A women with mass $\mathbf{50}\mathbf{}\mathbf{kg}$is standing on the rim of a large disk that is rotating at $\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{rev}\mathbf{/}\mathbf{s}$about an axis through its center. The disk has mass $\mathbf{110}\mathbf{}\mathbf{kg}$and radius $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$. Calculate the magnitude of the total angular momentum of the woman-disk system. (Assume that you can treat the woman as a point.)

The mass of the woman is$m=50\text{kg}$.

The angular speed of the rim is as follows:

localid="1667976336551" $$\begin{gathered} \omega =\text{0.80}\text{rev/s} \\ =\text{0.80}\text{rev/s}\times \frac{\text{2}\pi \text{rad}}{\text{1}} \\ =\text{5.02}\text{rad/s} \end{gathered}$$

The mass of the disk is $m_d=\text{110}\text{kg}$.

The radius of the disk is $r_d=\text{4.0}\text{m}$.

The law of conservation of angular momentum states that the product of moment of inertial and the angular velocity of a system remains constant.

${\text{I}}_{\mathbf{i}}{\mathbf{\omega }}_{\mathbf{i}}\mathbf{=}{\mathbf{I}}_{\mathbf{f}}{\mathbf{\omega }}_{\mathbf{f}}\mathbf{=}\mathbf{constant}$

The disk has a moment of inertial which is given as-

Thus, disk’s moment of inertial is$\text{880}{\text{kgm}}^{\text{2}}$.

The woman’s moment of inertia is given as:

Thus, the women’s moment of inertia is $\text{800}{\text{kgm}}^{\text{2}}$.

The moment of inertia of the system is solved as follows:

$$\begin{gathered} I_t=I_D+I_{\omega } \\ =\text{800}{\text{kgm}}^{\text{2}}+\text{880}{\text{kgm}}^{\text{2}} \\ =\text{1680}{\text{kgm}}^{\text{2}} \end{gathered}$$

The angular momentum can be obtained as follows:

$$\begin{gathered} L=I_t\omega \\ =\text{1680}{\text{kgm}}^{\text{2}}\times \text{5.02}\text{m}/\text{s} \\ =\text{8433.6}{\text{kgm}}^{\text{2}}\text{/s} \end{gathered}$$

Thus, for the woman-disk system, the angular momentum is $\text{8433.6}{\text{kgm}}^{\text{2}}\text{/s}$.