an ultracentrifuge accelerates from rest to 100,000 rpm in 2.90 min. (a) what is its angular acceleration in rad/s2? rad/s2 (b) what is the tangential acceleration, in m/s2, of a point 11.30 cm from the axis of rotation? m/s2 (c) what is the radial acceleration, in m/s2, of this point at full rpm? m/s2 (d) express this radial acceleration as a multiple of g.

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16-12-2022
Physics

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an ultracentrifuge accelerates from rest to 100,000 rpm in 2.90 min. (a) what is its angular acceleration in rad/s2? rad/s2 (b) what is the tangential acceleration, in m/s2, of a point 11.30 cm from the axis of rotation? m/s2 (c) what is the radial acceleration, in m/s2, of this point at full rpm? m/s2 (d) express this radial acceleration as a multiple of g.

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kumargaurav1012000 kumargaurav1012000 · Answer 1 · 2022-12-22T00:54:40-05:00

A final angular velocity of 100,000 rpm is reached by an ultra-centrifuge after starting at angular velocity zero. Radiant per square second (rad/s2) is the SI unit for angular acceleration (also known as rotary acceleration).

The angular acceleration of a body whose angular velocity changes uniformly to 1rad/s during the course of 1 s is 1 rad/s2. A = AR, where A is the angular acceleration and r is the circle's radius, is the equation for tangential acceleration. The rate of variation in the matter's tangential velocity along a circular path is known as tangential acceleration. From a standstill, an ultracentrifuge accelerates to 100000 rpm in 2 minutes. Where v is the speed and r is the radius of curvature at an instant, the amount of radial acceleration at any instant is equal to v2/r.

We are given these following data:

ω = 100000rev/min * (2πrad/rev) * (1min/60s) = 10,500 rad/s

t = 1.7 min = 102 s

r = 0.1055 m

(a) ω = ωo + at

10,500 rad/s = 0rad/s + α*102s

α = 102.94 rad/s²

(b) a_t = α r = 102.94 rad/s² * 0.1055 m = 10.86 m/s²

(c) a_r = ω² r = (10500rad/s)² * 0.1055 m = 1.16 x 10^7 m/s²

(d) which is also equal to1.16 x 10^7 / 9.8 = 1.18 x 10^6 g

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