Chapter 10, Problem 67P

Two astronauts (Fig. P10.67), each having a mass of 75.0 kg, are connected by a 10.0-m rope of negligible mass. They are isolated in space, orbiting their center of mass at speeds of 5.00 m/s. Treating the astronauts as particles, calculate

1. (a) the magnitude of the angular momentum of the two-astronaut system and
2. (b) the rotational energy of the system. By pulling on the rope, one astronaut shortens the distance between them to 5.00 m.
3. (c) What is the new angular momentum of the system?
4. (d) What are the astronauts’ new speeds?
5. (e) What is the new rotational energy of the system?
6. (f) How much chemical potential energy in the body of the astronaut was converted to mechanical energy in the system when he shortened the rope?

Figure P10.67 Problems 67 and 68.

To determine

The magnitude of angular momentum of the two astronaut system.

Answer to Problem 67P

The magnitude of angular momentum of the two astronaut system is $\underset{\_}{3.75\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}}$.

Explanation of Solution

Consider the figure given below.

Write the expression for the magnitude of angular momentum.

  $\begin{array}{c}\left|\overrightarrow{L}\right|=m\left|\overrightarrow{r}\times \overrightarrow{v}\right|\\ =mr\upsilon \sin \theta \end{array}$        (I)

Here, $m$ is the mass of the astronaut, $r$ is the distance from center of mass to the astronaut, $\upsilon$ is the speed.

From figure $r$ and $\upsilon$ are perpendicular to each other, hence equation (I) will be.

  $\begin{array}{c}L=mr\upsilon \sin \left(90°\right)\\ =mr\upsilon \end{array}$        (II)

Conclusion:

Substitute, $2\times 7.50\text{kg}$ for $m$, $5.00\text{m}$ for $r$, $5.00\text{m}/\text{s}$ for $\upsilon$ in equation (II) to find the magnitude of angular momentum.

  $\begin{array}{c}L=2\times 7.50\text{kg}\times 5.00\text{m}/\text{s}\times 5.00\text{m}\\ =3.75\times {10}^3{\text{kg×m}}^2/\text{s}\end{array}$

Therefore, the magnitude of angular momentum of the two astronaut system is $\underset{\_}{3.75\times {10}^3\text{kg}\cdot {\text{m}}^2/\text{s}}$.

To determine

The rotational kinetic energy of the system.

Answer to Problem 67P

The rotational kinetic energy of the system is $\underset{\_}{1.88\text{kJ}}$.

Explanation of Solution

The total rotational kinetic energy is the sum of the kinetic energy of two astronauts.

Write the expression for the rotational kinetic energy.

  $K_i=\frac{1}{2}{m}_1{\upsilon }_1{}_i^2+\frac{1}{2}{m}_2{\upsilon }_2{}_i^2$        (III)

Here, $m_1$ is the mass of the first astronaut, $m_2$ is the mass of the second astronaut, ${\upsilon }_{1i}$ is the initial orbiting speed of first astronaut, and ${\upsilon }_{2i}$ is the initial orbiting speed of the second astronaut.

Conclusion:

The mass, and speed of two astronauts are same.

Substitute, $7.50\text{kg}$ for $m_1$, and $m_2$, $5.00\text{m}/\text{s}$ for ${\upsilon }_{1i}$, and ${\upsilon }_{2i}$ in equation (III) to find the rotational kinetic energy of the system.

  $\begin{array}{c}{K}_i=\frac{1}{2}\left(7.50\text{kg}\right){\left(5.00\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(7.50\text{kg}\right){\left(5.00\text{m}/\text{s}\right)}^2\\ =2\times \frac{1}{2}\left(7.50\text{kg}\right){\left(5.00\text{m}/\text{s}\right)}^2\\ =1.88\text{kJ}\end{array}$

Therefore, the rotational kinetic energy of the system is $\underset{\_}{1.88\text{kJ}}$.

To determine

The angular momentum when one of the astronaut shortens the distance between them to $5.00\text{m}$.

Answer to Problem 67P

The angular momentum when one of the astronaut shortens the distance between them to $5.00\text{m}$ is $\underset{\_}{3750\text{kg}\cdot {\text{m}}^2/\text{s}}$.

