Chapter 10, Problem 57SP

Figure 10-13 shows four masses that are held at the corners of a square by a very light frame. What is the moment of inertia of the system about an axis perpendicular to the page (a) through A and (b) through B?

Fig. 10-13

To determine

The moment of inertia of the system shown in Fig. 10.13, about an axis perpendicular to the page through A.

Answer to Problem 57SP

Solution:

$0.14\text{ kg}\cdot {\text{m}}^2$

Explanation of Solution

Given data:

The system of four masses are shown as follows:

Formula used:

Write the expression of moment of inertia of a mass system about an axis is given as follows.

$I=m_1{r}_1{}^2+m_2{r}_2{}^2+m_3{r}_3{}^2+m_4{r}_4{}^2$

Here, $I$ is the moment of inertia, $m_1,m_2,m_3,and{m}_4$ are the point masses and $r_1,r_2,r_3,\text{ and }{r}_4$ are the perpendicular distances of the masses from the given axis.

Write the expression of Pythagoras theorem in right angled triangle to find the distance of point mass from the axis.

$x^2+y^2=l^2$

Here, $x$ is the length of perpendicular, $y$ is the length of base, and $l$ is the length of hypotenuse.

Explanation:

Redraw the diagram of the system show provided in the question:

From fig. 1, infer that all the point masses are at equal distance from the axis perpendicular to page through A. Let the distance of point masses from the axis be $r$.

The expression of distance of point masses from axis perpendicular to page through A is,

$r^2+r^2=l^2$

Here, $l$ is the length of the side of square.

Substitute $80\text{ cm}$ for $l$

$r^2+r^2={\left(80\text{ cm}\right)}^2$

Solve for $r$.

$\begin{array}{c}2r^2={\left(80\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)}^2\\ r^2=\frac{0.64}{2}{\text{ m}}^2\\ r=\sqrt{\frac{0.64}{2}}\text{m}\\ =0.57\text{ m}\end{array}$

The expression of moment of inertia about a given axisfor a mass system is as follows,

$I=m_1{r}_1{}^2+m_2{r}_2{}^2+m_3{r}_3{}^2+m_4{r}_4{}^2$

Since, the perpendicular distances of all the point masses from the given axis are equal, therefore, the expression of the distance of all point masses from the given axis is,

$r_1=r_2=r_3=r_4=r$

Substitute $r$ for $r_1,r_2,r_3,and{r}_4$ in the expression of moment of inertia.

$\begin{array}{c}I=m_1{r}^2+m_2{r}^2+m_3{r}^2+m_4{r}^2\\ =\left(m_1+m_2+m_3+m_4\right)r^2\end{array}$

Substitute $150\text{ g}$ for $m_1$, $70\text{ g}$ for $m_2$, $150\text{ g}$ for $m_3$, $70\text{ g}$ for $m_4$, and $0.57\text{ m}$ for $r$

$\begin{array}{c}I=\left(m_1+m_2+m_3+m_4\right)r^2\\ =\left(150\text{ g}+70\text{ g}+150\text{ g}+70\text{ g}\right){\left(0.57\text{ m}\right)}^2\\ =\left(440\text{ g}\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\right){\left(0.57\text{ m}\right)}^2\\ =0.14\text{ kg}\cdot {\text{m}}^2\end{array}$

Conclusion:

The moment of inertia of the system about an axis perpendicular to the page through A is $0.14\text{ kg}\cdot {\text{m}}^2$.

To determine

The moment of inertia of the system shown in Fig. 10-13, about an axis perpendicular to the page through B.

Answer to Problem 57SP

Solution:

$\text{0}\text{.21 kg}\cdot {\text{m}}^2$

Explanation of Solution

Given data:

The system of four masses that are held at the corners of a square by a very light frame as shown in fig. 1.

Formula used:

Write the expression of moment of inertia about a given axis.

