Register to reply 
Deriving the moment of inertia for a sphere 
Share this thread: 
#1
Dec207, 04:44 AM

P: 47

1. The problem statement, all variables and given/known data
Derive the moment of inertia for a solid sphere with a uniform mass 2. Relevant equations [Tex]I= \sum mr^{2}[/tex] 3. 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. 


#2
Dec207, 04:45 AM

P: 47

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
Dec207, 04:51 AM

Sci Advisor
HW Helper
P: 4,300

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] 


#4
Dec207, 04:52 AM

Mentor
P: 8,317

Deriving the moment of inertia for a sphere



#5
Dec207, 03:32 PM

P: 1,874

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. 


Register to reply 
Related Discussions  
Moment of inertia of a sphere  Introductory Physics Homework  8  
Deriving Moment of Inertia  Introductory Physics Homework  2  
Moment of Inertia of a Sphere  Introductory Physics Homework  13  
Moment of inertia of a sphere  Introductory Physics Homework  2  
Moment of inertia of a sphere  Calculus & Beyond Homework  2 