p0250

# p0250 - x 2 ) 3 / 2 ² = ∓ R 2 ( R 2 − x 2 ) 3 / 2 So,...

This preview shows page 1. Sign up to view the full content.

134 CHAPTER 2. PARTICLE KINEMATICS 2.50 Chapter 2, Problem 50 Problem: A particle moves along a circular path, x 2 + y 2 = R 2 ,whe re R is a radius of the circle. Using Cartesian coordinates, compute the radius of curvature of the particle’s path. Solution: By definition, the radius of curvature is ρ = ± 1+( dy/dx ) 2 = 3 / 2 d 2 y/dx 2 The y coordinate on the circumference of a circle with radius R is y = ± 0 R 2 x 2 Thus, the first and second derivatives of y with respect to x are as follows. dy dx = ± 1 2 D R 2 x 2 i 1 / 2 ( 2 x )= x R 2 x 2 d 2 y dx 2 = ^ 1 R 2 x 2 1 2 x ( R 2 x 2 ) 3 / 2 ( 2 x ) ² = ^ R 2 x 2 ( R 2 x 2 ) 3 / 2 + x 2 ( R 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: x 2 ) 3 / 2 ² = ∓ R 2 ( R 2 − x 2 ) 3 / 2 So, the radius of curvature is ρ = 1 + x 2 / D R 2 − x 2 i 3 / 2 ∓ R 2 / ( R 2 − x 2 ) 3 / 2 = ∓ R 3 / D R 2 − x 2 i 3 / 2 R 2 / ( R 2 − x 2 ) 3 / 2 = ∓ R Note that ρ is negative on the upper half of the circle, which has concave curvature relative to the x axis. By contrast, ρ is positive on the lower part of the circle, which has convex curvature relative to the x axis....
View Full Document

## This note was uploaded on 10/15/2010 for the course AME 301 at USC.

Ask a homework question - tutors are online