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Ede4ka [16]
3 years ago
10

Human centrifuges are used to train military pilots and astronauts in preparation for high-g maneuvers. A trained, fit person we

aring 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
Physics
2 answers:
jeka57 [31]3 years ago
7 0

(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

denis-greek [22]3 years ago
6 0

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.

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