TOLC-med 2022 Q 48

Practice Question Solving
1

A laboratory centrifuge rotates at 6000 rpm. Called g the acceleration of gravity, what is worth, approximately as a function of g (consider g = 10 m/s²), the module of the centripetal acceleration of a sample that rotates at a distance of 12.5 cm from the center of rotation?
A) 5000 g
B) 80 g
C) 125 g
D) 630 g
E) 1250 g

Hi everyone!

this is from this years test, the answer is A), i’m not sure where i went wrong?

a centrifuge has uniform circular motion:
so ac = r * angular velocity²
r = 0,125m
10 rpm is about = 1 rad/s
6000 rpm = 600 rad/s

ac = 0,125 * 600² = 45000

i’m not sure what they mean by “as a function of g”, this is what i could think of to fit the answer: 45000 = 4500 * 10 = 4500 * g

i still don’t know how to find 5000, any ideas?

2

angular speed = 2 x pi x f where frequency = 100 rounds per second (converted minutes to seconds)
= (2 x pi)x 100 = about 628 rad/s (I used 3.14 as pi)
a = r x w^2
= 0.125 x 628^2 = 49,298 approximately 50,000
50,000/10 = 5000 g

They put g at the end so the answer A, 5000 g = 50,000. I think the 10rpm = 1 rad/s is not really used for the IMAT and the Italian IMAT, instead we convert it using a full round = 2pi. Hope this helps!!

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