Calculating Moment of Inertia for a Curved Rod with Respect to a Specific Axis

[Iqish]

Homework Statement

A thin, homogeneous, curved rod with
the radius of curvature 𝑅 and mass 𝑚 are in
𝑥𝑦 plane. Determine the moment of inertia,
𝐼𝑥'𝑥 ', with respect to the 𝑥 'axis, which goes
through the body's center of mass G

Relevant Equations

Steiners sats:𝐼𝑥𝑥 = 𝐼𝑥′𝑥′ + 𝑚𝑦^2𝐺

Given:Thin, homogeneous, curved rod with radius of curvature 𝑅 See figure to the down.

Find: The moment of inertia 𝐼𝑥′𝑥 ′ with respect to 𝑥′- the axis passing through the center of mass (point 𝐺).
Can someone who can help me ?

 

[Master1022]

Homework Statement:: A thin, homogeneous, curved rod with
the radius of curvature 𝑅 and mass 𝑚 are in
𝑥𝑦 plane. Determine the moment of inertia,
𝐼𝑥'𝑥 ', with respect to the 𝑥 'axis, which goes
through the body's center of mass G
Relevant Equations:: Steiners sats:𝐼𝑥𝑥 = 𝐼𝑥′𝑥′ + 𝑚𝑦^2𝐺

Given:Thin, homogeneous, curved rod with radius of curvature 𝑅 See figure to the down.
View attachment 258878
Find: The moment of inertia 𝐼𝑥′𝑥 ′ with respect to 𝑥′- the axis passing through the center of mass (point 𝐺).
Can someone who can help me ?

Have you learned about the parallel and perpendicular axis theorems?

Just an initial thought that I have on this. So it looks quite difficult to actually calculate the moment of inertia about the required axis, so it suggests to me that we will be shifting it to that axis (via parallel axis theorem). Perhaps, we can find I_{Oz} (out of the page) , turn that into I_{x} via perpendicular axis theorem and then shift it to the required axis (would require us to know the location of G, but that seems easier to find than calculating about that axis. Otherwise, maybe we find I_{x} and then shift?

(Note, we would be 'reverse' shifting it as the parallel axis theorem is defined for an axis relative to centroid axis, would just mean that we subtract the m d^2, rather than add it)