Chapter 10, Problem 10.27P

To determine

The inertia tensor for a uniform, thin hollow cone such as an ice cream cone of mass $M$ height $h$ and $R$ radius.

Answer to Problem 10.27P

The moment of inertia tensor of the hollow cone which spins about its pointed end is $\underset{\_}{\frac{1}{4}M\left[\begin{array}{ccc}\left(R^2+2h^2\right)& 0& 0\\ 0& \left(R^2+2h^2\right)& 0\\ 0& 0& 2R^2\end{array}\right]}$.

Explanation of Solution

Write the moment of inertia of a body about an axis.

    $I={\displaystyle {\int }_{}^{}{r}^{2}dm}$        (I)

Here, $I$ is the moment of inertia, $r$ is the distance of the mass distribution from the axis of rotation.

The moment of inertia of a body about an axis depends on the mass and the distance of the mass distribution from the axis of rotation.

Consider a hollow cone which is spinning about the $z$ axis. The moment of inertia of the cone about the z axis is,

    $I_{\text{zz}}={\displaystyle \int d}VQ\left(x^2+y^2\right)$        (II)

Here, $dV$ is the volume element in cylindrical coordinate and $Q$ is the mass density.

According to the given figure, the polar coordinates of any point on a cone in three dimensional space is,

    $\begin{array}{l}x=\rho \sin \theta \cos \varphi \\ y=\rho \sin \theta \sin \varphi \\ z=\rho \cos \theta \end{array}$        (III)

Here, $\rho =\sqrt{x^2+y^2}$ and $\theta$ is the half of the angle of vertex.

From the above figure,

    $\begin{array}{l}\sin \theta =\frac{R}{\sqrt{R^2+h^2}}\\ \cos \theta =\frac{h}{\sqrt{R^2+h^2}}\end{array}$        (IV)

The volume of small element of cone I cylindrical coordinate is,

    $dV=d\varphi \rho d\rho \sin \theta$        (V)

Use equation (V) and (III) in equation (II),

    $\begin{array}{c}{I}_{\text{ac}}=Q{\displaystyle {\int }_0^{2\pi }d}\varphi {\displaystyle {\int }_0^{\sqrt{R^2+h^2}}\rho }d\rho \sin \theta \left[{\left(\rho \sin \theta \cos \varphi \right)}^2+{\left(\rho \sin \theta \sin \varphi \right)}^2\right]\\ =Q{\left[\varphi \right]}_0^{2\pi }{\displaystyle {\int }_0^{\sqrt{R^2}+h^2}\rho }d\rho \sin \theta \left[\left({\rho }^2{\sin }^2\theta {\cos }^2\varphi \right)+\left({\rho }^2{\sin }^2\theta {\sin }^2\varphi \right)\right]\\ =Q\left[2\pi -0\right]{\displaystyle {\int }_0^{\sqrt{R^2+h^2}}\rho }d\rho \sin \theta \left({\rho }^2{\sin }^2\theta \right)\left({\cos }^2\varphi +{\sin }^2\varphi \right)\\ =Q\left[2\pi \right]{\displaystyle {\int }_0^{\sqrt{R^2+h^2}}{\rho }^3{\sin }^3\theta d\rho }\end{array}$

Use equation (IV) in the above equation,

    $\begin{array}{c}{I}_{zz}=2\pi Q{\displaystyle {\int }_0^{\sqrt{R^2+h^2}}{\rho }^3}{\left(\frac{R}{\sqrt{R^2+h^2}}\right)}^{3}d\rho \\ =2\pi Q\frac{R^3}{{\left(R^2+h^2\right)}^{\frac{3}{2}}}{\displaystyle {\int }_0^{\sqrt{R^2}+{\hslash }^2}{\rho }^3}d\rho \\ =2\pi Q\frac{R^3}{{\left(R^2+h^2\right)}^{\frac{3}{2}}}{\displaystyle {\int }_0^{\sqrt{R^2}+h^2}{\rho }^3}d\rho \\ =2\pi Q\frac{R^3}{{\left(R^2+h^2\right)}^{\frac{3}{2}}}{\left[\frac{{\rho }^{3+1}}{3+1}\right]}_0^{\sqrt{R^2+h^2}}\end{array}$

