Question

A woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.80 rev/s about an axis through its center. The disk has mass 270 kg and radius 4.0 m.

Calculate the magnitude of the total angular momentum of the woman–disk system. (Assume that you can treat the woman as a point.)

Express your answer to two significant figures and include the appropriate units.

Answer 1

Concepts and reason

The concepts used to solve this problem are the moment of inertia and angular momentum.

Use the concept of rotating body to calculate the moment of inertia.

Use the concepts of moment of inertia and angular velocity to calculate the angular momentum.

Fundamentals

The expression for angular moment of inertia is given below:

$I = m{r^2}$

Here, $I$ is the moment of inertia, $m$ is the mass of the object, and $r$ is the distance of the object from the rotation axis.

The expression for angular momentum is as follows:

$L = I\omega$

Here, $L$ is the angular momentum, $I$ is the momentum of inertia, and $\omega$ is the angular velocity.

The expression for the moment of inertia of the disk is given below:

${I_d} = \frac{1}{2}{m_d}{r^2}$

Here, ${I_d}$ is the moment of inertia of the disk, ${m_d}$ is the mass of the disk, and $r$ is the radius of the disk.

The moment of inertia of the large disk is given below:

${I_d} = \frac{1}{2}{m_d}{r^2}$

Substitute $270\,{\rm{kg}}$for ${m_d}$ and $4.0\,{\rm{m}}$ for $r$.

$\begin{array}{c}\\{I_d} = \frac{1}{2}\left( {270\,{\rm{kg}}} \right){\left( {4.0\,{\rm{m}}} \right)^2}\\\\ = 2160\,{\rm{kg}} \cdot {{\rm{m}}^2}\\\end{array}$

The moment of inertia of the woman is given below:

${I_w} = {m_w}{r^2}$

Here, ${I_w}$ is the moment of inertia of the woman, ${m_w}$ is the mass of the woman, and r is the radius of the disk.

Substitute $50\,{\rm{kg}}$ for ${m_w}$ and $4.0\,{\rm{m}}$ for $r$.

$\begin{array}{c}\\{I_w} = \left( {50\,{\rm{kg}}} \right){\left( {4.0\,{\rm{m}}} \right)^2}\\\\ = 800\,{\rm{kg}} \cdot {{\rm{m}}^2}\\\end{array}$

The total moment of inertia of the woman-disk system is as follows:

${I_{total}} = {I_d} + {I_w}$

Substitute $4320\,{\rm{kg}} \cdot {{\rm{m}}^2}$ for ${I_d}$ and $800\,{\rm{kg}} \cdot {{\rm{m}}^2}$ for ${I_w}$.

$\begin{array}{c}\\{I_{total}} = 2160\,{\rm{kg}} \cdot {{\rm{m}}^2} + 800\,{\rm{kg}} \cdot {{\rm{m}}^2}\\\\ = 2960\,{\rm{kg}} \cdot {{\rm{m}}^2}\\\end{array}$

The expression for angular momentum is given below:

$L = {I_{total}}\omega$

Substitute $2960\,{\rm{kg}} \cdot {{\rm{m}}^2}$ for ${I_{total}}$ and $0.80\,{\rm{rev}} \cdot {{\rm{s}}^{ - 1}}$ for $\omega$.

$\begin{array}{c}\\L = \left( {2960\,{\rm{kg}} \cdot {{\rm{m}}^2}} \right)\left( {0.80\,{\rm{rev}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {\frac{{2\pi {\rm{rad}}}}{{1.0\,{\rm{rev}}}}} \right)\\\\ = \left( {2960\,{\rm{kg}} \cdot {{\rm{m}}^2}} \right)\left( {5.024\,{\rm{rad}} \cdot {{\rm{s}}^{ - 1}}} \right)\\\\ = 14871.0\,{\rm{kg}} \cdot {{\rm{m}}^2} \cdot {{\rm{s}}^{ - 1}}\\\\ \approx 1.5 \times {10^4}\,{\rm{kg}} \cdot {{\rm{m}}^2} \cdot {{\rm{s}}^{ - 1}}\\\end{array}$

Ans:

The magnitude of the angular momentum of the woman-disk system is $1.5 \times 1{0^4}\,kg \cdot {m^2} \cdot {s^{ - 1}}$.