# Homework Help: Moment of Inertia of a Conical shell

1. Dec 2, 2011

### garis

1. The problem statement, all variables and given/known data

Given that the conical shell has uniform density and thickness is made of one sheet, has mass M, height h, and base radius R, derive the moment of inertia about its axis of symmetry.

2. Relevant equations
I = MR^2 for a ring about its central axis.
I = ∫dI - my approach

3. The attempt at a solution

For a conical shell, I am equating the change in the moment of inertia as a series of rings, where dI = dm(R)^2. R = y, and since y/z = r/h, y = (rz)/h = R.

dm = σdA = M(2∏(rz)/h)dz/(∏r^2 + √(r^2+h^2))

Therefore I = ∫[[M(2∏(rz)/h)[(rz)/h]^2]dz]/(∏r^2 + √(r^2+h^2)).

I know that is not the correct dm, as it is not including the base, while the base is included within the density function. I am able to derive the moment of inertia for a conical shell with an open base just fine (just take out the ∏r^2 from the denominator in the integral), and have found it to be [MR^2]/√8. I am aware that the moment of inertia for a conical shell is [MR^2]/2, yet I do not know quite how to approach this. It would be easier if I were able to set up a density function, and utilize a triple integral for Izz, but I had issues with that approach as well. How else could I approach this?

2. Dec 2, 2011

### garis

Should I instead calculate the moment of inertia for a solid cone, but without merging the volume and the integral; instead taking the derivative of the result of the integral with respect to R and the volume with respect to R separately, then combining them and simplifying? Such an approach worked for finding the moment of inertia of a spherical shell.

3. Dec 2, 2011

### garis

I have worked it out to be (3/5)MR^2. Is that correct? I ask because I am now questioning whether I = 0.5MR^2 for a conical shell is correct; because I found that formula from an old calc book, that was asking to prove that is the inertia when you are given a density function. It occurs to me now that such an inertia may vary depending on whether the density is uniform or not. Since it involved a non-constant density function, it wasn't uniform. My mistake. I think I see where I went wrong now. I must admit, it has been over a year since I took calc3. DiffyQ certainly didn't reverse the atrophy of my skills of deriving moments of inertia. Anyway, if anyone has consistently calculated a different result for the moment of inertia, I'd be happy to discuss your approach.

Last edited: Dec 2, 2011
4. Jan 11, 2012

### SamSpeight

Hey,
I have recently come to this problem myself. After finding the moment of inertia for a spherical shell of radius R by starting with a solid sphere of radius R and taking out a solid sphere of radius R-r and them taking the limit r tends to 0, I thought I would try the same approach with this problem (finding the moment of inertia of a conical shell).
The result I got was the same as you stated: (3/5)MR^2
I also was led to believe that the correct result was (1/2)MR^2 with a constant density function. I have been going over my working to try and spot something not quite right, but can't seem to do so. I've also thought about the concept of the method and can't seem to spot why this wouldn't be valid. With the conical shell I started with a solid cone of radius R and height h and took out a smaller cone of radius R-r and h-x, and again let the values r and x tend to 0.
One thought that has crossed my mind is (something you mentioned) whether this would be a conical shell with a base or without a base (and if the value we seem to have both been led to believe is for a conical shell with or without a base). If our value of (3/5)MR^2 is without a base and the value (1/2)MR^2 is with a base, would it not just be the case of adding on the moment of inertia for a disc (as the moment of inertia is defined as the sum of [m_i]*[r_i]^2)? However, their masses would be different so it would not just be the case of adding (3/5)MR^2 to (1/2)MR^2 (the latter being I for a uniform disc about an axis through the centre). I will go ahead and try to work this idea through and see if it comes out as (1/2)MR^2.
Look forward to hearing your ideas on this and ideas on any other methods you may have.
Sam

5. Jan 11, 2012

### SamSpeight

Thing is, (3/5)MR^2 > (1/2)MR^2, so whatever the mass of the disc, surely the total moment of inertia can only get greater than (3/5)MR^2 and never be reduced to (1/2)MR^2?