Question

2. A 75.0 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 15.3 rpm. Answer the following: a. What is the angular velocity of the centrifuge in rad/s? b. What is the linear velocity of the astronaut at the outer edge of the centrifuge? c. What is the centripetal acceleration of the astronaut at the end of the centrifuge? d. How many g’s does the astronaut experience? e. What is the centripetal force and net torque experienced by the astronaut? Give magnitudes and directions.

Answer 1

a. The angular velocity of the centrifuge is 1.597 rad/s.
b. The linear velocity of the astronaut at the outer edge of the centrifuge is 15.97 m/s.
c. The centripetal acceleration of the astronaut at the end of the centrifuge is 25.568 m/s^2.
d. The astronaut experiences 2.615 g's.
e. The centripetal force experienced by the astronaut is 1922.6 N and the net torque is 0 Nm.

Step-by-step explanation:

a. Angular velocity: Angular velocity, represented by the symbol ω, is the rate at which an object rotates around a fixed axis. To convert from rpm to rad/s, we can use the formula:
angular velocity in rad/s = (angular velocity in rpm) × (π/30)

Given: angular velocity in rpm = 15.3
angular velocity in rad/s = 15.3 × (π/30) = 15.3 × (3.14/30) = 1.597 rad/s.

b. Linear velocity: Linear velocity, represented by the symbol v, is the distance traveled per unit of time in a straight line. The linear velocity of the astronaut at the outer edge of the centrifuge is equal to the product of the angular velocity and the radius of the centrifuge:
linear velocity = angular velocity × radius

Given: angular velocity = 1.597 rad/s
radius = 10.0 m
linear velocity = 1.597 × 10.0 = 15.97 m/s.

c. Centripetal acceleration: Centripetal acceleration, represented by the symbol a, is the acceleration experienced by an object moving in a circular path. It can be calculated using the formula:
centripetal acceleration = (angular velocity)^2 × radius

Given: angular velocity = 1.597 rad/s
radius = 10.0 m
centripetal acceleration = (1.597)^2 × 10.0 = 25.568 m/s^2.

d. g's: The acceleration due to gravity on Earth is approximately equal to 9.8 m/s^2. To determine the number of g's the astronaut experiences, we can divide the centripetal acceleration by the acceleration due to gravity:
g's = centripetal acceleration / acceleration due to gravity

Given: centripetal acceleration = 25.568 m/s^2
acceleration due to gravity = 9.8 m/s^2
g's = 25.568 / 9.8 = 2.615 g's.

e. Centripetal force and net torque: The centripetal force and net torque experienced by the astronaut can be calculated using the formulas:
centripetal force = mass × centripetal acceleration
net torque = moment of inertia × angular acceleration

Given: mass of the astronaut = 75.0 kg
centripetal acceleration = 25.568 m/s^2
moment of inertia = mass × radius^2 ÷ 2 (assuming the astronaut is a point mass located at the outer edge of the centrifuge)
moment of inertia = 75.0 × (10.0)^2 ÷ 2 = 3750 kg m^2
angular acceleration = 0 (since the astronaut is spinning at a constant angular velocity)

centripetal force = 75.0 × 25.568 = 1922.6 N
net torque = 3750 × 0 = 0 Nm.

Answer 2

Since one revolution around a circle is 2 pi radian; hence 1 rpm equals 2 pi radians per minute. And because a minute has 60 seconds, 1 rpm equals 1/60 revolution per second. Therefore, we have 2 pie/60 * 15.3 = 0.2513 rps. The linear velocity v = wr where w is the angular velocity in rad/s and r is the distance. So we have 0.2513 * 10.0 = 2.513 rad/s The centripetal acceleration is given by a = w^2 r = (2.513)^2 * 10 = 63.15 rad/s2 The centripetal force F = mass * centripetal acc = 75 * 63.15 = 4736.25 N The torque = centripetal force * distance = 4736 * 63.15 = 299078.4 Nm Two forces acts on the astronaut. The normal force and acceleration due to gravity.