Solve Bungee Jump Issue: Derive Expression for x Double Dot, Show mg >= kh/4

Click For Summary
SUMMARY

The discussion focuses on deriving an expression for the acceleration (x double dot) of a bungee jumper, considering the forces acting on the jumper, including gravitational force (mg) and spring force (kx/h). The key conclusion is that the maximum allowable acceleration is limited to 3g, and the condition mg must be greater than or equal to kh/4 is established. Participants clarify that the use of elastic energy is unnecessary for this derivation, emphasizing a force-based approach instead.

PREREQUISITES
  • Understanding of Newton's Second Law of Motion
  • Familiarity with spring mechanics and Hooke's Law
  • Knowledge of basic calculus, particularly derivatives
  • Concept of gravitational force and its implications in dynamics
NEXT STEPS
  • Study the derivation of equations of motion for spring systems
  • Learn about the application of Newton's laws in non-constant acceleration scenarios
  • Explore the relationship between force, mass, and acceleration in dynamic systems
  • Investigate the implications of maximum acceleration limits in safety-critical applications
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators seeking to clarify concepts related to forces and motion in spring systems.

Aihara
Messages
3
Reaction score
0

Homework Statement


Typical bungee jumper. Rope has natural length h, spring constant k, man has mass m.

The safety limit for the max acceleration of the man is 3g. Derive an expression for x double dot, and show mg *greater or same than* kh/4

Homework Equations


Well I'm sure everyone knows them
I'm confused about Elastic Energy (kx^2) and Elastic Potential Energy (kx^2/2l), which do I use?


The Attempt at a Solution


I'm got four pages of working but not get where I want.
The acceleration (a) isn't constant, so are we trying to get an expression for a in terms of time or string extension...

At a random point after the rope has started extending mg - T = ma. so mg - kx/h = ma
so rearrange for acceleration... but then the x won't cancel and I can't show what I was asked too...

Thanks
 
Physics news on Phys.org
Welcome to PF!

Aihara said:
At a random point after the rope has started extending mg - T = ma. so mg - kx/h = ma
so rearrange for acceleration... but then the x won't cancel and I can't show what I was asked too...

Thanks

Hi Aihara! Welcome to PF! :smile:

(You won't need energy, only force)

why should the x cancel? :confused:

a = x'', so you should get an equation relating x'' and x …

then solve it. :smile:
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
6K
Replies
9
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 26 ·
Replies
26
Views
5K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
5K