Centrifugal Force

Many physicists claim that centrifugal force does not exist, only centripetal force, which pulls the body toward the center of revolution. Test pilots might disagree. In any case, it is the centrifugal force balancing with the gravitational force that keeps an orbiting body in continuous free-fall.

Centrifugal force = mass x velocity squared divided by radius

To understand the units in this equation, read the slug section below. When you multiply all the units out, centrifugal force should fall out of the equation in pounds.

Example: Two spaceplanes are connected by a 100 ft collapsible dorsal tunnel. At what speed must they rotate about their common center of gravity to generate artificial Mars gravity onboard each spaceplane? Mars surface gravity is 37% that at the surface of Earth.

Solution : Realize, from Newton's Second Law, that centrifugal force is really centrifugal acceleration x mass. Secondly, realize that centrifugal acceleration is equivalent to gravitational acceleration, which on Earth is 32.174 ft/s2.

F = ma = mv2 /r c c ac = v2 /r 0.c37ge = v2 /50ft v = (50 ft x 0.37 x 32.174 ft/s2 )1/2 v = 24.397 ft/s Speed at rim

Circumference of rotation = C = 2pr = 2p(50 ft) = 314.16 ft

Rotation rate = Speed at rim (ft/s)/Circumference of rotation (ft/rev) = rev/s

Rotation rate = (24.397 ft/s)/(314.16 ft/rev) = 0.077658 rev/s

Revolution period (sec/rev) = 1/Rotation rate = 1/0.077658 rev/sec = 12.877 sec/rev

With this information, the crew may want to either (a) lengthen the distance between their craft or (b) slow down the rotation rate and settle for a lower artificial g-field.

Was this article helpful?

0 0

Post a comment