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Alex
2 years ago
12

A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 10.0 m

) and spun at a constant angular velocity of 16.3 rpm. Answer the following:
What is the angular velocity of the centrifuge in ?

What is the linear velocity of the astronaut at the outer edge of the centrifuge?

What is the centripetal acceleration of the astronaut at the end of the centrifuge?

How many g’s does the astronaut experience?

What is the centripetal force and net torque experienced by the astronaut? Give magnitudes and directions.
Physics
1 answer:
vovangra [49]2 years ago
7 0

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

Angular velocity is 1.3823 rad/s

Linear velocity is given by

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

Centripetal force is given by

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

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Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

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dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

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θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

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