Inertia tensor of cone around its apex

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Homework Help Overview

The discussion revolves around calculating the principal moments of inertia (Ixx, Iyy, Izz) for a cone using triple integration in cylindrical coordinates. The original poster is working with symbolic variables in MATLAB and is encountering issues with the integration results, particularly with the term for Izz.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to perform triple integrals to derive the moments of inertia and questions the correctness of their results, especially regarding the integration limits for z. Some participants suggest verifying the integration boundaries and consider performing the integrals manually.

Discussion Status

Participants are actively discussing the integration boundaries and the setup of the problem. There is a recognition of an error in the original poster's limits for z, with a suggestion to reconsider the relationship between height and radius in the context of the cone's geometry.

Contextual Notes

The original poster mentions using a textbook example that simplifies the problem by assuming equal height and radius, which may have led to confusion regarding the integration limits.

Noxuz
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Im trying to calculate the principals moments of inertia (Ixx Iyy Izz) for the inertia tensor by triple integration using cylindrical coordinates in MATLAB.

inertiatensor.png


Code:
% Symbolic variables
syms r z theta R h M; % R (Radius) h(height) M(Mass)  

% Ixx
unox = int((z^2+(r*sin(theta))^2)*r,z,r,h); % First integration
dosx = int(unox,r,0,R); % Second Integration
tresx = int(dosx,theta,0,2*pi); % Third integration

Ix = tresx / ((pi*R^2*h)/3); % Division by the volume of a cone (Mass since densitiy is 1)
Ix = Ix*M  % Setting up in terms of mass M

% Iyy
unoy = int((z^2+(r*cos(theta))^2)*r,z,r,h);
dosy = int(unoy,r,0,R);
tresy = int(dosy,theta,0,2*pi);

Iy = tresy / ((pi*R^2*h)/3);
Iy = Iy*M

% Izz
unoz = int(r^2*r,z,r,h);
dosz = int(unoz,r,0,R);
tresz = int(dosz,theta,0,2*pi);

Iz = tresz / ((pi*R^2*h)/3);
Iz = Iz*M

I plug in test values in both the formulas from the image and the code and they seem correct except from Iz in which the height (h) still appears and terms like 3h-2r instead of h appear, Am I missing something in the integrations? Thank you in advance.
 

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Did you try doing the integrals by hand? (They are not difficult.)

Please show your work in determining the integration boundaries.
 
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Orodruin said:
Did you try doing the integrals by hand? (They are not difficult.)

Please show your work in determining the integration boundaries.

In cylindrical coordinates for a solid cone the only non-trivial boundarie would be on the first integration, since its a solid of revolution, theta goes grom 0 to 2*pi and r goes from 0 to R (radius of cone).
z would go from z = r (x^2 + y^2) to z = h (height of cone)
 
Noxuz said:
z would go from z = r (x^2 + y^2) to z = h (height of cone)
This is incorrect and you have not really provided an argument for why it should be the case. I suggest you take a second look at it.
 
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Orodruin said:
This is incorrect and you have not really provided an argument for why it should be the case. I suggest you take a second look at it.

Thank you, I found the error, the inferior limit of the integral for z isn't r, its (H/R) * r, The textbook I was reading gives an example for a cone of equal heigth and radius so it ignored the slope in the integral.
 

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