Why is this method not valid-moments?

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[SOLVED] Why is this method not valid-moments?

I'm trying to find the moment of inertia for a solid sphere, I've seen and understand the correct method, but I can't see anything wrong with this method, except the answer.
[tex]I=\int r^2 dm[/tex]
[tex]m=\delta V=\frac{4}{3}\delta\pi r^3[/tex]
[tex]dm=4\delta\pi r^2 dr[/tex]
Thus,
[tex]I=4\delta\pi\int r^4dr[/tex]
Which after some work, turns out to be, incorrectly,
[tex]I=\frac{3}{5}r^2m[/tex].
The correct answer is,
[tex]I=\frac{2}{5}r^2m[/tex].
Thanks in advance.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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I'm trying to find the moment of inertia for a solid sphere, I've seen and understand the correct method, but I can't see anything wrong with this method, except the answer.
[tex]I=\int r^2 dm[/tex]
[tex]m=\delta V=\frac{4}{3}\delta\pi r^3[/tex]
[tex]dm=4\delta\pi r^2 dr[/tex]
Thus,
[tex]I=4\delta\pi\int r^4dr[/tex]
Which after some work, turns out to be, incorrectly,
[tex]I=\frac{3}{5}r^2m[/tex].
The correct answer is,
[tex]I=\frac{2}{5}r^2m[/tex].
Thanks in advance.

[tex]m=\delta V=\frac{4}{3}\delta\pi r^3[/tex]

How'd you arrive at this?
 
  • #3
85
2
[tex]m=\delta V=\frac{4}{3}\delta\pi r^3[/tex]

How'd you arrive at this?
I did this by: [itex]\delta[/itex] is the density. Mass=density x volume, [itex]m=\delta V[/itex]. The volume of a sphere is: [itex]V=\frac{4}{3}\pi r^3[/itex].
So,
[tex]m=\frac{4}{3}\delta \pi r^3[/tex].
Take the derivative:
[tex]\frac{dm}{dr}=4\delta\pi r^2[/tex],
rearrange:
[tex]dm=4\delta\pi r^2 dr[/tex].
 
  • #4
2,111
16
It could be smart to denote the radius of the sphere with R, and use r for the integration variable. Using the same symbol causes confusion.

The true mistake however is that the r in the original formula is not the same r as the integration variable. The r in the original formula is the component of the [tex]\bar{r}[/tex] vector on the normal plane of the rotation axis.

Would these look good? I cannot guarantee they are right :wink:

[tex]
I = \int (r\sin(\theta))^2 dm
[/tex]

[tex]
dm = 2\pi\delta r^2\sin(\theta) d\theta\; dr
[/tex]

[tex]
0<r<R,\quad 0<\theta<\pi
[/tex]
 
  • #5
D H
Staff Emeritus
Science Advisor
Insights Author
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682
I'm trying to find the moment of inertia for a solid sphere, I've seen and understand the correct method, but I can't see anything wrong with this method, except the answer.
Here is what is wrong:
[tex]I=\int r^2 dm[/tex]
That [itex]r[/itex] is the distance of some differential quantity of mass [itex]dm[/itex] from the axis of rotation ...
[tex]dm=4\delta\pi r^2 dr[/tex]
... while that [itex]r[/itex] is the distance of some differential quantity of mass [itex]dm[/itex] from the center of the sphere.
 
  • #6
tiny-tim
Science Advisor
Homework Helper
25,832
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[tex]m=\delta V=\frac{4}{3}\delta\pi r^3[/tex]
[tex]dm=4\delta\pi r^2 dr[/tex]
Hi gamesguru! :smile:

You're integrating over spherical shells of constant radius.

But Moment of Inertia is about an axis, never about a point.

So you need to integrate over cylindrical shells of constant radius. :smile:
 
  • #7
85
2
Thanks D H, I see the problem now. And I see why using this method with a cylinder, by chance, works.
 

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