Deriving the moment of inertia for a sphere

In summary: As for your work, I can't tell what you are doing or what the notation means, so I can't tell you what you are doing wrong. But you have to get the integral in the right form, and I don't think you have. Your integral is over the hemisphere of some radius, and that is not right.I think the problem is that you're using the wrong formula for the moment of inertia. The correct formula isI = \int r^2 dmwhere r is the distance from the axis of rotation and dm is the mass element. In polar coordinates, this becomesI = \int\int\int r^2 \rho r^2 dVwhere r^2 = x
  • #1
amolv06
46
0

Homework Statement



Derive the moment of inertia for a solid sphere with a uniform mass

Homework Equations



[Tex]I= \sum mr^{2}[/tex]

The Attempt at a Solution



I decided to change everything to polar coordinates. Since the polar coordinate substitution is

[tex]\int\int\int_{v} Fr^{2}sin(\phi)drd\phi d\theta[/tex]

I figured that all you should do is plug in the moment of inertia equation into the integral giving you:[tex]m\int^{2\pi}_{0}\int^{\pi}_{0}\int^{R}_{0} r^{4}sin(\phi)drd\phid\theta =\frac{4\pi R^{5}}{5}[/tex]

however this does not seem to be the correct answer. Can anyone tell me what I'm doing wrong? Thanks in advance for your time and any help.
 
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  • #2
For some reason the images are coming up in the wrong places, and one image is actually missing. I don't know how to fix that. Sorry.
 
  • #3
The last image is probably not showing because it has a syntax error: there is something like
d\phid\theta
in it, and LaTeX doesn't know the symbol \phid. Insert a space and it will work.
In the first formula you quoted, what is this F and where does the sine come from? Also, did you put in the mass density
[tex]\rho = \frac{m}{V} = \frac{m}{\tfrac43 \pi R^3} [/tex]
correctly (I think not, which would account for the R^5 instead of R^2 as you should have found).

[edit]I think I see where the [itex]r^2 \sin\theta[/itex] comes from, is that the square of the shortest distance from a point on the sphere to some axis? I calculate that distance to be r sin(theta), so in the formula there is also a square missing probably, and the integration there is over the volume dV, the dr dtheta dphi only come in when you switch to polar coordinates[/edit]
 
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  • #4
amolv06 said:

Homework Statement



Derive the moment of inertia for a solid sphere with a uniform mass



Homework Equations



[tex]I= \sum mr^{2}[/tex]



The Attempt at a Solution



I decided to change everything to polar coordinates. Since the polar coordinate substitution is

[tex]\int\int\int_{v} Fr^{2}sin(\phi)drd\phi d\theta[/tex]

I figured that all you should do is plug in the moment of inertia equation into the integral giving you:


[tex]m\int^{2\pi}_{0}\int^{\pi}_{0}\int^{R}_{0} r^{4}sin(\phi)drd\phi d\theta =\frac{4\pi R^{5}}{5}[/tex]

however this does not seem to be the correct answer. Can anyone tell me what I'm doing wrong? Thanks in advance for your time and any help.

I've corrected your tex to make it easy to read. Your problem was that that the first tags were [ Tex] [ /tex] (plus the problems that compuchip mentioned). See the quote.
 
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  • #5
I too am not so sure about what you are doing. If you use an inertia tensor the formula goes as

[tex]I_{ij} = \int_V \rho(\mathbf{r}) (\delta_{ij} \sum_k x_k^2 -x_i x_j) dv[/tex]

so, for example, the first element would be

[tex]I_{11} = \int_V \frac{M}{4/3 \pi R^3} (x^2 + y^2 + z^2 - x^2) dv[/tex]

where you can use coordinate and jacobian (for the volume differential) conversions to get it into spherical form (which I think is what you meant because you certainly didn't show a polar jacobian). From symmetry all the off diagonals will be zero, and all the diagonals will be the same.
 

1. What is the formula for calculating the moment of inertia for a sphere?

The formula for calculating the moment of inertia for a sphere is given by I = (2/5) * mr2, where m is the mass of the sphere and r is the radius.

2. How is the moment of inertia for a sphere derived?

The moment of inertia for a sphere can be derived by considering the rotational motion of the sphere and applying the basic principles of calculus and physics.

3. What factors affect the moment of inertia for a sphere?

The moment of inertia for a sphere is affected by the mass and radius of the sphere. A larger mass or radius will result in a larger moment of inertia.

4. Can the moment of inertia for a sphere be negative?

No, the moment of inertia for a sphere cannot be negative. It is always a positive value as it represents the resistance of the sphere to changes in its rotational motion.

5. How is the moment of inertia for a hollow sphere different from that of a solid sphere?

The moment of inertia for a hollow sphere is smaller than that of a solid sphere with the same mass and radius. This is because the mass of a hollow sphere is further from the axis of rotation, resulting in a smaller moment of inertia.

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