Chapter 8, Problem 4P

(a)

To determine

The rotational inertia of the system of point particles shown in figure in question assuming the system is rotates about $x$ axis.

Answer to Problem 4P

The rotational inertia of the system of point particles about $x$ axis is $13,000\text{g}\cdot {\text{cm}}^2$.

Explanation of Solution

Write the expression for the rotational inertia about $x$ axis.

$I_x={\displaystyle \sum _{\text{i=}A}^C{m}_i{r}_i^2}$

Here, $I_x$ is the moment of inertia about $x$ axis, $m_i$ is the mass of $i^{\text{th}}$ particle and $r_i$ is distance of $i^{\text{th}}$ particle.

Expand above summation.

$I_x=m_A{r}_A^2+m_B{r}_B^2+m_C{r}_C^2$

Here, $m_{\text{A}}$ is the mass of particle at $A$ , $m_{\text{B}}$ is the mass of particle at $\text{B}$ , $m_{\text{C}}$ is the mass of particle at $C$ , $r_{\text{A}}$ is the perpendicular distance of point $A$ from $x$ axis, $r_{\text{B}}$ is the perpendicular distance of point $B$ from $x$ axis and $r_{\text{C}}$ is the perpendicular distance of point $C$ from $x$ axis.

Conclusion:

Substitute $200\text{g}$ for $m_{\text{A}}$ , $300\text{g}$ for $m_{\text{A}}$, $500\text{g}$ for $m_{\text{C}}$, $5.0\text{cm}$ for $r_{\text{A}}$ , $0\text{cm}$ for $r_{\text{B}}$ and $4.0\text{cm}$ for $r_{\text{C}}$ in above equation to get $I_x$.

$\begin{array}{c}{I}_x=\left(200\text{g}\right){\left(5.0\text{cm}\right)}^2+\left(300\text{g}\right){\left(0\text{cm}\right)}^2+\left(500\text{g}\right){\left(4.0\text{cm}\right)}^2\\ =13,000\text{g}\cdot {\text{cm}}^2\end{array}$

Therefore, the rotational inertia of the system of point particles about $x$ axis is $13,000\text{g}\cdot {\text{cm}}^2$.

(b)

To determine

The rotational inertia of the system of point particles shown in figure in question assuming the system is rotates about $y$ axis.

Answer to Problem 4P

The rotational inertia of the system of point particles about $y$ axis is $25,000\text{g}\cdot {\text{cm}}^2$.

Explanation of Solution

Write the expression for the rotational inertia about $y$ axis.

$I_y={\displaystyle \sum _{\text{i=}A}^C{m}_i{r}_i^2}$

Here, $I_x$ is the moment of inertia about $y$ axis, $m_i$ is the mass of $i^{\text{th}}$ particle and $r_i$ is distance of $i^{\text{th}}$ particle.

Expand above summation.

$I_y=m_{\text{A}}{r}_{\text{A}}^2+m_{\text{B}}{r}_{\text{B}}^2+m_{\text{C}}{r}_{\text{C}}^2$

Here, $m_{\text{A}}$ is the mass of particle at $A$ , $m_{\text{B}}$ is the mass of particle at $\text{B}$ , $m_{\text{C}}$ is the mass of particle at $C$ , $r_{\text{A}}$ is the perpendicular distance of point $A$ from $x$ axis, $r_{\text{B}}$ is the perpendicular distance of point $B$ from $x$ axis and $r_{\text{C}}$ is the perpendicular distance of point $C$ from $y$ axis.

Conclusion:

Substitute $200\text{g}$ for $m_{\text{A}}$ , $300\text{g}$ for $m_{\text{A}}$, $500\text{g}$ for $m_{\text{C}}$, $3.0\text{cm}$ for $r_{\text{A}}$ , $6.0\text{cm}$ for $r_{\text{B}}$ and $5.0\text{cm}$ for $r_{\text{C}}$ in above equation to get $I_y$.

$\begin{array}{c}{I}_y=\left(200\text{g}\right){\left(3.0\text{cm}\right)}^2+\left(300\text{g}\right){\left(6.0\text{cm}\right)}^2+\left(500\text{g}\right){\left(5.0\text{cm}\right)}^2\\ =25,000\text{g}\cdot {\text{cm}}^2\end{array}$

Therefore, the rotational inertia of the system of point particles about $y$ axis is $25,000\text{g}\cdot {\text{cm}}^2$.

(c)

To determine

The rotational inertia of the system of point particles shown in figure in question assuming the system is rotates about $z$ axis.

Answer to Problem 4P

The rotational inertia of the system of point particles about $z$ axis is $38,000\text{g}\cdot {\text{cm}}^2$.

Explanation of Solution

Write the expression for the rotational inertia about $z$ axis.

$I_z={\displaystyle \sum _{\text{i=}A}^C{m}_i{r}_i^2}$

Here, $I_z$ is the moment of inertia about z axis, $m_i$ is the mass of $i^{\text{th}}$ particle and $r_i$ is distance of $i^{\text{th}}$ particle.

Expand above summation.

