Chapter 5, Problem 54P

At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.70 g? Assume the spaceship’s diameter is 32 m, and give your answer as the time needed for one revolution. (See Question 9, Fig 5-33 )

To determine

The time required for one revolution of the cylindrical spaceship.

Answer to Problem 54P

Solution:

9.59 s

Explanation of Solution

Given:

Diameter of the cylindrical spaceship, $d=32\text{m}$

Radius of the cylindrical spaceship, $r=1\text{6}\text{m}$

Simulated gravity, $=0.7g$

Formula Used:

According to the Newton’s second law of motion,

  $F=ma$

The centripetal acceleration is given by

  $F_c=\frac{m{v}^2}{r}$

Where, m is the mass, v is the velocity, a is the acceleration and r is the radius.

Calculation:

From the above, we can write

  $ma=\frac{m{v}^2}{r}$

  $a=\frac{v^2}{r}$

The velocity of the spaceshipcan be obtained by

  $v=\frac{2\pi r}{T}$

Therefore, the accelerationof the spaceship is

  $a=\frac{{\left(\frac{2\pi r}{T}\right)}^2}{r}$

  $a=\frac{4{\pi }^{2}r}{T^2}$

  $a=\frac{4{\pi }^2\times 16}{T^2}$

Equating the above acceleration to the given simulated gravity,

  $0.7g=\frac{4{\pi }^2\times 16}{T^2}$

  $\begin{array}{c}{T}^2=\frac{4{\pi }^2\times 16}{0.7\times 9.81}\\ =91.984\end{array}$

  $T=9.59\text{s}$

Conclusion:

Therefore, the time required for one revolution of the cylindrical spaceship is 9.59 seconds.