Finding the inertia of a tilted cylinder

[undereducated]

1. Find the inertia of a tilted cylinder about its center of mass.

2. Pretty much i want to start with the point masses and prove it from there so the equation to use would be I = mr²

3. Heres where I'm at. I figure the first thing I need is to find the inertia of a tilted ring about its center of mass which I can use to find a tilted disk and build the cylinder out of these applying the parallel axis theorem. But for life of me I can't figure out the equation for a tilted ring. Regular ring would be I = integral from 0 to Pi of ((M / 2 Pi R) * R d-theta * R². The problem with the tilt is that the radius from the axis of rotation would be variable from The radius of the ring to the radius of the ring times sin x if x is the angle of tilt. I need to find some way to relate the variable radius with theta as it traces out the ring. I don't want it all worked out for me just a hint on the right way to approach it.

 

[kuruman]

Draw axes x,y in he plane of the ring and z perpendicular to the plane of the ring. Without loss of generality, your axis of rotation can be represented by a unit vector in the xz plane as

$$\hat{a}=sin\theta \hat{i}+cos \theta \hat{k}$$

Note that the axis of rotation is tilted by angle θ with respect to the perpendicular to the ring plane.

The position vector of an element dm on the ring is

$$\vec{R}=Rcos\phi \hat{i}+Rsin \phi \hat{j}$$

To calculate the moment of inertia, you need the perpendicular distance r from dm to the axis of rotation. You get this from the Pythagorean theorem (R is the hypotenuse of the right triangle)

$$r^2=R^2-(\hat{a}\cdot \vec{R})^2$$

I will stop here and let you proceed on your own.

 

[kerimek]

I would suggest using the moment of inertia formula for a thin ring, which is I = MR^2. This formula assumes that the ring is rotating about its center of mass, which is not the case for a tilted ring. However, by applying the parallel axis theorem, we can find the moment of inertia about the tilted axis of rotation by adding the moment of inertia about the center of mass (MR^2) and the product of the mass (M) and the distance between the center of mass and the tilted axis of rotation squared (d^2). This gives us the equation I = MR^2 + Md^2.

To find the moment of inertia of a tilted disk, we can divide the disk into infinitesimally thin rings and integrate the moment of inertia formula for each ring. This will give us the equation I = integral from 0 to R of (MR^2 + Md^2) dR, where R is the radius of the disk and d is the distance between the center of mass and the tilted axis of rotation.

Finally, to find the moment of inertia of a tilted cylinder, we can use the same approach by dividing the cylinder into infinitesimally thin disks and integrating the moment of inertia formula for each disk. This will give us the equation I = integral from 0 to H of (integral from 0 to R of (MR^2 + Md^2) dR) dH, where H is the height of the cylinder and R is the radius of the disk. This equation takes into account the varying distance between the tilted axis of rotation and the center of mass as we move up the height of the cylinder.

I hope this helps guide you in the right direction. Remember to always use the correct formula for the specific shape and to apply the parallel axis theorem when needed. Good luck with your calculations!