Relative kinematics and radius of curvature

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SUMMARY

The discussion focuses on calculating the radius of curvature for a particle's trajectory in a rotating disk system. The particle's position is defined by the equation r = 8t² j, while the disk rotates around the Z-axis with a constant angular speed of ω = 3 k. The user successfully derived the velocity and acceleration of the particle relative to an observer on the disk, yielding v = 6 i + 8 j and a = 48 i - 2 j at t = 1/2. The user seeks clarification on how to compute the second derivative of y with respect to x to find the radius of curvature.

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Homework Statement


(Only need help for b) I think but I'll post the whole problem)
(All values are SI units)
(O; i, j, k) orthonormal basis.
A particle moves following this law: r = 8t2 j
A disk with radius 2 in the plane XOY rotates around Z with constant angular speed: ω = 3 k.
At the time the particle is 2m away from the origin, find:
a) Velocity and acceleration for the particle with respect to an observer placed at the edge of the disk, linked to it.
b) Radius of curvature for the trajectory of the particle with respect to the observer.

Homework Equations


I got rid of a) I think:
For O being the observer and P the particle I found that:
OP = 2 [ cos(3t) i + (4t2 - sin(3t)) j ]
v = 24t2 i + 16t j
a = 96t i + (16 - 72t2) j
t = 1/2 therefore:
v = 6 i + 8 j
a = 48 i - 2j

The Attempt at a Solution


For b) I think I should use:
1/ρ = cos3(3t)*d2y/dx2
But how do I find d2y/dx2?
Or if this is not how it's done please correct me.

Thanks a lot for your help.

PS: The problem is originally written in Spanish so please bear with me if I mistranslated any word.
 
Last edited:
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Bump. Could someone give me hand? Thanks a lot.
 

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