Calculate the angular momentum for a rotating disk, sphere, and rod: (a) A uniform disk of mass $\mathbf{13}\mathbf{}\mathit{k}\mathit{g}$, thickness $\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathit{m}$and radius$\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}\mathit{m}$is located at the origin, oriented with its axis along the$\mathit{y}$ axis. It rotates clockwise around its axis when viewed form above (that is, you stand at a point on the $\mathbf{+}\mathit{y}$ axis and look toward the origin at the disk). The disk makes one complete rotation every$\mathbf{0}\mathbf{.}\mathbf{6}\mathbf{}\mathit{s}$ . What is the rotational angular momentum of the disk? What is the rotational kinetic energy of the disk? (b) A sphere of uniform density, with mass$\mathbf{22}\mathbf{}\mathit{k}\mathit{g}$ and radius$\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{}\mathit{m}$ is located at the origin and rotates around an axis parallel with the$\mathit{x}$ axis. If you stand somewhere on the $\mathbf{+}\mathit{x}$axis and look toward the origin at the sphere, the sphere spins counterclockwise. One complete revolution takes$\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{}\mathit{s}$ .What is the rotational angular momentum of the sphere? What is the rotational kinetic energy of the sphere? (c) A cylindrical rod of uniform density is located with its center at the origin, and its axis along the$\mathit{z}$ axis. Its radius is$\mathbf{0}\mathbf{.}\mathbf{06}\mathbf{}\mathit{m}$ its length is$\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{}\mathit{m}$ and its mass is $\mathbf{}\mathbf{5}\mathbf{}\mathit{k}\mathit{g}$It makes one revolution every $\mathbf{0}\mathbf{.}\mathbf{03}\mathbf{}\mathit{s}$If you stand on the $\mathbf{+}\mathit{x}$axis and look toward the origin at the rod, the rod spins clockwise. What is the rotational angular momentum of the rod? What is the rotational kinetic energy of the rod?

The rotating analogue of linear momentum is angular momentum (also known as moment of momentum or rotational momentum). Because it is a conserved quantity—the total angular momentum of a closed system remains constant—it is a significant quantity in physics.

In terms of its moment of inertia, the rotational angular momentum of an object rotating with angular speed is as follows:

$L_{rot}=I\omega$

Here,$I$is the moment of inertia of the object.

Angular velocity of the disk is calculated in $rad/s$as follows,

$\begin{array}{rcl}\omega & =& Irev/0.6s\\ & =& \left(Irev/0.6s\right)\left(\frac{2\pi rad}{1rev}\right)\\ & =& 10.47rad/s\\ & & \end{array}$

Moment of inertia of a uniform disk is calculated as follows:

$I_{disk}=\frac{1}{2}M{R}^2$

Substitute $13kg$for $M$and $0.02m$for $R$

$\begin{array}{rcl}{I}_{disk}& =& \frac{1}{2}M{R}^2\\ & =& \frac{1}{2}\left(13kg\right){\left(0.2m\right)}^2\\ & =& 0.26kg·m^2\\ & & \end{array}$

Now, the expression for the rotational energy momentum of the disk is,

$L_{not}=I_{disk}\omega$

Substitute $10.47rad/s$for $\omega$and $0.26kg·m^2$for $I_{disk}$.

$\begin{array}{rcl}{L}_{not}& =& I_{disk}\omega \\ & =& \left(0.26kg·m^2\right)\left(10.47rad/s\right)\\ & =& 2.72kg·m^2/s\\ & & \end{array}$

Therefore, the rotational angular momentum of the disk is$2.72kg·m^2/s$

The formula to find the rotational kinetic energy of an object is,

$k_{not}=\frac{1}{2}I{\omega }^2$

Substitute $10.47rad/s$for $\omega$and $0.26kg·m^2$for $I_{disk}$

$\begin{array}{rcl}{K}_{dot}& =& \frac{1}{2}{I}_{disk}{\omega }^2\\ & =& \frac{1}{2}\left(0.26kg·m^2\right){\left(10.47rad/s\right)}^2\\ & =& 14.25J\\ & & \end{array}$

Therefore, the rotational kinetic energy of the disk of$14.25J$ .

