Normal and centrifugal force for arbitrary curves

Click For Summary
SUMMARY

This discussion focuses on deriving equations for normal and centripetal forces acting on a ball rolling down a parametric curve defined as g:[0,t]->IR², where g(t)=(x(t),y(t)). The key equation provided for calculating the radius of curvature is radius(t) = ((x'(t))² + (y'(t))²)^(3/2) / |(x'(t) y''(t)) - (y'(t) x''(t))|. Understanding these forces is crucial for determining the conditions under which the ball may leave the curve.

PREREQUISITES
  • Understanding of parametric curves and their derivatives
  • Knowledge of gravitational forces and dynamics
  • Familiarity with concepts of normal and centripetal forces
  • Basic calculus, particularly differentiation
NEXT STEPS
  • Research the derivation of normal force equations in non-linear motion
  • Study centripetal force calculations in varying gravitational fields
  • Explore applications of curvature in physics, specifically in motion analysis
  • Learn about the implications of radius of curvature in real-world scenarios
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of motion along curves in gravitational fields.

Gavroy
Messages
232
Reaction score
0
Hi, i have a curve g:[0,t]->IR² with g(t)=(x(t),y(t)) in a homogenous gravitational field and i want to look at a ball rolling down this curve. therefore i want to derive some equations in order to calculate the normal force and the centripetal force at each point of this curve in order to see where the ball "leaves" the curve. therefore i am looking for an equation that gives me both forces just by using the parametrization of my curve. is there an equation for this?
 
Physics news on Phys.org
This might help, the radius of curvature for a parametric curve is:

radius(t) = ( (x'(t))2 + (y'(t))2 )3/2 / | (x'(t) y''(t)) - (y'(t) x''(t)) |
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
2
Views
3K