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?

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**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

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