Inertia tensor of cone around its apex

  • #1
Noxuz
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Thread moved from the technical forums, so no Homework Template is shown
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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Answers and Replies

  • #2
Orodruin
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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.
 
  • #3
Noxuz
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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.

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)
 
  • #4
Orodruin
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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.
 
  • #5
Noxuz
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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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