In Figure 11.102, the uniform solid disk has mass $\mathbf{0}\mathbf{.}\mathbf{4}\mathit{k}\mathit{g}$(moment of inertia$\mathit{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{M}{\mathit{R}}^{\mathbf{2}}$). At the instant shown, the angular velocity is 20 rad/sinto the page. (a) At this instant, what are the magnitude and direction of the angular momentum about the center of the disk? (b) At a time$\mathbf{0}\mathbf{.}\mathbf{2}\mathit{s}$later, what are the magnitude and direction of the angular momentum about the center of the disk? (c) At this later time, what are the magnitude and direction of the angular velocity?

It is the property of a rotating body given by the product of the moment of inertia and the angular velocity of the rotating object. It is a vector quantity, which implies that here along with magnitude, the direction is also considered.

b. Due to the presence of an external force a torque is generated and it is expressed as follows:

$\overrightarrow{\tau }=\frac{d\overrightarrow{L}}{dt}$

Here, $d\overrightarrow{L}$represents the change in angular momentum.

The expression for torque is given as follows:

$\overrightarrow{\tau }=\overrightarrow{r}\times \overrightarrow{F}$

Here, $\overrightarrow{r}$is radius of the disk and $\overrightarrow{F}$is the applied force.

The force is applied perpendicular to the radius vector thus the torque is maximum and it is directed along the negative$z-$axis.

${\tau }_2=rF$

Substitute $8cm$for $R$and $10N$for$F.$

$\begin{array}{rcl}{\tau }_2& =& -\left(8cm\right)\left(\frac{{10}^{-2}m}{1cm}\right)\left(10N\right)\\ & =& -0.8N.\\ & & \end{array}$

Rearrange the expression$\overrightarrow{\tau }=\frac{d\overrightarrow{L}}{dt}$for $d\overrightarrow{L}.$

$\begin{array}{rcl}d\overrightarrow{L}& =& \overrightarrow{\tau }dt\\ {\overrightarrow{L}}_2-{\overrightarrow{L}}_1& =& \overrightarrow{\tau }dt\\ {\overrightarrow{L}}_2& =& {\overrightarrow{L}}_1+\overrightarrow{\tau }dt\\ & & \end{array}$

Here, ${\overrightarrow{L}}_2$represents the angular momentum of the disk at an instant of time$0.2s$later.

Substitutekg.meter square per second for for$\overrightarrow{\tau }$and $0.2s$for $dt.$

Therefore, the magnitude of the angular momentum at a time$0.2s$later is 0.1856 kg.meter squre per second and it is directed along the negative$z-$direction.

c.The angular velocity of the disk after is given as follows:

$L_2=\left(\frac{1}{2}M{R}^2\right){\omega }_2$

Substitute -0.1856 kg.meter squre per second for $L_2,0.4kg$for$M,$and $8cm$for$R.$

$-0.1856=\left(\frac{1}{2}(0.4kg){\left(8cm\right)}^2{\left(\frac{{10}^{-2}m}{1cm}\right)}^2\right){\omega }_2$

$\begin{array}{rcl}{\omega }_2& =& \frac{-0.1856}{0.00128}\\ & =& -145\\ & & \end{array}$

Therefore, the magnitude of the angular velocity of the disk at time $0.2s$later is 145 rad/s and it is directed along the negative $z-$axis.