Chapter 12, Problem 46A

The two-pod system in the previous question rotates 1.2 RPM to provide artificial gravity for its occupants. If one of the pods pulls in 100 m of cable (bringing the pods closer together), what will be the system’s new rotation rate?

To determine

The new rotation rate of the two-pod system.

Answer to Problem 46A

The new rotation rate of the two-pod system is $1.5\text{ rpm}$ .

Explanation of Solution

Given info:

The rotation is ${\omega }_1=1.2\text{ rpm}$ .

The mass of the space pods is $m=1000\text{ kg}$ .

The length of the cable is $l=900\text{ m}$ .

The pod pulls at a distance of $100\text{ m}$ .

Formula used:

The expression for the rotational inertia as follows:

  $I=m{r}^2$

Here, $m$ is the mass, $v$ is the tangential speed, and $r$ is the distance.

The expression for the law of conservation of momentum as follows:

  $I_1{\omega }_1=I_2{\omega }_2$

Here, $I_2$ is the rotational inertia and ${\omega }_2$ is the angular velocity.

Calculation:

The distance from one end of the cable to its center as follows:

  $\begin{array}{l}{r}_1=\frac{l}{2}\\ r_1=\frac{900\text{ m}}{2}\\ r_1=450\text{ m}\end{array}$

The distance from one end of the cable to its center after pulls up as follows:

  $\begin{array}{l}{r}_2=\frac{900\text{ m}-100\text{ m}}{2}\\ r_2=400\text{ m}\end{array}$

The new rotation rate by the law of conservation of momentum as follows:

  $\begin{array}{l}{I}_1{\omega }_1=I_2{\omega }_2\\ m{r}_1^2{\omega }_1=m{r}_2^2{\omega }_2\\ {\left(450\text{ m}\right)}^2\left(1.2\text{ rpm}\right)={\left(400\text{ m}\right)}^2{\omega }_2\\ {\omega }_2=1.5\text{ rpm}\end{array}$

Conclusion:

Thus, the new rotation rate of the two-pod system is $1.5\text{ RPM}$ .