Chapter 19.2, Problem 1PP

Determine the angular momentum of the 100-kg disk or rod about point G and about point O

To determine

The angular momentum of the disk or rod about point $G$ and point $O$.

Answer to Problem 1PP

1. (a) The angular momentum of the disk or rod about point $G$ and point $O$ is $\text{600}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)\text{and}1800\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)$ respectively.
2. (b) The angular momentum of the disk or rod about point $G$ and point $O$ is $\text{300}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)\text{and}1200\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)$ respectively.
3. (c) The angular momentum of the disk or rod about point $G$ and point $O$ is $\text{800}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)\text{and}2400\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)$ respectively.
4. (d) The angular momentum of the disk or rod about point $G$ and point $O$ is $\text{400}\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{ccw}\right)\text{and}700\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{ccw}\right)$ respectively.

Explanation of Solution

Given:

The mass of the disk, $m=100\text{kg}$.

Conclusion:

(a)

Express the angular momentum of the disk about point $G$.

$H_G=\left[\frac{1}{2}m{r}^2\right]\omega$ (I)

Here, radius of disk is $r$ and angular velocity is $\omega$.

Substitute $\text{100}\text{kg}$ for $m$, $\text{2}\text{m}$ for $r$ and $\text{3}\text{rad}/\text{s}\left(\text{cw}\right)$ for $\omega$ in Equation (I).

$\begin{array}{c}{H}_G=\left[\frac{1}{2}\left(100\right){\left(2\right)}^2\right]\left(3\right)\\ =600\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)\end{array}$

Hence, the angular momentum of the disk about point $G$ is $600\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)$.

Express the angular momentum of the disk about point $O$.

$H_O=\left[\frac{1}{2}m{r}^2+m{r}^2\right]\omega$ (II)

Substitute $\text{100}\text{kg}$ for $m$, $\text{2}\text{m}$ for $r$ and $\text{3}\text{rad}/\text{s}\left(\text{cw}\right)$ for $\omega$ in Equation (II).

$\begin{array}{c}{H}_O=\left[\frac{1}{2}\left(100\right){\left(2\right)}^2+100{\left(2\right)}^2\right]\left(3\right)\\ =\left(200+400\right)3\\ =1800\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)\end{array}$

Hence, the angular momentum of the disk about point $O$ is $1800\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)$.

(b)

Express the angular momentum of the rod about point $G$.

$H_G=\left[\frac{1}{12}m{l}^2\right]\omega$ (III)

Here, length of the rod is $l$.

Substitute $\text{100}\text{kg}$ for $m$, $\text{3}\text{m}$ for $l$ and $\text{4}\text{rad}/\text{s}\left(\text{cw}\right)$ for $\omega$ in Equation (III).

$\begin{array}{c}{H}_G=\left[\frac{1}{12}\left(100\right){\left(3\right)}^2\right]\left(4\right)\\ =300\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)\end{array}$

Hence, the angular momentum of the rod about point $G$ is $300\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)$.

Express the angular momentum of the rod about point $O$.

$H_O=\left[\frac{1}{12}m{l}^2+m{\left(l/2\right)}^2\right]\omega$ (IV)

Substitute $\text{100}\text{kg}$ for $m$, $\text{3}\text{m}$ for $l$ and $\text{4}\text{rad}/\text{s}\left(\text{cw}\right)$ for $\omega$ in Equation (IV).

$\begin{array}{c}{H}_O=\left[\frac{1}{12}\left(100\right){\left(3\right)}^2+100{\left(3/2\right)}^2\right]\left(4\right)\\ =\left(75+225\right)4\\ =1200\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)\end{array}$

Hence, the angular momentum of the rod about point $O$ is $1200\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)$.

(c)

Express the angular momentum of the disk about point $G$.

$H_G=\left[\frac{1}{2}m{r}^2\right]\omega$ (V)

Substitute $\text{100}\text{kg}$ for $m$, $\text{2}\text{m}$ for $r$ and $4\text{rad}/\text{s}\left(\text{cw}\right)$ for $\omega$ in Equation (V).

$\begin{array}{c}{H}_G=\left[\frac{1}{2}\left(100\right){\left(2\right)}^2\right]\left(4\right)\\ =800\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)\end{array}$

Hence, the angular momentum of the disk about point $G$ is $800\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)$.

Express the angular momentum of the disk about point $O$.

$H_O=\left[\frac{1}{2}m{r}^2+m{r}^2\right]\omega$ (VI)

Substitute $\text{100}\text{kg}$ for $m$, $\text{2}\text{m}$ for $r$ and $4\text{rad}/\text{s}\left(\text{cw}\right)$ for $\omega$ in Equation (VI).

$\begin{array}{c}{H}_O=\left[\frac{1}{2}\left(100\right){\left(2\right)}^2+100{\left(2\right)}^2\right]\left(4\right)\\ =\left(200+400\right)4\\ =2400\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)\end{array}$

Hence, the angular momentum of the disk about point $O$ is $2400\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{cw}\right)$.

(d)

Express the angular momentum of the rod about point $G$.

$H_G=\left[\frac{1}{12}m{l}^2\right]\omega$ (VII)

Substitute $\text{100}\text{kg}$ for $m$, $\text{4}\text{m}$ for $l$ and $\text{3}\text{rad}/\text{s}\left(\text{ccw}\right)$ for $\omega$ in Equation (VII).

$\begin{array}{c}{H}_G=\left[\frac{1}{12}\left(100\right){\left(4\right)}^2\right]\left(3\right)\\ =400\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{ccw}\right)\end{array}$

Hence, the angular momentum of the rod about point $G$ is $400\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{ccw}\right)$.

Express the angular momentum of the rod about point $O$.

$H_O=\left[\frac{1}{12}m{l}^2+m{\left(l_{OG}\right)}^2\right]\omega$ (VIII)

Here, length between $O\text{and}G$ is $l_{OG}$.

Substitute $\text{100}\text{kg}$ for $m$, $\text{4}\text{m}$ for $l$, $\text{1}\text{m}$ for $l_{OG}$ and $\text{3}\text{rad}/\text{s}\left(\text{ccw}\right)$ for $\omega$ in Equation (VIII).

$\begin{array}{c}{H}_O=\left[\frac{1}{12}\left(100\right){\left(4\right)}^2+100{\left(1\right)}^2\right]\left(3\right)\\ =\left(233.333\right)3\\ =700\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{ccw}\right)\end{array}$

Hence, the angular momentum of the rod about point $O$ is $700\text{kg}\cdot {\text{m}}^{\text{2}}/\text{s}\left(\text{ccw}\right)$.