Inertia tensor of cone around its apex

In summary: 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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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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  • #2
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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  • #3
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)
 
  • #4
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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  • #5
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.
 

1. What is the inertia tensor of a cone around its apex?

The inertia tensor of a cone around its apex is a mathematical representation of the cone's resistance to rotational motion. It is a 3x3 symmetric matrix that describes how an object's mass is distributed around its rotational axis.

2. How is the inertia tensor of a cone around its apex calculated?

The inertia tensor of a cone around its apex can be calculated using the cone's mass, height, and radius. The formula for the inertia tensor is complex and involves integration of the cone's mass distribution.

3. What is the significance of the inertia tensor of a cone around its apex?

The inertia tensor of a cone around its apex is important in understanding the rotational behavior of the cone. It is used in engineering and physics to analyze the stability and dynamics of rotating cones, such as in gyroscopic applications.

4. Does the shape of the cone affect its inertia tensor around its apex?

Yes, the shape of the cone does affect its inertia tensor around its apex. The distribution of mass within the cone and its dimensions, such as height and radius, determine the values of the inertia tensor.

5. How does the inertia tensor of a cone around its apex differ from that of other shapes?

The inertia tensor of a cone around its apex is unique to its shape and cannot be directly compared to that of other shapes. However, it shares similarities with other objects that have a pointed apex, such as a pyramid or a sharp-tipped arrow.

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