Chapter 8, Problem 80AP

Two astronauts (Fig. P8.80), each haring a mass of 75.0 kg, are connected by a 10.0-m rope of negligible mass. They are isolated in space, moving in circles around the point halfway between them at a speed of 5.00 m/s. Treating the astronauts as particles, calculate (a) the magnitude of the angular momentum and (b) the rotational energy of the system. By pulling on the rope, the astronauts shorten the distance between them to 5.00 m. (c) What is the new angular momentum of the system? (d) What are their new’ speeds? (e) What is the new rotational energy of the system? (f) How much work is done by the astronauts in shortening the rope?

Figure P8.80 Problems 80 and 81

To determine

The angular momentum of the astronauts.

Answer to Problem 80AP

The angular momentum of the astronauts is $3.75\times {10}^3\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}$

Explanation of Solution

Given Info:

The mass of the astronauts is $75\text{kg}$, the speed of the astronauts is $5.00\text{m}\cdot {\text{s}}^{-1}$, the distance between them is $10.0\text{m}$.

The formula to calculate angular momentum is given by

$L_i=m{v}_i{r}_i$

• $L_i$ is the initial angular momentum,
• m is the mass of astronauts
• $v_i$ is initial the speed of the astronauts
• $r_i$ is the initial distance between the astronaut from the centre of mass.

Substitute $75\text{kg}$ for m, $5.00\text{m}\cdot {\text{s}}^{-1}$ for $v_i$ and $10.0\text{m}$ for $r_i$ in the above expression to calculate $L_i$.

$\begin{array}{c}{L}_i=\left(75\text{kg}\right)\left(5.00\text{m}\cdot {\text{s}}^{-1}\right)\left(10.0\text{m}\right)\\ =3750\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}\\ =3.75\times {10}^3\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}\end{array}$

Conclusion:

Therefore the angular momentum of the astronauts is $3.75\times {10}^3\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}$

To determine

The rotational energy of the system.

Answer to Problem 80AP

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

Explanation of Solution

The formula to calculate rotational energy of the system is given by

$K{E}_i=m{v}_i^2$

• $K{E}_i$ is the initial rotational energy of the system.

Substitute $75\text{kg}$ for m, and $5.00\text{m}\cdot {\text{s}}^{-1}$ for $v_i$ in the above equation to calculate rotational kinetic energy of the system.

$\begin{array}{c}K{E}_i=\left(75\text{kg}\right){\left(5.00\text{m}\cdot {\text{s}}^{-1}\right)}^2\\ =1875\text{J}\left(\frac{1\text{kJ}}{{10}^{-3}\text{J}}\right)\\ \approx 1.88\text{kJ}\end{array}$

Conclusion: The rotational energy of the system is $1.88\text{kJ}$

To determine

The angular momentum of the system for the shortened distance.

Explanation of Solution

In the absence of external torque, the angular momentum of the system remains conserved.

According to conservation of angular momentum angular momentum must be conserved.

$L_i=L_f$

Therefore the initial angular momentum is same as final angular momentum

Conclusion:

Therefore the new angular momentum is $3.75\times {10}^3\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}$.

To determine

speed of the astronauts

Answer to Problem 80AP

The speed of the astronauts is $10\text{m}\cdot {\text{s}}^{-1}$.

Explanation of Solution

Given Info:

The formula to calculate angular speed of the system is

$v_f=\frac{L_f}{m{r}_f}$

• $r_f$ is the new distance between astronauts

Substitute $3750\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}$ for $L_f$ , $75\text{kg}$ for m and $5.0\text{m}$ for $r_f$ to calculate $v_f$

$\begin{array}{c}{v}_f=\frac{3750\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}}{\left(75\text{kg}\right)\left(5.0\text{m}\right)}\\ =10\text{m}\cdot {\text{s}}^{-1}\end{array}$

Conclusion:

Therefore the new speed of the astronauts is $10\text{m}\cdot {\text{s}}^{-1}$

To determine

The new rotational energy of the system.

Answer to Problem 80AP

The new rotational energy of the system is $7.5\text{kJ}$

Explanation of Solution

Given Info:

The formula to calculate rotational energy of the system is given by

$K{E}_f=m{v}_f^2$

Substitute $75\text{kg}$ for m, and $10\text{m}\cdot {\text{s}}^{-1}$ for $v_f$ in the above equation to calculate new rotational kinetic energy

$\begin{array}{c}K{E}_f=\left(75\text{kg}\right){\left(10\text{m}\cdot {\text{s}}^{-1}\right)}^2\\ =7500\text{J}\left(\frac{1\text{kJ}}{{10}^3\text{J}}\right)\\ =7.5\text{kJ}\end{array}$

Conclusion: The rotational energy of the system is $7.5\text{kJ}$

To determine

The work done by the astronauts.

Answer to Problem 80AP

The work done by the astronaut is $5.62\text{kJ}$.

Explanation of Solution

Given Info:

According to work energy theorem work done is change in kinetic energy here kinetic energy is rotational energy.

The formula to calculate rotational energy of the system is given by

$W=K{E}_f-K{E}_i$

Substitute $7.5\text{kJ}$ for $K{E}^{\prime }$ and $1.88\text{kJ}$ for KE to calculate W.

$\begin{array}{c}W=7.5\text{kJ}-1.88\text{kJ}\\ =562\text{J}\left(\frac{1\text{kJ}}{{10}^3\text{J}}\right)\\ =5.62\text{kJ}\end{array}$

Thus the work done is $5.62\text{kJ}$

Conclusion: The work done by the astronaut is $5.62\text{kJ}$