Engr Help

Center of Curvature (Locating center of curvature ˗ example continued)

At t = 1 sec,

v = ˗ 3 i ˗ j so v = √ 10

a = d²r_p/dt² = 2 j

and a_n = a • e_n = 2 j • ([ ˗ i + 3 j ] /√10 = 6 / √ 10

Use a_n = v² / ρ or ρ = v² / a_n

So ρ = 10 / (6/ √ 10) = (5/3) √ 10 ft (radius of curvature at t = 1 sec)

And at t = 1 sec,

r_c = r_p + ρ e_n = ˗ 3i ˗ 4 j + (5/3) √ 10 [ ˗ i + 3 j ] /√10 = i ( ˗ 3 ˗ 5/3) + j (˗ 4 + 5 )

r_c = ˗ 4.66666 i + j ft (result for vector location of the center of curvature)

Note: As a check for plane motion, the curvature of the path can be calculated using

the formula for curvature from calculus:

1/ ρ = | {(dx/dt)(d²y/dt²) ˗ d²x/dt²)(dy/dt)} / { ( dx/dt)² + dy/dt)² } ^3/2 |

This equation does not apply for motion in three dimensions.

Here x = ˗ 3 t and y = t² ˗ 3 t ˗ 2

Which yields the same result that ρ = (5/3) √ 10

Also note that for a space curve the radius of curvature should be calculated using ρ = v² / a_n .