Question

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?

Answer 1

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

Answer 2

can you explain how to find the acceleration at a point halfway out to the end?

Answer 3

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.

Answer 4

So how do you get the first part? "{

Answer 5

3.2 / 2