Question

An ultracentrifuge accelerates from rest to 100,000 rpm in 2.25 min. (Enter the magnitudes.)

(a) What is the average angular acceleration in rad/s2?

_________rad/s2

(b) What is the tangential acceleration (in m/s2) of a point 6.25 cm from the axis of rotation?

_________m/s2

(c) What is the centripetal acceleration in m/s2 and multiples of g of this point at full rpm?

ac in m/s2 ______________m/s2

ac as a multiple of g _______________g

(d) What is the total distance traveled (in m) by a point 6.25 cm from the axis of rotation of the ultracentrifuge?

_____________________m

Answer 1

Part A.

Given that:

w = 100000 rpm = 100000*2*pi/60 = 10471.97 rad/sec

Avg angular acceleration is given by:

alpha = dw/dt = (wf - wi)/dt

alpha = (10471.97 - 0)/(2.25*60)

alpha = 77.57 rad/sec^2

Part B.

tangential acceleration is given by:

at = alpha*R = 77.57*6.25*10^-2

at = 4.85 m/s^2

Part C.

Centripetal acceleration is given by:

ac = w^2/R

ac = 10471.97^2*6.25*10^-2

ac = 6.85*10^6 m/sec^2

as a multiple of 'g'

ac = 6.85*10^6*g/9.81

ac = 6.98*10^5*g

Part D.

Using 2nd rotational kinematic equation:

theta = wi*t + (1/2)*alpha*t^2

theta = 0*2.25*60 + (1/2)*77.57*(2.25*60)^2

theta = 706856.625 rad

Now total distance traveled will be:

d = R*theta

d = 6.25*10^-2*706856.625

d = 44178.5 m = 4.42*10^4 m

Please Upvote.