Register to reply 
Moment of inertia tensor of hollow cone 
Share this thread: 
#1
Mar1412, 02:28 AM

P: 35

1. The problem statement, all variables and given/known data
Find the inertia tensor for a uniform, thin hollow cone, such as an ice cream cone, of mass M, height h, and base radius R, spinning about its pointed end. 2. Relevant equations [itex]I_{zz} = \sum x^{2}+y^{2}[/itex] [itex]\rho = \sqrt{x^{2}+y^{2}}[/itex] 3. The attempt at a solution I first tried to think of this as a bunch of little rings [itex]area = 2\pi\rho dz[/itex] as rho is changing as height increases i defined rho as [itex]\rho=\frac{z R}{h}[/itex] but how would i set up this integral? My book shows a few examples of doing it with a solid but i don't know how to do it with an area. so far i have [itex]\int \frac{2\pi R^{3}}{h^{3}} z^{3}dz[/itex] I know that that the [itex]I_{zz}= \frac{MR^{2}}{2}[/itex] just need help getting started 


#2
Mar1412, 05:42 AM

Sci Advisor
HW Helper
Thanks
P: 26,148

Hi Lawrencel2!



#3
Apr412, 04:10 PM

P: 35

Another way to think about the moment of inertia of a conical shell is this: If the thickness of the shell is infinitesimally small then the rings do not overlap. This is just like splitting a solid disc up unto little rings. As Lawrencel2 mentioned, the moment of inertia of the hollow cone is MR^2/2, which is also that of a solid disc.
Now here's my problem. How does one do this when the walls of the cone are not thin (say inner radius b and outer radius a)? First, I'd like to explain how I did this for a hollow sphere because I think the same approach should work here. My approach for the sphere was to change the limits for the radius to be from b to a instead of from 0 to R. When the integration is said and done you end up with [itex]I_{zz}[/itex]=[itex]\frac{2M(a^{5}b^{5})}{5(a^{3}b^{3})}[/itex] This can be factored to give [itex]I_{zz}[/itex]=[itex]\frac{2M(a^{4}+a^{3}b+a^{2}b^{2}+a b^{3}+b^{4})}{(a^{2}+ab+b^{2})}[/itex] If you then let the inner radius b equal the outer radius a you will get the familiar result that a thin hollow sphere has MoI equal to 2/3MR^{2} Now, the same approach for the hollow cone. Its the same set up for a solid cone Integrate[[itex]r^{3} dr d[/itex][itex]\phi[/itex] dz] using cylindrical polar coordinates, but now the limits on r will be from bz/h to az/h. I end up with [itex]\frac{1}{10}[/itex] [itex]\rho[/itex] [itex]π h(a^{4}b^{4})[/itex] Where [itex]\rho[/itex] is the density and h is the height. Plugging in M/V for the density I get (the volume of a conical shell is V= [itex]\frac{π h(a^{3}b^{3})}{(3a}[/itex] [itex]\frac{3 M a}{(a^{3}b^{3} π h}[/itex] [itex]\frac{1}{10}[/itex] [itex]\rho[/itex] π h[itex](a^{4}b^{4})[/itex] or [itex]\frac{2Ma(a^{3}+a^{2}b+a b^{2}+a b^{3})}{(a^{2}+ab+b^{2})}[/itex] By the way, how the heck does this volume formula V= [itex]\frac{π h(a^{3}b^{3})}{(3a}[/itex] for a cone come about?! I looked it up and I'm puzzled by it. This answer is not correct and does not reduce to the thin conical shell The correct expression for the MoI (assuming Wolfram Alpha is correct) is [itex]\frac{3(a^{5}b^{5})}{(10(a^{3}b^{3})}[/itex] Which does reduce to the correct thin walled answer. Can anyone point out where my mistake(s) are? 


#4
Apr812, 05:19 PM

P: 35

Moment of inertia tensor of hollow cone
Another question: how can I correctly set up the other moment of inertia integrals [itex]I_{xx}[/itex] and [itex]I_{yy}[/itex] (they will be the same because of the symmetry) for a hollow cone? I have no problem doing this if the cone is solid, but I'm having trouble figuring it out for a hollow one.
Thanks in advance! 


Register to reply 
Related Discussions  
Inertia Tensor  Rotating Cone  Advanced Physics Homework  1  
Inertia Tensor of a Hollow Sphere and of a Slender Rod  Introductory Physics Homework  6  
Magnetic moment of uniformly charged hollow cone  Introductory Physics Homework  0  
Moment of inertia of a cone  Introductory Physics Homework  6  
Moment of Inertia of (Right) Cone  Calculus & Beyond Homework  4 