$I_z=m_{\text{A}}{r}_{\text{A}}^2+m_{\text{B}}{r}_{\text{B}}^2+m_{\text{C}}{r}_{\text{C}}^2$

Here, $m_{\text{A}}$ is the mass of particle at $A$ , $m_{\text{B}}$ is the mass of particle at $\text{B}$ , $m_{\text{C}}$ is the mass of particle at $C$ , $r_{\text{A}}$ is the perpendicular distance of point $A$ from $x$ axis, $r_{\text{B}}$ is the perpendicular distance of point $B$ from $x$ axis and $r_{\text{C}}$ is the perpendicular distance of point $C$ from $z$ axis.

Conclusion:

Substitute $200\text{g}$ for $m_{\text{A}}$ , $300\text{g}$ for $m_{\text{A}}$, $500\text{g}$ for $m_{\text{C}}$, ${\left(3.0\text{cm}\right)}^2+{\left(5.0\text{cm}\right)}^2$ for $r_{\text{A}}$ , $6.0\text{cm}$ for $r_{\text{B}}$ and ${\left(5.0\text{cm}\right)}^2+{\left(4.0\text{cm}\right)}^2$ for $r_{\text{C}}$ in above equation to get $I_z$.

$\begin{array}{c}{I}_z=\left(200\text{g}\right)\left[{\left(3.0\text{cm}\right)}^2+{\left(5.0\text{cm}\right)}^2\right]+\left(300\text{g}\right){\left(6.0\text{cm}\right)}^2+\left(500\text{g}\right)\left[{\left(5.0\text{cm}\right)}^2+{\left(4.0\text{cm}\right)}^2\right]\\ =38,000\text{g}\cdot {\text{cm}}^2\end{array}$

Therefore, the rotational inertia of the system of point particles about $z$ axis is $38,000\text{g}\cdot {\text{cm}}^2$.

(d)

To determine

The $x$ and $y$ coordinates of center of mass of system.

Answer to Problem 4P

The $x$ and $y$ coordinates of center of mass of system is $\left(x,y\right)=\left(-1.3\text{cm,}-\text{1}\text{.0}\text{cm}\right)$.

Explanation of Solution

Write the expression for the $x$ coordinate of center of mass of system.

$x_{\text{CM}}=\frac{m_{\text{A}}{x}_{\text{A}}+m_{\text{B}}{x}_{\text{B}}{\text{+m}}_{\text{C}}{x}_{\text{C}}}{m_{\text{A}}+m_{\text{B}}+m_{\text{C}}}$ (I)

Here, $x_{\text{CM}}$ is the $x$ coordinate of center of mass system, $x_{\text{A}}$ is the $x$ coordinate of center of mass, $x_{\text{B}}$ is the $x$ coordinate of center of mass of $\text{B}$ and $x_{\text{C}}$ is the $x$ coordinate of center of mass of

C.

Write the expression for the $y$ coordinate of center of mass of system.

$y_{\text{CM}}=\frac{m_{\text{A}}{y}_{\text{A}}+m_{\text{B}}{y}_{\text{B}}{\text{+m}}_{\text{C}}{y}_{\text{C}}}{m_{\text{A}}+m_{\text{B}}+m_{\text{C}}}$ (II)

Here, $y_{\text{CM}}$ is the $y$ coordinate of center of mass system, $y_{\text{A}}$ is the $y$ coordinate of center of mass, $y_{\text{B}}$ is the $y$ coordinate of center of mass of $\text{B}$ and $y_{\text{C}}$ is the $y$ coordinate of center of mass of

C.

Conclusion:

Substitute $200\text{g}$ for $m_{\text{A}}$ , $300\text{g}$ for $m_{\text{A}}$, $500\text{g}$ for $m_{\text{C}}$, $-3.0\text{cm}$ for $x_{\text{A}}$ , $6.0\text{cm}$ for $x_{\text{B}}$ and $-5.0\text{cm}$ for $x_{\text{C}}$ in equation (I) to get $x_{\text{CM}}$.

$\begin{array}{c}{x}_{\text{CM}}=\frac{\left(200\text{g}\right)\left(-3.0\text{cm}\right)+\left(300\text{g}\right)\left(6.0\text{cm}\right)\text{+}\left(500\text{g}\right)\left(-5.0\text{cm}\right)}{200\text{g}+300\text{g}+500\text{g}}\\ =-1.3\text{cm}\end{array}$

Substitute $200\text{g}$ for $m_{\text{A}}$ , $300\text{g}$ for $m_{\text{A}}$, $500\text{g}$ for $m_{\text{C}}$, $5.0\text{cm}$ for $x_{\text{A}}$ , $0\text{cm}$ for $x_{\text{B}}$ and $-4.0\text{cm}$ for $x_{\text{C}}$ in equation (II) to get $y_{\text{CM}}$.

$\begin{array}{c}{x}_{\text{CM}}=\frac{\left(200\text{g}\right)\left(5.0\text{cm}\right)+\left(300\text{g}\right)\left(0\text{cm}\right)\text{+}\left(500\text{g}\right)\left(-4.0\text{cm}\right)}{200\text{g}+300\text{g}+500\text{g}}\\ =-1.0\text{cm}\end{array}$

Therefore, the $x$ and $y$ coordinates of center of mass of system is $\left(x,y\right)=\left(-1.3\text{cm,}-\text{1}\text{.0}\text{cm}\right)$.