Posted by Steph on Friday, November 26, 2010 at 4:58pm.
A laboratory centrifuge on earth makes n rpm and produces an acceleration of 3.20 g at its outer end.
A) What is the acceleration (in g's, i.e., acceleration divided by g) at a point halfway out to the end?
B) This centrifuge is now used in a space capsule on the planet Mercury, where g_mercury is 0.378 what it is on earth. How many rpm (in terms of n) should it make to produce 7 g_mercury at its outer end?

physics  Damon, Friday, November 26, 2010 at 5:10pm
Ac = v^2/R = w^2 r
divide r by two > divide Ac by to
7*.378 = 2.646 g
so
Ac final = 2.646/3.2 = .827 Ac original
w^2 r final = .827 w^2 r original
r did not change
w final = sqrt (.827) w original
n final = .91 n original 
physics  Steph, Friday, November 26, 2010 at 6:34pm
can you explain how to find the acceleration at a point halfway out to the end?

physics  Damon, Friday, November 26, 2010 at 7:12pm
sure,
Ac = w^2, the angular velocity squared
times the radius.
If you cut the radius in half, go halfway out, then you cut the acceleration in half since w is the same. 
physics  Jessica, Saturday, November 27, 2010 at 1:06pm
So how do you get the first part? "{

physics  Damon, Saturday, November 27, 2010 at 5:57pm
3.2 / 2