Chapter 10, Problem 10.22P

(a)

To determine

The moment of inertia tensor for rotation about a corner of the cube.

Answer to Problem 10.22P

The moment of inertia tensor for rotation about a corner of the cube is $\underset{\_}{2m{a}^2\left(\begin{array}{ccc}4& -1& -1\\ -1& 4& -1\\ -1& -1& 4\end{array}\right)}$.

Explanation of Solution

The diagram for the inertia tensor for a cube is given by figure 1.

Write the expression for the inertia tensor for a rigid body.

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

Here, $I$ is the inertia tensor, $I_{xy},I_{xz},I_{yx},,I_{yz},I_{zx},I_{zy}$ is the product of inertia, $I_{xx},I_{yy},I_{zz}$ is the moment of inertia.

Write the expression for the moment of inertia of the rigid body about the x-axis.

$\begin{array}{c}{I}_{xx}=m{\displaystyle \sum _{\alpha }\left(y_{\alpha }^2+z_{\alpha }^2\right)}\\ =m\left[\begin{array}{l}\left(y_1^2+z_1^2\right)+\left(y_2^2+z_2^2\right)+\left(y_3^2+z_3^2\right)+\left(y_4^2+z_4^2\right)+\\ \left(y_5^2+z_5^2\right)+\left(y_6^2+z_6^2\right)+\left(y_7^2+z_7^2\right)+\left(y_8^2+z_8^2\right)\end{array}\right]\end{array}$        (II)

Here, $m$ is the mass of the particle, $y_1,y_2,y_3,y_4,y_5,y_6,y_7,y_8$ are the y-distances, $z_1,z_2,z_3,z_4,z_5,z_6,z_7,z_8$ are the z-distances.

Write the expression for the product of moment of inertia of the rigid body about the xy-plane.

$\begin{array}{c}{I}_{xy}=-m{\displaystyle \sum _{\alpha }\left(x_{\alpha }{y}_{\alpha }\right)}\\ =m\left[\begin{array}{l}\left(x_1{y}_1\right)+\left(x_2{y}_2\right)+\left(x_3{y}_3\right)+\left(x_4{y}_4\right)+\\ \left(x_5{y}_5\right)+\left(x_6{y}_6\right)+\left(x_7{y}_7\right)+\left(x_8{y}_8\right)\end{array}\right]\end{array}$        (III)

Here, $m$ is the mass of the particle, $x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8$ are the x-distances.

Conclusion:

Substitute, $0$ for $\left(y_1^2+z_1^2\right)$, $0$ for $\left(y_2^2+z_2^2\right)$, $a^2$ for $\left(y_3^2+z_3^2\right)$, $a^2$ for $\left(y_4^2+z_4^2\right)$, $2a^2$ for $\left(y_5^2+z_5^2\right)$, $2a^2$ for $\left(y_6^2+z_6^2\right)$, $a^2$ for $\left(y_7^2+z_7^2\right)$, $a^2$ for $\left(y_8^2+z_8^2\right)$ in equation (II) to find $I_{xx}$.

$\begin{array}{c}{I}_{xx}=m\left[0+0+a^2+a^2+2a^2+2a^2+a^2+a^2\right]\\ =8m{a}^2\end{array}$

Similarly by using symmetry, the moment of inertia about the y and z axes is,

$I_{yy}=8m{a}^2$

$I_{zz}=8m{a}^2$

Substitute, $0$ for $x_1{y}_1$, $0$ for $x_2{y}_2$, $0$ for $x_3{y}_3$, $0$ for $x_4{y}_4$, $0$ for $x_5{y}_5$, $a^2$ for $x_6{y}_6$, $a^2$ for $x_7{y}_7$, $0$ for $x_8{y}_8$ in equation (III) to find $I_{xy}$.

