What is the equation for the length of a falling chain through a hole?

Click For Summary
SUMMARY

The discussion focuses on deriving the equation for the length of a falling chain through a hole, specifically y(t) = A eγt + B e−γt, where γ is a constant. The problem involves a flexible chain of mass M = 0.7 kg and length L = 3.0 m, with an initial length L0 = 0.5 m hanging through a hole. The conservation of energy principle is applied, leading to the potential energy equation U(y) = (-Mg/(2L))(y2). The goal is to calculate the time when half of the chain has fallen through the hole.

PREREQUISITES
  • Understanding of conservation of energy principles in physics
  • Familiarity with calculus, particularly integration
  • Knowledge of differential equations
  • Basic mechanics of motion and forces
NEXT STEPS
  • Study the derivation of potential energy in systems involving gravity
  • Learn about solving second-order differential equations
  • Explore applications of exponential functions in physical systems
  • Investigate similar problems involving chains or flexible bodies in motion
USEFUL FOR

Students studying classical mechanics, physics educators, and anyone interested in solving problems related to motion and energy conservation in flexible systems.

kraigandrews
Messages
105
Reaction score
0

Homework Statement



A flexible chain of mass M and length L lies on a frictionless table, with a very short portion of its length L0 hanging through a hole. Initially the chain is at rest. Find a general equation for y(t), the length of chain through the hole, as a function of time.
(Hint: Use conservation of energy. The answer has the form y(t) = A eγt+ B e−γt where γ is a constant.)
Calculate the time when 1/2 of the chain has gone through the hole.
Data: M = 0.7 kg; L = 3.0 m; L0 = 0.5 m.

Homework Equations



E=U(y)+.5m(dy/dt)^2
dm=M/L

The Attempt at a Solution



U(y)=integral of(-gydm)
dm=M/L
then integral of(-g(M/L)ydm)=(-Mg/(2L))(y^2)=U(y)

then i plug into E(which is constant by conversation of energy)=U(y)+.5M(dy/dt)^2

don't know where to go from there.
thanks.
 
Physics news on Phys.org
Google "sliding chain brown university"
 
Thank you
 

Similar threads

  • · Replies 30 ·
2
Replies
30
Views
3K
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
3K