Chapter 10, Problem 10.28P

To determine

Show that $i$ has the form similar to the axisymmetric body.

Answer to Problem 10.28P

The inertia tensor for the prism is $\underset{\_}{\frac{M}{3}\left[\begin{array}{ccc}\frac{M}{4}\left(2a^2+L^2\right)& -\frac{M}{6}\left(L^2+6a^2\right)& -\frac{M}{24}\left(L^2+4a^2\right)\\ -\frac{M}{6}\left(L^2+6a^2\right)& \frac{M}{4}\left(2a^2+L^2\right)& -\frac{M}{8}\left(8a^2+L^2\right)\\ -\frac{M}{24}\left(L^2+4a^2\right)& -\frac{M}{8}\left(8a^2+L^2\right)& M{a}^2\end{array}\right]}$.

Explanation of Solution

Write the expression for the moment of inertia tensor.

    $I=\left[\begin{array}{lll}{I}_{xx}\hfill & I_{xy}\hfill & I_{xz}\hfill \\ I_{yx}\hfill & I_{yy}\hfill & I_{yz}\hfill \\ I_{zx}\hfill & I_{zy}\hfill & I_{zz}\hfill \end{array}\right]$        (I)

Here, $I_{xx,}{I}_{yy},I_{zz}$ are the moment of inertia along respective axis.

Figure 1 shows a equilateral triangular prism having length $L$ and the length of each side of triangle of prism is $2a$.

The height of the prism is,

    $\begin{array}{c}h=\sqrt{\left(2a^2\right)-{\left(a\right)}^2}\\ =\sqrt{3}a\end{array}$

The area of the triangular prism base is,

    $\begin{array}{c}A=\frac{1}{2}\left(\text{base}\right)\left(\text{height}\right)\\ =\frac{1}{2}\left(2a\right)\left(\sqrt{3}a\right)\\ =\sqrt{3}{a}^2\end{array}$

The volume of the triangular prism is, $V=AL$ so that the volume density of the triangular prism is given by,

    $\begin{array}{c}\rho =\frac{M}{V}\\ =\frac{M}{AL}\end{array}$

Here, $V$ is the volume of the prism, $M$ is the mass of the prism.

Substitute $\sqrt{3}{a}^2$ for $A$

    $\rho =\frac{M}{\sqrt{3}{a}^{2}L}$        (II)

The range of y is $0\le y\le \sqrt{3}a$ so that the range of $x$ at any value of $y$ is,

    $-\frac{y}{3}\le x\le \frac{y}{\sqrt{3}}$

The moment of inertia about an axis parallel to the zz axis of the prism is,

    $\begin{array}{c}{I}_{\text{az}}=2\rho {\displaystyle {\int }_{z=0}^{L}d}z{\displaystyle {\int }_{y=0}^{\sqrt{2}a}d}y{\displaystyle {\int }_{x=0}^{\frac{1}{\sqrt{3}}a}\left(x^2+y^2\right)}dx\\ =2\rho {\displaystyle {\int }_{z=0}^{L}d}z{\displaystyle {\int }_{y=0}^{\sqrt{3}a}d}y{\left[\frac{x^3}{3}+x{y}^2\right]}_{x=0}^{\frac{1}{\sqrt{3}y}}\\ =2\rho {\displaystyle {\int }_{z=0}^{L}d}z{\displaystyle {\int }_{y=0}^{\sqrt{5a}}d}y\left[\frac{1}{3}{\left(\frac{1}{\sqrt{3}}y\right)}^3+\left(\frac{1}{\sqrt{3}}y\right)y^2\right]\\ =2\rho {\displaystyle {\int }_{z=0}^{L}d}z{\displaystyle {\int }_{y=0}^{\sqrt{3}a}d}y\left[\left(\frac{1}{9\sqrt{3}}+\frac{1}{\sqrt{3}}\right)y^3\right]\end{array}$

Further solve will give,

    $\begin{array}{c}{I}_{tz}=\frac{2}{\sqrt{3}}\left(\frac{1}{9}+1\right)\rho {\displaystyle {\int }_{z=0}^{L}d}z{\displaystyle {\int }_{y=0}^{\sqrt{3}a}d}y{y}^3\\ =2\left(\frac{10}{9\sqrt{3}}\right)\rho {\displaystyle {\int }_{z=0}^{L}d}z{\left[\frac{y^4}{4}\right]}_{y=0}^{\sqrt{3}a}\\ =2\left(\frac{10}{9\sqrt{3}}\right)\left[\frac{{(\sqrt{3}a)}^4}{4}\right]\rho {\displaystyle {\int }_{z=0}^{L}d}z\\ =2\left(\frac{10}{9\sqrt{3}}\right)\left[\frac{9a^4}{4}\right]\rho (L)\end{array}$

