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Physics 211, Quiz 20

Question 1: A space station of R = 50 m rotates with uniform circular motion, so that a person on its outer rim feels 1 gee of centripetal acceleration.

Then, one day, the captain fires rockets which accelerate the station at alpha = 0.02 rad/s^2 for t = 10 sec.

What is the centripetal acceleration at the outer rim after the engines have fired?

  Answer: The original centripetal acceleration is

                                     
                  a(centripetal)  =  9.8 m/s^2   =  R * omega^2

          Therefore, the initial angular velocity is 

                                  9.8 m/s^2
                  omega   =  sqrt(---------)  = 0.443 rad/s
                                    50 m

          When the captain fires the engines, he increases the angular
          velocity to

                  omega   =  0.443 rad/s  +  (0.02 rad/s^2) * (10 s)

                          =  0.443 rad/s  +  0.20 rad/s

                          =  0.643 rad/s

          So the centripetal acceleration on the rim is now

                  a(centripetal)  =  R * omega^2
  
                                  =  (50 m) * (0.643 rad/s)^2

                                  =  20.7 m/s^2

Creative Commons License Copyright © Michael Richmond. This work is licensed under a Creative Commons License.