$\begin{array}{c}{I}_{xy}=-m\left[0+0+0+0+0+a^2+a^2+0\right]\\ =-2m{a}^2\end{array}$

Similarly by using symmetry, the product of moment of inertia about the yz and xz plane is,

$I_{xz}=-2m{a}^2$

$I_{yz}=-2m{a}^2$

Substitute $8m{a}^2$ for $I_{xx}$, $8m{a}^2$ for $I_{yy}$, $8m{a}^2$ for $I_{zz}$, $-2m{a}^2$ for $I_{xy}$, $-2m{a}^2$ for $I_{xz}$, $-2m{a}^2$ for $I_{yx}$, $-2m{a}^2$ for $I_{yz}$, $-2m{a}^2$ for $I_{zx}$, $-2m{a}^2$ for $I_{zy}$ in equation (I) to find $I$.

$\begin{array}{c}I=\left(\begin{array}{ccc}8m{a}^2& -2m{a}^2& -2m{a}^2\\ -2m{a}^2& 8m{a}^2& -2m{a}^2\\ -2m{a}^2& -2m{a}^2& 8m{a}^2\end{array}\right)\\ =2m{a}^2\left(\begin{array}{ccc}4& -1& -1\\ -1& 4& -1\\ -1& -1& 4\end{array}\right)\end{array}$

Thus, the moment of inertia tensor for rotation about a corner of the cube is $\underset{\_}{2m{a}^2\left(\begin{array}{ccc}4& -1& -1\\ -1& 4& -1\\ -1& -1& 4\end{array}\right)}$.

(b)

To determine

The moment of inertia tensor for rotation about the center of the cube.

Answer to Problem 10.22P

The moment of inertia tensor for rotation about the center of the cube is $\underset{\_}{4m{a}^2\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)}$.

Explanation of Solution

The diagram for the inertia tensor for a cube is given by figure 2.

Write the expression for the inertia tensor for a rigid body.

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

Here, $I$ is the inertia tensor, $I_{xy},I_{xz},I_{yx},,I_{yz},I_{zx},I_{zy}$ is the product of inertia, $I_{xx},I_{yy},I_{zz}$ is the moment of inertia.

Write the expression for the moment of inertia of the rigid body about the x-axis.

$\begin{array}{c}{I}_{xx}=m{\displaystyle \sum _{\alpha }\left(y_{\alpha }^2+z_{\alpha }^2\right)}\\ =m\left[\begin{array}{l}\left(y_1^2+z_1^2\right)+\left(y_2^2+z_2^2\right)+\left(y_3^2+z_3^2\right)+\left(y_4^2+z_4^2\right)+\\ \left(y_5^2+z_5^2\right)+\left(y_6^2+z_6^2\right)+\left(y_7^2+z_7^2\right)+\left(y_8^2+z_8^2\right)\end{array}\right]\end{array}$        (II)

Here, $m$ is the mass of the particle, $y_1,y_2,y_3,y_4,y_5,y_6,y_7,y_8$ are the y-distances, $z_1,z_2,z_3,z_4,z_5,z_6,z_7,z_8$ are the z-distances.

Write the expression for the product of moment of inertia of the rigid body about the xy-plane.

$\begin{array}{c}{I}_{xy}=-m{\displaystyle \sum _{\alpha }\left(x_{\alpha }{y}_{\alpha }\right)}\\ =m\left[\begin{array}{l}\left(x_1{y}_1\right)+\left(x_2{y}_2\right)+\left(x_3{y}_3\right)+\left(x_4{y}_4\right)+\\ \left(x_5{y}_5\right)+\left(x_6{y}_6\right)+\left(x_7{y}_7\right)+\left(x_8{y}_8\right)\end{array}\right]\end{array}$        (III)

Here, $m$ is the mass of the particle, $x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8$ are the x-distances.