$I=m_1{r}_1{}^2+m_2{r}_2{}^2+m_3{r}_3{}^2+m_4{r}_4{}^2$

Here, $I$ is the moment of inertia, $m_1,m_2,m_3,\text{ and }{m}_4$ are the point masses and $r_1,r_2,r_3,\text{ and }{r}_4$ are the perpendicular distances of the masses from the given axis.

Write the expression of Pythagoras theorem in right angled triangle to find the distance of point mass from the axis.

$x^2+y^2=l^2$

Here, $x$ is the length of perpendicular, $y$ is the length of base, and $l$ is the length of hypotenuse.

Explanation:

From fig. 1, infer that the distance of point mass $m_1,\text{ 150 g}$ and $m_2,\text{ 70 g}$ are at equal distance from the axis perpendicular to page through B and the axis B is at midway from the two masses.

The expression of distance of point mass, $m_1$ and $m_2$ from axis perpendicular to page through B is,

$r_1=r_2=\frac{l}{2}$

Here, $l$ is the length of the side of square.

Substitute $80\text{ cm}$ for $l$

$\begin{array}{c}{r}_1=r_2=\frac{\left(80\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)}{2}\\ =40\times {10}^{-2}\text{ m}\\ =\text{0}\text{.4 m}\end{array}$

The expression of Pythagoras theorem in right angled triangle to find the distance of point mass from the axis B.

$x^2+y^2=z^2$

Here, $z$ is the distance of point mass $m_3,\text{ 150 g}$ and $m_4,\text{ 70 g}$ from the axis B.

Substitute $80\text{ cm}$ for $x$ and $40\text{ cm}$ for $y$

${\left(80\text{ cm}\right)}^2+{\left(40\text{ cm}\right)}^2=z^2$

Solve for $z$.

$\begin{array}{c}z=\sqrt{{\left(80\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)}^2+{\left(40\text{ cm}\left(\frac{{10}^{-2}\text{ m}}{1\text{ cm}}\right)\right)}^2}\\ =\sqrt{{\left(0.8\text{ m}\right)}^2+{\left(0.4\text{ m}\right)}^2}\\ =\sqrt{0.64+0.16}\text{m}\\ =\sqrt{0.8}\text{ m}\end{array}$

Further, solve the expression.

$z=0.89\text{m}$

The expression of moment of inertia about a given axis is,

$I=m_1{r}_1{}^2+m_2{r}_2{}^2+m_3{r}_3{}^2+m_4{r}_4{}^2$

Substitute $r$ for $r_1,r_2$ and $z$ for $r_3,r_4$

$\begin{array}{c}I=m_1{r}^2+m_2{r}^2+m_3{z}^2+m_4{z}^2\\ =\left(m_1+m_2\right)r^2+\left(m_3+m_4\right)z^2\end{array}$

Substitute $150\text{ g}$ for $m_1$, $70\text{ g}$ for $m_2$, $150\text{ g}$ for $m_3$, $70\text{ g}$ for $m_4$, $0.4\text{ m}$ for $r$, and $0.89\text{ m}$ for $z$

$\begin{array}{c}I=\left(150\text{ g}+70\text{ g}\right){\left(0.4\text{m}\right)}^2+\left(150\text{ g}+70\text{ g}\right){\left(0.89\text{ m}\right)}^2\\ =\left(220\text{ g}\left(\frac{{10}^{-3}\text{ kg}}{1\text{g}}\right)\right)\left(0.16{\text{ m}}^2\right)+\left(220\text{ g}\left(\frac{{10}^{-3}\text{ kg}}{1\text{g}}\right)\right)\left(0.79{\text{ m}}^2\right)\\ =0.0352\text{ kg}\cdot {\text{m}}^2+0.1738\text{ kg}\cdot {\text{m}}^2\\ =0.21\text{ kg}\cdot {\text{m}}^2\end{array}$

Conclusion:

The moment of inertia of the system about an axis perpendicular to the page through B is $0.21\text{ kg}\cdot {\text{m}}^2$.