What is the angular momentum vector of the diameter rotating disk?

The expression for the angular velocity is as follow:

$\omega =2\pi N\mathrm{rad}$

Here, $N$is the number of turns.

The frequency is

$f=600rpm$

The expression for the moment of inertia is as follow:

$I=\frac{1}{2}M{R}^2$

Here, $M$is mass and $R$is the radius.

The expression for the angular momentum is as follow:

$L=I\omega$

The expression for the relation bewteenn the diameter and the radius is as follow:

$R=\frac{d}{2}$

Here,$R$is the radius and$d$is the diameter

Convert the $rpm$into $rps$as follow :

$f=600\mathrm{rpm}\left(\frac{1\min }{60\mathrm{s}}\right)$

$=10rps$

Solve the angular velocity by substituting $10rps$for $N$in the equation $\omega =2\pi N\mathrm{rad}$

$\omega =2\pi \left(10\mathrm{rps}\right)\mathrm{rad}$

$=62.83\mathrm{rad}/\mathrm{s}$

Calculate the value of the radius by substituting $4.0$cm and for $d$in the equation $R=\frac{d}{2}$

$R=\frac{4.0\mathrm{cm}}{2}$

$=\frac{4.0\mathrm{cm}\left(\frac{{10}^{-2}\mathrm{m}}{1.00\mathrm{cm}}\right)}{2}$

$=2\times {10}^{-2}\mathrm{m}$

Now, substitute $2.0kg$for $M$and $2\times {10}^{-2}\mathrm{m}$and $R$in the equation $I=\frac{1}{2}M{R}^2$

$I=\frac{1}{2}\left(2.0\mathrm{kg}\right){\left(2\times {10}^{-2}\mathrm{m}\right)}^2$

$=4\times {10}^{-4}\mathrm{kg}\cdot {\mathrm{m}}^2$

Now, substitute $=4\times {10}^{-4}\mathrm{kg}\cdot {\mathrm{m}}^2$for $I$and $62.83\mathrm{rad}/\mathrm{s}$for $\omega$in the equation $L=l\omega$

$L=\left(4\times {10}^{-4}\mathrm{kg}\cdot {\mathrm{m}}^2\right)(62.83\mathrm{rad}/\mathrm{s})$

$=0.025\mathrm{kg}\cdot {\mathrm{m}}^2\cdot {\mathrm{s}}^{-1}$