Angular speed of the sphere in $rad/s$is,

$\begin{array}{rcl}\omega & =& 1rev/0.5s\\ & =& \left(1rev/0.5s\right)\left(\frac{2\pi rad}{1rev}\right)\\ & =& 12.57rad/s\\ & & \end{array}$

Moment of inertia of a sphere of uniform density is calculated as follows

$I_{sphere}=\frac{2}{5}M{R}^2$

Substitute $22kg$for $M$and $0.7m$ for $R$

$\begin{array}{rcl}{I}_{sphere}& =& \frac{2}{5}M{R}^2\\ & =& \frac{2}{5}\left(22kg\right){\left(0.7m\right)}^2\\ & =& 4.312kg·m^2\\ & & \end{array}$

The rotational angular momentum of the sphere is calculated as follows,

$L_{not}=I_{sphere}\omega$

Substitute $12.57rad/s$for $\omega$and $4.312kg·m^2$for $I_{sphere}$.

$\begin{array}{rcl}{L}_{rot}& =& I_{sphere}\omega \\ & =& \left(4.312kg·m^2\right)\left(12.57rad/s\right)\\ & =& 54.2kg·m^2/s\\ & & \end{array}$

Therefore, the rotational angular momentum of the sphere is$54.2kg·m^2/s$

The rotational kinetic energy of the sphere is calculated as follows.

$K_{not}=\frac{1}{2}{I}_{sphere}{\omega }^2$

Substitute $12.57rad/s$for $\omega$and $4.312kg·m^2$ for $I_{sphere}$

$$\begin{gathered} =\frac{1}{2}\left(4.312kg·m^2\right){\left(12.57rad/s\right)}^2 \\ =340J \end{gathered}$$

Therefore, the rotational kinetic energy of the sphere is$340J$ .

Angular speed of the cylinder rod in $rad/s$ is,

$\begin{array}{rcl}\omega & =& 1rev/0.03s\\ & =& \left(1rev/0.03s\right)\left(\frac{2\pi rad}{1rev}\right)\\ & =& 209.33rad/s\\ & & \end{array}$

Moment of inertia of a cylindrical rod of uniform density is calculated as follows:

$I_{sphere}=\frac{1}{12}M{I}^2$

Substitute $5kg$for $M$and $0.7m$for $I$

$\begin{array}{rcl}{I}_{cylinder}& =& \frac{1}{12}M{I}^2\\ & =& \frac{1}{12}\left(5kg\right){\left(0.7m\right)}^2\\ & =& 0.204kg·m^2\\ & & \end{array}$

The rotational angular momentum of the cylindrical rod can be calculated as follows.

$L_{rot}=I_{cylinder}\omega$

Substitute $209.33rad/s$for $\omega$and $0.204kg·m^2$ for $I_{cylinder}$.

$\begin{array}{rcl}{L}_{rot}& =& I_{cylinder}\omega \\ & =& \left(0.204kg·m^2\right)\left(209.33rad/s\right)\\ & =& 42.7kg·m^2/s\\ & & \end{array}$

Therefore, the rotational angular momentum of the cylinder rod is$42.7kg·m^2/s$ .

The rotational kinetic energy of the cylindrical rod is calculated as follows,

$K_{rot}=\frac{1}{2}{I}_{cylinder}{\omega }^2$

Substitute$209.33rad/s$for$\omega$ and$0.204kg.m^2$ for$I_{cylinder}$.

$\begin{array}{rcl}{K}_{rot}& =& \frac{1}{2}{I}_{cylinder}{\omega }^2\\ & =& \frac{1}{2}\left(0.204kg·m^2\right){\left(209.33rad/s\right)}^2\\ & =& 4469.5J\\ & & \end{array}$

Therefore, the rotational kinetic energy of the cylindrical rod is$4469.5J$ .