A uniform rod rotates in a horizontal plane about a vertical axis through one end. The rod is 6.00 m long, weighs 240 rev/minand rotates atCalculate (a) its rotational inertia about the axis of rotation? (b) the magnitude of its angular momentum about that axis?

Length of rod is 6.00 m

Weight is 10.0 N,

Rotational speed is 240 rev/min .

Using the formula $\mathbf{L}\mathbf{=}\mathbf{I\omega }$and parallel axis theorem,localid="1661154101186" $\mathbf{}\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{com}}\mathbf{+}{\mathbf{Md}}^{\mathbf{2}}$, find rotational inertia about the axis of rotation and the magnitude of angular momentum about that axis.

Formula is as follow:

localid="1661154104672" $\mathbf{}\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{com}}\mathbf{+}{\mathbf{Md}}^{\mathbf{2}}$

Where, I is moment of inertia and M is mass and d is distance.

According to parallel axis theorem,

$\begin{array}{rcl}I& =& I_{com}+M{d}^2\\ I& =& \frac{1}{12}M{L}^2+M{\left(\frac{L}{2}\right)}^2\\ I& =& \frac{1}{12}M{L}^2+\frac{1}{4}M{L}^2\\ I& =& \frac{1}{3}M{L}^2\\ I& =& \frac{1}{3}\left(\frac{\text{10.0}}{\text{9.8}}\text{kg}\right){\left(\text{6.00}\text{m}\right)}^{\text{2}}\\ I& =& 12.2kg.m^2\\ & & \end{array}$

Hence, rotational inertia about the axis of rotation is 12.2 kg.m².

It is given that,

$f=240rev/min$

So,

$$\begin{gathered} \omega =\left(\text{240}\frac{\text{rev}}{\text{min}}\right)\left(\frac{\text{2}\pi }{\text{60}}\right) \\ \omega =25.1rad/s \end{gathered}$$

Therefore,

$$\begin{gathered} L=I\omega \\ L=\left(12.2kg.m^2\right)\left(25.1rad/s\right) \\ L=306kg.m^2/s \end{gathered}$$

Hence, the magnitude of angular momentum about the axis of rotation is 306 kg.m²/s.

Therefore, using the parallel axis theorem, the rotational inertia can be found about the given axis. Also, using the formula for the angular momentum in terms of rotational inertia and angular velocity, the angular momentum of the object can be found.