Conclusion:

Substitute, $\left(a/2\right)$ for $y_1$, $\left(a/2\right)$ for $y_2$, $\left(a/2\right)$ for $y_3$, $\left(a/2\right)$ for $y_4$, $-\left(a/2\right)$ for $y_5$, $-\left(a/2\right)$ for $y_6$, $-\left(a/2\right)$ for $y_7$, $-\left(a/2\right)$ for $y_8$, $\left(a/2\right)$ for $z_1$, $\left(a/2\right)$ for $z_2$, $-\left(a/2\right)$ for $z_3$, $-\left(a/2\right)$ for $z_4$, $-\left(a/2\right)$ for $z_5$, $-\left(a/2\right)$ for $z_6$, $\left(a/2\right)$ for $z_7$, $\left(a/2\right)$ for $z_8$ in equation (II) to find $I_{xx}$.

$\begin{array}{c}{I}_{xx}=m\left[\begin{array}{l}\left({\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2\right)+\left({\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2\right)+\left({\left(\frac{a}{2}\right)}^2+{\left(\frac{-a}{2}\right)}^2\right)+\left({\left(\frac{a}{2}\right)}^2+{\left(-\frac{a}{2}\right)}^2\right)+\\ \left({\left(-\frac{a}{2}\right)}^2+{\left(-\frac{a}{2}\right)}^2\right)+\left({\left(-\frac{a}{2}\right)}^2+{\left(-\frac{a}{2}\right)}^2\right)+\left({\left(-\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2\right)+\left({\left(-\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2\right)\end{array}\right]\\ =4m{a}^2\end{array}$

Here, $a$ is the side of the cube.

Similarly by using symmetry, the moment of inertia about the y and z axes is,

$I_{yy}=4m{a}^2$

$I_{zz}=4m{a}^2$

Substitute, $\left(a/2\right)$ for $y_1$, $\left(a/2\right)$ for $y_2$, $\left(a/2\right)$ for $y_3$, $\left(a/2\right)$ for $y_4$, $-\left(a/2\right)$ for $y_5$, $-\left(a/2\right)$ for $y_6$, $-\left(a/2\right)$ for $y_7$, $-\left(a/2\right)$ for $y_8$, $\left(a/2\right)$ for $x_1$, $-\left(a/2\right)$ for $x_2$, $-\left(a/2\right)$ for $x_3$, $\left(a/2\right)$ for $x_4$, $\left(a/2\right)$ for $x_5$, $-\left(a/2\right)$ for $x_6$, $-\left(a/2\right)$ for $x_7$, $\left(a/2\right)$ for $z_8$ in equation (III) to find $I_{xy}$.

$\begin{array}{c}{I}_{xy}=-m\left[\begin{array}{l}\left\{\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\right\}+\left\{\left(-\frac{a}{2}\right)\left(\frac{a}{2}\right)\right\}+\left\{-\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\right\}+\left\{\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\right\}+\\ \left\{\left(\frac{a}{2}\right)\left(-\frac{a}{2}\right)\right\}+\left\{\left(-\frac{a}{2}\right)\left(-\frac{a}{2}\right)\right\}+\left\{\left(-\frac{a}{2}\right)\left(-\frac{a}{2}\right)\right\}+\left\{\left(\frac{a}{2}\right)\left(-\frac{a}{2}\right)\right\}\end{array}\right]\\ =0\end{array}$

Similarly by using symmetry, the product of moment of inertia about the yz and xz plane is,

$I_{xz}=0$

$I_{yz}=0$

Substitute $4m{a}^2$ for $I_{xx}$, $4m{a}^2$ for $I_{yy}$, $4m{a}^2$ for $I_{zz}$, $0$ for $I_{xy}$, $0$ for $I_{xz}$, $0$ for $I_{yx}$, $0$ for $I_{yz}$, $0$ for $I_{zx}$, $0$ for $I_{zy}$ in equation (I) to find $I$.

$\begin{array}{c}I=\left(\begin{array}{ccc}4m{a}^2& 0& 0\\ 0& 4m{a}^2& 0\\ 0& 0& 4m{a}^2\end{array}\right)\\ =4m{a}^2\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\end{array}$

Thus, the moment of inertia tensor for rotation about the center of the cube is $\underset{\_}{4m{a}^2\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)}$.