Question

Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. A trained, fit person wearing a g-suit can withstand accelerations up to about 9g (88.2 m/s2) without losing consciousness. HINT (a) If a human centrifuge has a radius of 5.15 m, what angular speed (in rad/s) results in a centripetal acceleration of 9g? rad/s (b) What linear speed (in m/s) would a person in the centrifuge have at this acceleration? m/s

Answer 1

(a) $4.14 rad/s^2$

The relationship beween centripetal acceleration and angular speed is

$a=\omega^2 r$

where

$\omega$ is the angular speed

r is the radius of the circular path

Here we gave

$a = 9g = 88.2 m/s^2$ is the centripetal acceleration

r = 5.15 m is the radius

Solving for $\omega$, we find:

$\omega = \sqrt{\frac{a}{r}}=\sqrt{\frac{88.2 m/s^2}{5.15 m}}=4.14 rad/s^2$

(b) 21.3 m/s

The relationship between the linear speed and the angular speed is

$v=\omega r$

where

v is the linear speed

$\omega$ is the angular speed

r is the radius of the circular path

In this problem we have

$\omega=4.14 rad/s$

r = 5.15 m

Solving the equation for v, we find

$v=(4.14 rad/s)(5.15 m)=21.3 m/s$

Answer 2

Explanation:

(a) Centripetal acceleration, $a=9g=88.2\ m/s^2$

Radius, r = 5.15 m

Let $\omega$ is the angular speed. The relation between the angular speed and angular acceleration is given by :

$a=\omega^2 r$

$\omega=\sqrt{\dfrac{a}{r}}$

$\omega=\sqrt{\dfrac{88.2}{5.15}}$

$\omega=4.13\ rad/s$

(b) Let v is the linear speed of the person in the centrifuge have at this acceleration. It is given by :

$v=r\times \omega$

$v=5.15\times 4.13$

v = 21.26 m/s

Hence, this is the required solution.