$\begin{array}{c}{I}_{zz}=2\pi Q\frac{R^3}{{\left(R^2+h^2\right)}^{\frac{3}{2}}}{\left[\frac{{\rho }^4}{4}\right]}_0^{\sqrt{R^2+h^2}}\\ =2\pi Q\frac{R^3}{{\left(R^2+h^2\right)}^{\frac{3}{2}}}\left[\frac{{(\sqrt{R^2+h^2})}^4}{4}-0\right]\\ =\frac{1}{2}\pi Q{R}^3\sqrt{R^2+h^2}\end{array}$        (VI)

The mass of the cone is given by,

    $M=Q{\displaystyle {\int }_{}^{}dV}$        (VII)

Use equation (V)in equation (VI),

    $M=Q{\displaystyle \int d}\varphi \rho d\rho \sin \theta$        (VIII)

Substitute $\frac{R}{\sqrt{R^2+h^2}}$ for $\sin \theta$ in equation (VIII),

    $\begin{array}{c}M=Q{\displaystyle {\int }_0^{2\pi }d}\varphi {\displaystyle {\int }_0^{\sqrt{R^2+h^2}}\rho }d\rho \left(\frac{R}{\sqrt{R^2+h^2}}\right)\\ =Q{[}_{\varphi }^{]}\left(\frac{R}{\sqrt{R^2+h^2}}\right){\left[\frac{{\rho }^{1+1}}{1+1}\right]}_0^{\sqrt{R^2+h^2}}\\ =Q[2\pi -0]\left(\frac{R}{\sqrt{R^2+h^2}}\right)\left(\frac{1}{2}\right)\left[{(\sqrt{R^2+h^2})}^2-0\right]\\ =Q\pi R\sqrt{R^2+h^2}\end{array}$

Solve above equation for $Q$,

    $Q=\frac{M}{\pi R\sqrt{R^2+h^2}}$        (IX)

Use equation (IX) in equation (VI),

    $\begin{array}{c}{I}_{zz}=\frac{1}{2}\pi \left[\frac{M}{\pi R\sqrt{R^2+h^2}}\right]R^3\sqrt{R^2+h^2}\\ =\frac{1}{2}M{R}^2\end{array}$        (X)

He moment of inertia of the cone about the x axis is,

    $I_{xx}={\displaystyle \int d}VQ\left(x^2+z^2\right)$

Use equation (IV) and (V) in the above equation,

    $\begin{array}{l}{I}_{xx}=Q{\displaystyle {\int }_0^{2\pi }d}\varphi {\displaystyle {\int }_0^{\sqrt{R^2+h^2}}\rho }d\rho \sin \theta \left[{\left(\rho \sin \theta \cos \varphi \right)}^2+{\left(\rho \cos \theta \right)}^2\right]\\ =Q{\displaystyle {\int }_0^{2\pi }d}\varphi {\displaystyle {\int }_0^{\sqrt{R^2+h^2}}{\rho }^3}d\rho {\sin }^3\theta \left({\cos }^2\varphi \right)+Q{\displaystyle {\int }_0^{2\pi }d}\varphi {\displaystyle {\int }_0^{\sqrt{R^2+{\mu }^2}}\sin }\theta {\cos }^2\theta {\rho }^{3}d\rho \\ =Q{\displaystyle {\int }_0^{2\pi }d}\varphi {\displaystyle {\int }_0^{\sqrt{R^2+h^2}}{\rho }^3}d\rho {\sin }^3\theta \left(\frac{1+\cos 2\varphi }{2}\right)+Q{\displaystyle {\int }_0^{2\pi }d}\varphi {\displaystyle {\int }_0^{\sqrt{R^2+h^2}}\sin }\theta {\cos }^2\theta {\rho }^{3}d\rho \\ =Q{\displaystyle {\int }_0^{2\pi }\left(\frac{1+\cos 2\varphi }{2}\right)}d\varphi {\displaystyle {\int }_0^{\sqrt{R^2}+{\hslash }^2}{\rho }^3}d\rho {\sin }^3\theta +Q{\displaystyle {\int }_0^{2\pi }d}\varphi {\displaystyle {\int }_0^{\sqrt{R^2+h^2}}\sin }\theta {\cos }^2\theta {\rho }^{3}d\rho \end{array}$

