Moment of inertia and axis of symmetry

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  • #1
bon
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Homework Statement



Derive an expression for the moment of inertia about the axis of symmetry for
a cylinder of mass M , length L and radius a, where the mass density decreases as a
function of distance from the axis as 1/r

Homework Equations





The Attempt at a Solution



1) am i right in thinking this would just be the same as the MOI of a disk of same mass?

and is the answer 2pi/3 a^3 or alternatively 1/3 M/L a^3
 

Answers and Replies

  • #2
656
2

Homework Statement



Derive an expression for the moment of inertia about the axis of symmetry for
a cylinder of mass M , length L and radius a, where the mass density decreases as a
function of distance from the axis as 1/r

Homework Equations





The Attempt at a Solution



1) am i right in thinking this would just be the same as the MOI of a disk of same mass?

and is the answer 2pi/3 a^3 or alternatively 1/3 M/L a^3

No. If most of the mass is closer to the axis of rotation the moment of inertia would be less than for mass evenly distributed throughout assuming the mass and radius of both cylinders are the same.

Could you show your work? The moment of inertia must come out to units of kg*m^2, so I was wondering about the first answer...
 
Last edited:
  • #3
bon
559
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No. If most of the mass is closer to the axis of rotation the moment of inertia would be less than for mass evenly distributed throughout assuming the mass and radius of both cylinders are the same.

Could you show your work?

Ok so you can just do it for a disk..as the MOI of the cylinder is just the same as a disk. Look:

I = integral of r^2 dm

=integral of r^2 density dx dy

density = 1/r

so = integral of r rdrdtheta (dxdy = rdrdtheta)

If you compute the double integral you get my answer..
 
  • #4
656
2
Ok so you can just do it for a disk..as the MOI of the cylinder is just the same as a disk. Look:

I = integral of r^2 dm

=integral of r^2 density dx dy

density = 1/r

so = integral of r rdrdtheta (dxdy = rdrdtheta)

If you compute the double integral you get my answer..

I got your second answer using sigma = sigma o *a/r basically the same thing.
dm = sigma o*a/r dA... dA = 2pi*rdr... works the same.

And if the math is right, it makes sense as 1/3Ma^2 is smaller than a uniform cylinder 1/2Ma^2

ohh and I left out the L so the a^3 on top would be more correct given the question.
 
Last edited:

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