At its Ames Research Center, NASA uses its large "20-G"centrifuge to test the effects of very large accelerations (“hyper gravity”) on test pilots and astronauts. In this device, an arm 8.84 m long rotates about one end in a horizontal plane, and an astronaut is strapped in at the other end. Suppose that he is aligned along the centrifuge’s arm with his head at the outermost end. The maximum sustained acceleration to which humans are subjected in this device is typically 12.5 g. (a) How fast must the astronaut’s head be moving to experience this maximum acceleration? (b) What is the difference between the acceleration of his head and feet if the astronaut is 2.00 m tall? (c) How fast in rpm rpm 1 rev/min² is the arm turning to produce the maximum sustained acceleration?

The given data can be listed below,

• The length of the arm of devise is, $R_h=8.84m$
• The acceleration of the device is, $a=12.5g$
• Acceleration due to gravity, g=9.8m/s²
• The height of the astronaut is, h = 2m

A spinning system experiences centripetal acceleration. The phrase is frequently used when the system rotates with a constant angular velocity or when the emphasis is drawn to a certain period of time when the angular velocity has a particular value.

According to the centripetal force equation, the velocity of the astronaut’s head is given by,

$\begin{array}{rcl}{a}_r& =& \frac{v^2}{R}\\ v& =& \sqrt{a_{r}R}\end{array}$

Here, a_r is the radial acceleration of the astronaut and R is the radius of the head.

Substitute known values in the above equation.

$\begin{array}{rcl}v& =& \sqrt{\left(12.5g\right)\left(8.84m\right)}\\ & =& \sqrt{\left(12.5\right)\left(9.8m/s^2\right)\left(8.84m\right)}\\ & =& 32.9m/s\end{array}$

Thus, the velocity of the astronaut head is 32.9m/s.

The radial distance from rotation axis to astronaut’s feet is given by,

$R_f=R_h-h$

Here, R_h is the length of arm and h is the height of the astronaut.

Substitute all the values in the above,

$\begin{array}{rcl}{R}_f& =& 8.84m-2m\\ & =& 6.84m\end{array}$

As the distance parameter is directly proportional to radial acceleration the ratio of acceleration of the astronaut’s head and feet is given by,

$\frac{a_{feet}}{a_{head}}=\frac{R_{feet}}{R_{head}}$

Here, R_feet is the length of the feet and R_head is the length of the head.

Substitute all the values in the above expression.

$\begin{array}{rcl}{a}_{feet}& =& \frac{R_{feet}}{R_{head}}\times a_{head}\\ & =& \frac{6.84m}{8.84m}\times 12.5g\\ & =& 9.67g\end{array}$

The difference between acceleration of feet and head is given by,

$∆a_{rad}=a_{head}-a_{feet}$

Here, a_head is the acceleration of head of the astronaut and a_feet is the acceleration of the feet of astronaut.

Substitute all the values in the above,

$\begin{array}{rcl}∆a_{rad}& =& 12.5g-9.67g\\ & =& 2.83g\left(\frac{9.8m/s^2}{g}\right)\\ & =& 27.7m/s^2\end{array}$

Thus, the difference between the acceleration of head and feet is 27.7 m/s².

The frequency for constant velocity of arm is given by,

$$\begin{gathered} v=\omega R \\ v=2\pi fR \\ f=\frac{v}{2\mathrm{\pi R}} \end{gathered}$$

Here, v is the velocity of arm and R is the length of arm.

Substitute all the values in the above,

$\begin{array}{rcl}f& =& \frac{32.9m/s}{2\left(3.14\right)\left(8.84m\right)}\\ & =& \frac{32.9m/s\left(60s/1min\right)}{6.28\left(8.84m\right)}\\ & =& 35.5rev/min\end{array}$

Thus, the frequency of arm to attain maximum acceleration is 35.5 rev/min.