Using equation (V),

    $\begin{array}{c}{I}_{xx}=\frac{Q}{2}{\left[\varphi +\frac{\sin 2\varphi }{2}\right]}_0^{2\pi }\left[\frac{R^3}{{\left(R^2+h^2\right)}^{\frac{3}{2}}}\right]{\left[\frac{{\rho }^{3+1}}{3+1}\right]}_0^{\sqrt{R^2+h^2}}\\ +Q{\left[\varphi \right]}_0^{2\pi }\left(\frac{R}{\sqrt{R^2+h^2}}\right){\left(\frac{h}{\sqrt{R^2+h^2}}\right)}^2{\left[\frac{{\rho }^{3+1}}{3+1}\right]}_0^{\sqrt{R^2+h^2}}\\ =\left[\begin{array}{c}\frac{Q}{2}\left[\left(2\pi +0\right)-\left(0+\frac{0}{2}\right)\right]\left[\frac{R^3}{4{\left(R^2+h^2\right)}^{\frac{3}{2}}}\right]{\left(R^2+h^2\right)}^2\\ +Q\left[2\pi -0\right]\frac{R{h}^2}{4{\left(R^2+h^2\right)}^{\frac{3}{2}}}{\left(R^2+h^2\right)}^2\end{array}\right]\\ =\frac{1}{4}Q\pi R\sqrt{R^2+h^2}\left[R^2+2h^2\right]\end{array}$

Use equation (IX) in the above equation,

    $\begin{array}{c}{I}_{xx}=\frac{1}{4}\left(\frac{M}{\pi R\sqrt{R^2+h^2}}\right)\pi R\sqrt{R^2+h^2}\left[R^2+2h^2\right]\\ =\frac{1}{4}M\left(R^2+2h^2\right)\end{array}$        (XI)

By the symmetry of the axes, the moment of inertia $I_{yy}$ about y axis is equal to the moment of inertia about x axis.

    $I_{yy}=\frac{1}{4}M\left(R^2+2h^2\right)$        (XII)

The moment of inertia about its pointed end of a hollow cone is,

    $I=\left[\begin{array}{ccc}{I}_{xx}& 0& 0\\ 0& I_{yy}& 0\\ 0& 0& I_{zz}\end{array}\right]$

Use equation (XI) (XII) and (X) in equation above,

    $\begin{array}{c}I=\left[\begin{array}{ccc}\frac{1}{4}M\left(R^2+2h^2\right)& 0& 0\\ 0& \frac{1}{4}M\left(R^2+2h^2\right)& 0\\ 0& 0& \frac{1}{2}M{R}^2\end{array}\right]\\ =\frac{1}{4}M\left[\begin{array}{ccc}\left(R^2+2h^2\right)& 0& 0\\ 0& \left(R^2+2h^2\right)& 0\\ 0& 0& 2R^2\end{array}\right]\end{array}$

Conclusion:

Therefore, The moment of inertia tensor of the hollow cone which spins about its pointed end is $\underset{\_}{\frac{1}{4}M\left[\begin{array}{ccc}\left(R^2+2h^2\right)& 0& 0\\ 0& \left(R^2+2h^2\right)& 0\\ 0& 0& 2R^2\end{array}\right]}$.