We continue our journey with moments of inertia by calculating the I for a solid cone about two different axes i.e through the axis that goes through the center of mass and the axis that goes perpendicular to its pointy edge.
5) Axis through the center of mass
\begin{align}
I_z &= \rho \left( \int r^2\, dV \right)\\
&= \rho \left( \int r^3 \, d\phi dr dz \right)\\
&= \rho 2\pi \left(\int_0^{\frac{zR}{L}} r^3 \, dr \,\int_0^ L\, dz\right)\\
&= \frac{\pi \rho R^4 L }{10} = \frac{3MR^2}{10}
\end{align}
Axis that is perpendicular to the point edge.
Now this interesting, and it worth slowing down the pace and considering. This problem can be done two ways: One is to realize that if we put in the cone upside(not necessary) then this axis is \(I_{yy}\) in the moment of inertia tensor; now calculate away the integrals. But supposing you know nothing about the inertial tensor, is there hope? Indeed, but it requires a little foray into the perpendicular axis theorem.
Let's suppose you have a two dimensional figure(located in the xy place). Then the perpendicular axis theorem says \(I_z\) = \(I_x + I_y\). The proof is quite simple and goes as follows:
\begin{align}
I_z &= \left(\int r^2 dm \right) = \left( \int x^2 + y^2 dm\right)\\
&= \left(\int x^2 dm\right) + \left( \int y^2 dm\right) = I_y + I_x \hspace{20mm}QED\\
\end{align}
Its most common application is for example finding \(I_x\) or \(I_y\) of a disk. Since we know \(I_z\) because it is easy to calculate, then finding \(I_x\) or \(I_y\) is trivial. Again you do not need integrals, the theorem saves you.
This theorem suggest another way of calculating our desired results. Here is the strategy:
1) There is symmetry in the x- y plane so \(I_x\) and \(I_y\) are the same and we calculated \(I_z\) in the problem above(axis through center of mass)
2) Starting from the perpendicular axis theorem we can add \(I_z\) to both sides to get that \(2I = I_x+I_y+I_z\). Since we know \(I_x = I_y\) then \( 2I_y + I_z = 2I\). Now all we need to do is calculate 2I. This is a generalization of the perpendicular axis theorem.
\begin{align}
2I &= 2\rho \left( \int (r^2 \cos^2 \phi + r^2 \sin^2\phi + z^2 ) r dr d\phi dz \right)\\
& = 4\pi \rho \left(\int_0^L dz \int_0^{\frac{zR}{L}}dr(r^3 + z^2r) \right)\\
& = 4\pi \rho \left( \frac{R^4L}{20}+ \frac{L^3R^2}{10}\right)\\
& = 12M\left( \frac{R^2}{20} + \frac{L^2}{10} \right)\\
& \text{plugging into previous results for } I_z \text{ we get } \frac{3M}{5}\left( \frac{R^2}{4} + L^2\right)
\end{align}
This theorem is very useful if you are in three dimensions and have symmetry in two or more axes. For example take the moment of inertia for a spherical shell, \( I_z, I_y \) and \(I_x\) are all the same so we have that \( 3I_z = 2 I\) and I is easy to calculate. "I"here is the moment of inertia about an axis in the \( 4^{th}\) dimension!!! Poooffff ( Your mind being blown away)
5) Axis through the center of mass
\begin{align}
I_z &= \rho \left( \int r^2\, dV \right)\\
&= \rho \left( \int r^3 \, d\phi dr dz \right)\\
&= \rho 2\pi \left(\int_0^{\frac{zR}{L}} r^3 \, dr \,\int_0^ L\, dz\right)\\
&= \frac{\pi \rho R^4 L }{10} = \frac{3MR^2}{10}
\end{align}
Axis that is perpendicular to the point edge.
Now this interesting, and it worth slowing down the pace and considering. This problem can be done two ways: One is to realize that if we put in the cone upside(not necessary) then this axis is \(I_{yy}\) in the moment of inertia tensor; now calculate away the integrals. But supposing you know nothing about the inertial tensor, is there hope? Indeed, but it requires a little foray into the perpendicular axis theorem.
Let's suppose you have a two dimensional figure(located in the xy place). Then the perpendicular axis theorem says \(I_z\) = \(I_x + I_y\). The proof is quite simple and goes as follows:
\begin{align}
I_z &= \left(\int r^2 dm \right) = \left( \int x^2 + y^2 dm\right)\\
&= \left(\int x^2 dm\right) + \left( \int y^2 dm\right) = I_y + I_x \hspace{20mm}QED\\
\end{align}
Its most common application is for example finding \(I_x\) or \(I_y\) of a disk. Since we know \(I_z\) because it is easy to calculate, then finding \(I_x\) or \(I_y\) is trivial. Again you do not need integrals, the theorem saves you.
This theorem suggest another way of calculating our desired results. Here is the strategy:
1) There is symmetry in the x- y plane so \(I_x\) and \(I_y\) are the same and we calculated \(I_z\) in the problem above(axis through center of mass)
2) Starting from the perpendicular axis theorem we can add \(I_z\) to both sides to get that \(2I = I_x+I_y+I_z\). Since we know \(I_x = I_y\) then \( 2I_y + I_z = 2I\). Now all we need to do is calculate 2I. This is a generalization of the perpendicular axis theorem.
\begin{align}
2I &= 2\rho \left( \int (r^2 \cos^2 \phi + r^2 \sin^2\phi + z^2 ) r dr d\phi dz \right)\\
& = 4\pi \rho \left(\int_0^L dz \int_0^{\frac{zR}{L}}dr(r^3 + z^2r) \right)\\
& = 4\pi \rho \left( \frac{R^4L}{20}+ \frac{L^3R^2}{10}\right)\\
& = 12M\left( \frac{R^2}{20} + \frac{L^2}{10} \right)\\
& \text{plugging into previous results for } I_z \text{ we get } \frac{3M}{5}\left( \frac{R^2}{4} + L^2\right)
\end{align}
This theorem is very useful if you are in three dimensions and have symmetry in two or more axes. For example take the moment of inertia for a spherical shell, \( I_z, I_y \) and \(I_x\) are all the same so we have that \( 3I_z = 2 I\) and I is easy to calculate. "I"here is the moment of inertia about an axis in the \( 4^{th}\) dimension!!! Poooffff ( Your mind being blown away)
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