2. Am 80.0 kg astronaut is training for accelerations that he will experience upon re-entry to earth’s gravity from space. He is placed in a centrifuge (r = 25.0 m) and spun at a constant angular velocity of 10.0 rpm (revolutions per minute). a. Find the linear velocity of the centrifuge in m/s. Show your work

b. Find the magnitude and direction of the centripetal acceleration when he is spinning at this constant velocity.

c. How many g’s is the astronaut experiencing? (at constant velocity)

d. Find the linear deceleration and torque required to bring the centrifuge (5000.0 kg) to a stop over a 5 minute time period.

a)   v = 26.2 m / s, b) acceleration is radial, a = 27.4 m / s², c)   a = 2.8 g and

d) a = - 8.73 10⁻² m / s²,  τ = 1.09 10⁴ N m

Explanation:

a) For this exercise we can use the relationships between rotational and linear motion

           v = w r

let's reduce the magnitudes to the SI system

          w = 10 rpm (2pi rad / 1 rev) 1 min / 60s) = 1,047 rad / s

          r = 25.0 m

let's calculate

          v = 1.047 25.0

          v = 26.2 m / s

b) When the body is rotating at constant speed, the relationship must be perpendicular to the speed, therefore the direction of acceleration is radial, that is, towards the center of the circle and its magnitude is

            a = v² / r

            a = 26.2²/25

            a = 27.4 m / s²

c) Let's look for the relationship between the centripetal acceleration and the acceleration due to gravity

           a / g = 27.4 / 9.8

           a / g = 2.8

           a = 2.8 g

d) let's find the deceleration and torque to stop the centripette in 5 min

           t = 5 min (60 s / 1min) = 300 s

let's use the rotational kinematics relations

           w = w₀ + α t

initial angular velocity is wo = 1,047 rad / s and the final as is stop do w = 0

           α = - w₀ / t

           α = - 1,047 / 300

           α = -3.49 10⁻³ rad / s²

angular and linear are related

           a = α r

           a = -3.49 10⁻³ 25

           a = - 8.73 10⁻² m / s²

the negative sign indicates that the acceleration is stopping the movement

torque is

           τ = F r

The force can be found with Newton's second law

          F = m a

we substitute

         τ = m a r

         τ = 5000.0   8.73 10⁻²  25

         τ = 1.09 10⁴ N m

c. magnitude and direction