A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe very large accelerations.
Part A) What is the acceleration at the end of a test tubethat is 10 cm from the axis of rotation?
Part B) For comparison, what is the magnitude of theacceleration a test tube would experience if dropped from a heightof 1.0 m and stopped in a 1.0-ms-long encounter with a hardfloor?

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moya1316 moya1316 · Answer 1 · 2021-03-13T11:43:14+01:00

Answer:

A) a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

a_c = v² / r

the angular and linear variables are related

v = w r

we substitute

a_c = w² r

let's reduce the magnitudes to the SI system

w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

a_c = 418.88² 0.1

a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

v² = v₀² - 2g (y- y₀)

v² = 0 - 2 9.8 (0 + 1)

v =√19.6

v = -4.427 m / s

now let's look for the applied steel to stop the test tube

v_f = v + a t

0 = v + at

a = -v / t

a = 4.427 / 0.001

a = 4.43 10³ m / s²