An ultracentrifuge accelerates from rest to 100,000 rpm in 1.20 min.

Question

(a) What is its angular acceleration in rad/s^2?
(b) What is the tangential acceleration, in m/s^2, of a point 13.30 cm from the axis of rotation?
(c) What is the radial acceleration, in m/s^2, of this point at full rpm?
(d) Express this radial acceleration as a multiple of g.

Answer 1

(a) First convert 100,000 rpm to rad/s:

ω = (100,000 rpm)(2π rad/1 min)(1 min/60 s) = 10472 rad/s

Now use the formula for angular acceleration:

α = ω/t = (10472 rad/s) / (1.20 min * 60 s/min) = 145.11 rad/s^2

Therefore, the angular acceleration is 145.11 rad/s^2.

(b) To find the tangential acceleration, we use the formula:

at = rα

where r = 13.30 cm = 0.1330 m

at = (0.1330 m)(145.11 rad/s^2) = 19.48 m/s^2

Therefore, the tangential acceleration is 19.48 m/s^2.

(c) The radial acceleration is given by:

ar = rω^2

at full rpm, ω = 100,000 rpm = 10472 rad/s

ar = (0.1330 m)(10472 rad/s)^2 = 1.41 x 10^6 m/s^2

Therefore, the radial acceleration at full rpm is 1.41 x 10^6 m/s^2.

(d) To express this radial acceleration as a multiple of g, divide by the acceleration due to gravity, g = 9.81 m/s^2:

(1.41 x 10^6 m/s^2) / (9.81 m/s^2) ≈ 143715

Therefore, the radial acceleration at full rpm is approximately 143715 times the acceleration due to gravity.