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Calculating the Inertia Tensor of a Homogeneous Sphere

  • #1

Homework Statement



Calculate the moments of Inertia I[itex]_{1}[/itex], I[itex]_{2}[/itex], I[itex]_{3}[/itex] for a homogenous sphere

Homework Equations



I[itex]_{jk}[/itex]=[itex]\int[/itex]x[itex]^{2}_{l}[/itex][itex]\delta[/itex][itex]_{ik}[/itex]-x[itex]_{i}[/itex]x[itex]_{k}[/itex]dV


The Attempt at a Solution



For I[itex]_{x}[/itex] i set up the equation using the above equation in cartesian coordinates and then i switched into polar coordinates and i get the following integral
[itex]\rho[/itex][itex]\int[/itex][itex]\int[/itex][itex]\int[/itex](r[itex]^{2}[/itex]-rsin([itex]\vartheta[/itex])cos([itex]\phi[/itex]))r[itex]^{2}[/itex]sin([itex]\vartheta[/itex])d[itex]\phi[/itex]d[itex]\vartheta[/itex]dr
with 0[itex]\leq[/itex]r[itex]\leq[/itex]R, 0[itex]\leq[/itex][itex]\phi[/itex][itex]\leq[/itex]2[itex]\pi[/itex], and 0[itex]\leq[/itex][itex]\vartheta[/itex][itex]\leq[/itex][itex]pi[/itex]

when i solve this integral i get I=[itex]\rho[/itex][itex]\frac{4}{5}[/itex][itex]\pi[/itex]R[itex]^{2}[/itex] and then setting [itex]\rho[/itex]= [itex]\frac{M}{4/3\piR^{3}
}[/itex]
so after simplifying i end up with I=[itex]\frac{3}{5}[/itex]MR[itex]^{2}[/itex]
But the answer i believe is [itex]\frac{2}{5}[/itex]MR[itex]^{2}[/itex] , so i dont really know where is went wrong.
 

Answers and Replies

  • #2
938
9
How did you end up with that integral? What are i an k in your expression? Remember that the principal moments of inertia are the diagonal terms in Iij. Start by making the easiest choice for i=j and then think if you have to calculate more.
 
  • #3
as i started to type i response i realized my integrand was essentially r[itex]^{2}[/itex]-x. when i should have been r[itex]^{2}[/itex]-x[itex]^{2}[/itex]. So what i should have is r[itex]^{4}[/itex]-r[itex]^{4}[/itex]sin([itex]\vartheta[/itex])[itex]^{3}[/itex]cos[itex]\phi[/itex][itex]^{2}[/itex]

If i integrate this then i get 2/5[itex]\pi[/itex][itex]^{2}[/itex]R[itex]^{5}[/itex]. which looks better.
 
  • #4
although if i then multiply this by rho i get 3/10[itex]\pi[/itex]MR[itex]^{2}[/itex]...
 
  • #5
938
9
as i started to type i response i realized my integrand was essentially r[itex]^{2}[/itex]-x. when i should have been r[itex]^{2}[/itex]-x[itex]^{2}[/itex]. So what i should have is r[itex]^{4}[/itex]-r[itex]^{4}[/itex]sin([itex]\vartheta[/itex])[itex]^{3}[/itex]cos[itex]\phi[/itex][itex]^{2}[/itex]

If i integrate this then i get 2/5[itex]\pi[/itex][itex]^{2}[/itex]R[itex]^{5}[/itex]. which looks better.
That's not right either, one sin(θ) should be in the integration measure. Perhaps it would be easier to calculate only the zz-component of I. Why is this enough?
 

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