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Finding the inertia of a tilted cylinder

  1. Oct 30, 2009 #1
    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.
     
  2. jcsd
  3. Oct 30, 2009 #2

    kuruman

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    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

    [tex]\hat{a}=sin\theta \hat{i}+cos \theta \hat{k}[/tex]

    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

    [tex]\vec{R}=Rcos\phi \hat{i}+Rsin \phi \hat{j}[/tex]

    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)

    [tex]r^2=R^2-(\hat{a}\cdot \vec{R})^2[/tex]

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