Explanation of Solution

Even if the distance between the astronauts changed, the tension of the rope not generating ant torque about the center of mass. Since there is no change in torque the angular momentum of two astronaut –rope system will be same as that of the initial case

Since there is no outside torque the angular momentum when one of the astronaut shortens the distance between them to $5.00\text{m}$ is $3750\text{kg}\cdot {\text{m}}^2/\text{s}$.

Conclusion:

Therefore, the angular momentum when one of the astronaut shortens the distance between them to $5.00\text{m}$ is $\underset{\_}{3750\text{kg}\cdot {\text{m}}^2/\text{s}}$.

To determine

The speed of the astronauts after shortening the distance.

Answer to Problem 67P

The speed of the astronauts after shortening the distance is $\underset{\_}{10.0\text{m}/\text{s}}$.

Explanation of Solution

Use equation (III) to find the new speed of the astronauts. The angular momentum of the system remains same even if he distance between the astronauts changes.

Conclusion:

Substitute, $2\times 7.50\text{kg}$ for $m$, $3750\text{kg}\cdot {\text{m}}^2/\text{s}$ for $L$, $2.50\text{m}$ for $r$ in equation (III) to obtain the speed of the astronauts.

  $\begin{array}{c}3750\text{kg}\cdot {\text{m}}^2/\text{s}=2\times 7.50\text{kg}\times 2.50\text{m}\times {\upsilon }_f\\ {\upsilon }_f=\frac{3750\text{kg}\cdot {\text{m}}^2/\text{s}}{2\times 7.50\text{kg}\times 2.50\text{m}}\\ =10.0\text{m}/\text{s}\end{array}$

Therefore, the speed of the astronauts after shortening the distance is $\underset{\_}{10.0\text{m}/\text{s}}$.

To determine

The new rotational kinetic energy of the system.

Answer to Problem 67P

The new rotational kinetic energy of the system is $\underset{\_}{7.50\text{kJ}}$.

Explanation of Solution

The total rotational kinetic energy is the sum of the kinetic energy of two astronauts.

Write the expression for the rotational kinetic energy.

  $K_f=\frac{1}{2}{m}_1{\upsilon }_1{}_f^2+\frac{1}{2}{m}_2{\upsilon }_2{}_f^2$        (IV)

Here, $m_1$ is the mass of the first astronaut, $m_2$ is the mass of the second astronaut, ${\upsilon }_{1f}$ is the final orbiting speed of first astronaut, and ${\upsilon }_{2f}$ is the final orbiting speed of the second astronaut.

Conclusion:

The mass, and speed of two astronauts are same.

Substitute, $7.50\text{kg}$ for $m_1$, and $m_2$, $10.0\text{m}/\text{s}$ for ${\upsilon }_{1f}$, and ${\upsilon }_{2f}$ in equation (IV) to find the new rotational kinetic energy of the system.

  $\begin{array}{c}{K}_i=\frac{1}{2}\left(7.50\text{kg}\right){\left(10.0\text{m}/\text{s}\right)}^2+\frac{1}{2}\left(7.50\text{kg}\right){\left(10.0\text{m}/\text{s}\right)}^2\\ =2\times \frac{1}{2}\left(7.50\text{kg}\right){\left(10.0\text{m}/\text{s}\right)}^2\\ =7.50\text{kJ}\end{array}$

Therefore, the rotational kinetic energy of the system is $\underset{\_}{7.50\text{kJ}}$.

To determine

The amount of chemical; energy converted to mechanical energy.

Answer to Problem 67P

The amount of chemical; energy converted to mechanical energy is $\underset{\_}{5.62\text{kJ}}$.

Explanation of Solution

The amount of chemical; energy converted to mechanical energy is equal to the work done by the astronaut. According to work energy theorem the work dine will be the change in rotational kinetic energy.

Write the expression for work done.

  $W=K_f-K_i$        (V)

Conclusion:

Substitute, $7.50\text{kJ}$ for $K_f$, and $1.88\text{kJ}$ for $K_i$ in equation (V) to find the change in chemical energy to mechanical energy.

  $\begin{array}{c}W=7.50\text{kJ}-1.88\text{kJ}\\ \text{=5}\text{.62}\text{kJ}\end{array}$

Therefore, the amount of chemical; energy converted to mechanical energy is $\underset{\_}{5.62\text{kJ}}$.