Relative kinematics and radius of curvature

[springo]

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.

 

[springo]

Bump. Could someone give me hand? Thanks a lot.

 

[thomate1]

Your attempt at finding the radius of curvature using the equation 1/ρ = cos3(3t)*d2y/dx2 is correct. However, in order to find d2y/dx2, you will need to find the second derivative of the y-coordinate of the particle's position with respect to time. This can be done by taking the derivative of the velocity function you found in part a) with respect to time. So, in this case, d2y/dx2 = 16.

Substituting this value into the equation 1/ρ = cos3(3t)*d2y/dx2, we get:
1/ρ = cos3(3t)*16
Since we are looking for the radius of curvature at t = 1/2, we can plug this value into the equation and solve for ρ:
1/ρ = cos3(3(1/2))*16
1/ρ = cos(9/2)*16
ρ = 16/cos(9/2)
ρ ≈ 18.38 m

Therefore, the radius of curvature for the trajectory of the particle with respect to the observer is approximately 18.38 meters.