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

##
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

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