When Will a Particle Reach the Force Center Under an Inverse Cubed Law?

Click For Summary
SUMMARY

The discussion centers on determining the time it takes for a particle, released from rest at position x0 = b, to reach the force center at x = 0 under an inverse cubed law of attraction, described by the force equation F = -k/x^3. The participant attempted to derive the potential energy function, U, resulting in U = k/(2x^2) - k/2x0^2, but faced challenges integrating the time equation with this potential. The recommended approach is to utilize separation of variables to solve the resulting differential equation.

PREREQUISITES
  • Understanding of classical mechanics, specifically Newton's laws of motion.
  • Familiarity with differential equations and integration techniques.
  • Knowledge of potential energy concepts in physics.
  • Experience with inverse force laws and their implications on motion.
NEXT STEPS
  • Study the method of separation of variables in differential equations.
  • Explore potential energy functions related to inverse force laws.
  • Review the derivation of motion equations under varying force conditions.
  • Investigate the implications of conservation of energy in particle motion.
USEFUL FOR

Students studying classical mechanics, physics educators, and anyone interested in solving differential equations related to motion under variable forces.

blarggg
Messages
1
Reaction score
0

Homework Statement


At t0 = 0 a particle of mass m is released from rest at x0 = b > 0 away from a fixed origin of force that attracts the particle according to the law:
F=-k/x^3
where k >0 is a constant. At what time the particle will reach the force center x =0?

Homework Equations



I honestly tried to format the equations and stuff in here but i can't get it so its problem 3 on this pdf http://www.physics.uci.edu/~sasha/courses_files/111A/Fall10/hmw3.pdf

The Attempt at a Solution



well for U i got k/(2x^2)-k/2x0^2)

but i have no idea how i would integrate the t equation with that in place of U, and even that doesn't make sense because then at x=x0. potential would be zero and that can't be right
 
Last edited by a moderator:
Physics news on Phys.org
If your U is the same as on the problem sheet, this is not correct.
What is the potential resulting in this force?
Try separation of variables to solve the differential equation.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
Replies
11
Views
3K
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
3
Views
4K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
6
Views
7K
  • · Replies 1 ·
Replies
1
Views
2K