Moment of inertia of solid sphere

  • Thread starter arunbg
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Sometime back, I had quite some fun trying to derive the expressions for MOI of different rigid bodies using calculus.
However, I had some trouble with the MOI of a solid sphere about an axis through its centre. The problem being that my final answer was markedly different from the original expression ie 2/5MR^2 ,if my memory is correct.
My answer ( it contained terms like pi) was somewhat close to the original answer but not close enough (at least for me).
So is the original expression just an approximation or do I have to recheck my work ?
Also can someone tell me what assumptions are to be made when calculating the MOI of a " thin rod " ?
 

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  • #2
arildno
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A "thin rod" has only length, neither breadth or width.

The formula given for the ball is correct; what did you use as your differential volume element when calculating the momentie of inertia?
 
  • #3
Hootenanny
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No, pi should disappear when you substitute in the density of the sphere. If you like, you could show your working and I will try to pick out where you have made a mistake. The moment of inertia of a sphere is exactly [itex]\frac{2}{5}Mr^2[/itex]

~H

Edit: Sorry arildno, didn't see your post
 
  • #4
arildno
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How impertinent of you, Hootenanny!

However, I can excuse you based on previous good behaviour.:smile:
 
  • #5
Hootenanny
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arildno said:
How impertinent of you, Hootenanny!

However, I can excuse you based on previous good behaviour.:smile:
Just because you can type faster than me...

~H
 
  • #6
arildno
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It isn't that fun with so much manual skill&practice. :frown:
 
  • #7
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Well, My approach was to consider the MOI of a diff. cross section of the sphere ( a disc) whose moment of inertia I had already derived and integrate from the centre to the top as radius decreases from say r to 0.
I am sorry , I can't post my working in this case, as it would probably take me over a year to do it in latex. I can tell you that it involved eventually the pretty lengthy definite integral of sin(x)^4 .Maybe I just have to recheck my work.

Is this the usual procedure (worked for me with all previous cases) or is there an easier method using perhaps spherical coordinates or the sort ?

And for the thin rod, you should assume linear mass distribution only, right ?
Edit :Okay, I got the answer for the rod, pretty simple with the assumptions in place
Great , someone moved the thread from the general physics section.
This is not coursework, it was something that I tried independently for my own amusement.
The purpose of the thread was to verify whether there was any approximation in the MOI of a solid sphere. Does that sound like coursework for a 12th grader ?
 
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  • #8
Hootenanny
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arunbg said:
Well, My approach was to consider the MOI of a diff. cross section of the sphere ( a disc) whose moment of inertia I had already derived and integrate from the centre to the top as radius decreases from say r to 0.
I am sorry , I can't post my working in this case, as it would probably take me over a year to do it in latex. I can tell you that it involved eventually the pretty lengthy definite integral of sin(x)^4 .Maybe I just have to recheck my work.

Is this the usual procedure (worked for me with all previous cases) or is there an easier method using perhaps spherical coordinates or the sort ?
The derivation I know does not contain any trigonometrical functions. You are correct in your approach using a thin disk(radius but no volume), however, I have noticed that you are integrating from 0 to r. This only gives you the moment of inertia of a hemisphere. You must integrate from r to -r. As you have stated that this is not a homework question I have no quarms about showing you a (fairly) complete solution.

The moment of inertia of a thin disc of radius y from the centre of rotation is given by [itex]\frac{1}{2}my^2[/itex]. The change is moment of inertia is due to the chnage in mass;

[tex]dI = \frac{1}{2}y^2 dm[/tex]

Writing change in mass in terms of density ([itex]\rho[/itex]) and volume (V);

[tex]dI = \frac{1}{2}y^2 \rho dV[/tex]

The volume of the disk (V) is given by the product of the area ([itex]\pi y^2[/itex]) and the height of the disk ([itex]dz[/itex]) (Z is the axis fo rotation), thus giving;

[tex]dI = \frac{1}{2}y^2 \rho\pi y^{2} dz = \frac{1}{2}\rho\pi y^{4} dz[/tex]

Writing y interms of the radius and the axis of rotation (z) (using pythag for a triangle between the radius of the disk, the axis of rotation and the radius of the sphere being the hyp);

[tex]y =\sqrt{r^{2} - z^{2}[/tex]

[tex]dI = \frac{1}{2}\rho\pi (r^{2} - z^{2})^{2} dz[/tex]

To find I we must integrate between the limits r and -r;

[tex]I = \int^{r}_{-r} \frac{1}{2}\rho\pi (r^{2} - z^{2})^{2} dz[/tex]

Evaluating this integral and substituing the density of a sphere into the resultant equation should give you the moment of inertia of a sphere. Feel free to point out any errors in my working as I'm sure there are a few.

~H
 
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  • #9
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Well hoot, I haven't solved your final integral but I guess it must be right.
But my method is a bit different.
I used the MOI of a disc allright, but for the part corresponding to z in your proof, I took an an angle dz which the upper and lower parts of the disc subtend at the centre. Then I found the thickness of cross section using the arc length formula and muliplied with the area of the "circle" to get volume.
I thought this method should also give a similar result, since as dz is very small, the cross section is almost equivalent to that of a disc, and at the time, thought this would give a better approximation as compared to the simple disc approach. Apparently it doesn't (though it's close).

And yeah, in my earlier post I forgot to mention that after integrating from 0 to r ,I would multiply by 2 owing to the symmetry of the figure, which is same as integrating from -r to +r .
And Hootenanny, thanks for your time .

Arun
 
  • #10
Hootenanny
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I think I have spotted where the two proofs diverge. You say that you multiply the arc length with the area of a circle, however, as it is an arc the cross-sectional area will not be constant throughout. Unless, I missunderstand you.

arunbg said:
And Hootenanny, thanks for your time .
No problem. I enjoy working through proofs anyway :biggrin:

~H
 
  • #11
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You say that you multiply the arc length with the area of a circle, however, as it is an arc the cross-sectional area will not be constant throughout.
It might be constant if the thickness of the cross section is infinitesimally small as I have assumed in taking r(dz).

Arun
 

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