.28✔ N Two cords slack initiall king
Simple harmonic motion
Simple harmonic motion is a type of periodic motion in which an object undergoes oscillatory motion. The restoring force exerted by the object exhibiting SHM is proportional to the displacement from the equilibrium position. The force is directed towards the mean position. We see many examples of SHM around us, common ones are the motion of a pendulum, spring and vibration of strings in musical instruments, and so on.
Simple Pendulum
A simple pendulum comprises a heavy mass (called bob) attached to one end of the weightless and flexible string.
Oscillation
In Physics, oscillation means a repetitive motion that happens in a variation with respect to time. There is usually a central value, where the object would be at rest. Additionally, there are two or more positions between which the repetitive motion takes place. In mathematics, oscillations can also be described as vibrations. The most common examples of oscillation that is seen in daily lives include the alternating current (AC) or the motion of a moving pendulum.
Trending now
This is a popular solution!
Step by step
Solved in 7 steps with 1 images