The value of $I_{zz}$ is,

    $I_e=\frac{5}{\sqrt{3}}{a}^4\rho L$        (III)

Use equation (II) in equation (III),

    $\begin{array}{c}{I}_{zz}=\frac{5}{\sqrt{3}}{a}^4\left(\frac{M}{\sqrt{3}{a}^{2}L}\right)L\\ =\frac{5}{3}M{a}^2\end{array}$

The moment of inertia of the prism around an axis passing through centroid is,

    $\begin{array}{c}{I}_0=M{d}^2\\ =M{\left(\frac{2}{\sqrt{3}}a\right)}^2\\ =\frac{4}{3}M{a}^2\end{array}$

Using the moment of inertia about an axis parallel to the z-axis passing through one corner of the prism and the moment of inertia of the prism around pan axis passing through centroid to find the required moment of inertia of the prism is,

    $I_{zz}=\frac{M{a}^2}{3}$        (IV)

The values of $i_{xx}$ and $I_{yy}$ is,

    $\begin{array}{c}{I}_{xx}=I_{yy}\\ =\frac{M}{18}\left(6a^2+L^2\right)\\ I_{xz}=I_{zx}\\ =-\frac{M}{72}\left(L^2+4a^2\right)\\ I_{yz}=I_{zy}\\ =-\frac{M}{24}\left(8a^2+L^2\right)\end{array}$

Thus the given inertia tensor becomes,

    $\begin{array}{c}I=\left[\begin{array}{ccc}\frac{M}{12}\left(2a^2+L^2\right)& -\frac{M}{18}\left(L^2+6a^2\right)& -\frac{M}{72}\left(L^2+4a^2\right)\\ -\frac{M}{18}\left(L^2+6a^2\right)& \frac{M}{12}\left(2a^2+L^2\right)& -\frac{M}{24}\left(8a^2+L^2\right)\\ -\frac{M}{72}\left(L^2+4a^2\right)& -\frac{M}{24}\left(8a^2+L^2\right)& \frac{M{a}^2}{3}\end{array}\right]\\ =\frac{M}{3}\left[\begin{array}{ccc}\frac{M}{4}\left(2a^2+L^2\right)& -\frac{M}{6}\left(L^2+6a^2\right)& -\frac{M}{24}\left(L^2+4a^2\right)\\ -\frac{M}{6}\left(L^2+6a^2\right)& \frac{M}{4}\left(2a^2+L^2\right)& -\frac{M}{8}\left(8a^2+L^2\right)\\ -\frac{M}{24}\left(L^2+4a^2\right)& -\frac{M}{8}\left(8a^2+L^2\right)& M{a}^2\end{array}\right]\end{array}$

Therefore, the inertia tensor for the prism is $\underset{\_}{\frac{M}{3}\left[\begin{array}{ccc}\frac{M}{4}\left(2a^2+L^2\right)& -\frac{M}{6}\left(L^2+6a^2\right)& -\frac{M}{24}\left(L^2+4a^2\right)\\ -\frac{M}{6}\left(L^2+6a^2\right)& \frac{M}{4}\left(2a^2+L^2\right)& -\frac{M}{8}\left(8a^2+L^2\right)\\ -\frac{M}{24}\left(L^2+4a^2\right)& -\frac{M}{8}\left(8a^2+L^2\right)& M{a}^2\end{array}\right]}$

Conclusion:

Therefore, the inertia tensor for the prism $\underset{\_}{\frac{M}{3}\left[\begin{array}{ccc}\frac{M}{4}\left(2a^2+L^2\right)& -\frac{M}{6}\left(L^2+6a^2\right)& -\frac{M}{24}\left(L^2+4a^2\right)\\ -\frac{M}{6}\left(L^2+6a^2\right)& \frac{M}{4}\left(2a^2+L^2\right)& -\frac{M}{8}\left(8a^2+L^2\right)\\ -\frac{M}{24}\left(L^2+4a^2\right)& -\frac{M}{8}\left(8a^2+L^2\right)& M{a}^2\end{array}\right]}$