Chapter 8, Problem 6P

(a)

To determine

The rotational inertia of the arrangement.

Answer to Problem 6P

The rotational inertia of the arrangement is $\underset{\_}{1.5\text{kg}\cdot {\text{m}}^2}$.

Explanation of Solution

The arrangement of the point mass is given in the Figure (a). Given that the mass of the point masses are $3.0\text{kg}$ each, and the side length of the square is $0.50\text{m}$.

Write the expression for the rotational inertia of the given system.

$I=m\left(r_{\text{A}}^2+r_{\text{B}}^2+r_{\text{C}}^2+r_{\text{D}}^2\right)$

Here, $m$ is the mass of each point, $r_{\text{A}}$, $r_{\text{B}}$, $r_{\text{C}}$, and $r_{\text{D}}$ are the distance between the axis of rotation and the point mass A, B, C, and D respectively.

In the given arrangement, the axis of rotation passes through B and C. Thus $r_{\text{B}}$ and $r_{\text{C}}$ are zero.

Conclusion:

Substitute $3.0\text{kg}$ for $m$, $0.50\text{m}$ for $r_{\text{A}}$, $0$ for $r_{\text{B}}$, $0$ for $r_{\text{C}}$, and $0.50\text{m}$ for $r_{\text{D}}$ to find $I$.

$\begin{array}{c}I=\left(3.0\text{kg}\right)\left({\left(0.50\text{m}\right)}^2+0^2+0^2+{\left(0.50\text{m}\right)}^2\right)\\ =1.5\text{kg}\cdot {\text{m}}^2\end{array}$

Therefore, the rotational inertia of the arrangement is $\underset{\_}{1.5\text{kg}\cdot {\text{m}}^2}$.

(b)

To determine

The rotational inertia of the arrangement.

Answer to Problem 6P

The rotational inertia of the arrangement is $\underset{\_}{0.75\text{kg}\cdot {\text{m}}^2}$.

Explanation of Solution

The arrangement of the point mass is given in the Figure (b). Given that the mass of the point masses are $3.0\text{kg}$ each, and the side length of the square is $0.50\text{m}$.

Write the expression for the rotational inertia of the given system.

$I=m\left(r_{\text{A}}^2+r_{\text{B}}^2+r_{\text{C}}^2+r_{\text{D}}^2\right)$

In the given arrangement, the axis of rotation passes through A and C. Thus $r_{\text{A}}$ and $r_{\text{C}}$ are zero. The distance $r_{\text{B}}$ and $r_{\text{D}}$ are $\left(0.50/\sqrt{2}\right)\text{m}$ each.

Conclusion:

Substitute $3.0\text{kg}$ for $m$, $0$ for $r_{\text{A}}$, $\left(0.50/\sqrt{2}\right)\text{m}$ for $r_{\text{B}}$, $0$ for $r_{\text{C}}$, and $\left(0.50/\sqrt{2}\right)\text{m}$ for $r_{\text{D}}$ to find $I$.

$\begin{array}{c}I=\left(3.0\text{kg}\right)\left(0^2+{\left(\frac{0.50}{\sqrt{2}}\text{m}\right)}^2+0^2+{\left(\frac{0.50}{\sqrt{2}}\text{m}\right)}^2\right)\\ =0.75\text{kg}\cdot {\text{m}}^2\end{array}$

Therefore, the rotational inertia of the arrangement is $\underset{\_}{0.75\text{kg}\cdot {\text{m}}^2}$.

(c)

To determine

The rotational inertia of the arrangement.

Answer to Problem 6P

The rotational inertia of the arrangement is $\underset{\_}{1.5\text{kg}\cdot {\text{m}}^2}$.

Explanation of Solution

The arrangement of the point mass is given in the Figure (c). Given that the mass of the point masses are $3.0\text{kg}$ each, and the side length of the square is $0.50\text{m}$.

Write the expression for the rotational inertia of the given system.

$I=m\left(r_{\text{A}}^2+r_{\text{B}}^2+r_{\text{C}}^2+r_{\text{D}}^2\right)$

In the given arrangement, the axis of rotation passes through the center of the square. The distance $r_{\text{A}}$, $r_{\text{B}}$, $r_{\text{C}}$, and $r_{\text{D}}$ are $\left(0.50/\sqrt{2}\right)\text{m}$ each.

Conclusion:

Substitute $3.0\text{kg}$ for $m$, $0.50\text{m}$ for $r_{\text{A}}$, $0$ for $r_{\text{B}}$, $0$ for $r_{\text{C}}$, and $0.50\text{m}$ for $r_{\text{D}}$ to find $I$.

$\begin{array}{c}I=\left(3.0\text{kg}\right)\left({\left(\frac{0.50}{\sqrt{2}}\text{m}\right)}^2+{\left(\frac{0.50}{\sqrt{2}}\text{m}\right)}^2+{\left(\frac{0.50}{\sqrt{2}}\text{m}\right)}^2+{\left(\frac{0.50}{\sqrt{2}}\text{m}\right)}^2\right)\\ =1.5\text{kg}\cdot {\text{m}}^2\end{array}$

Therefore, the rotational inertia of the arrangement is $\underset{\_}{1.5\text{kg}\cdot {\text{m}}^2}$.