Question

Can you please explain the steps when you answer ? A laboratory ultracentrifuge is designed to produce a centripetal acceleration of 0.35x10^6 g at a distance of 2.50 cm from the axis. What angular velocity in rev/min is required?

Answer 1

`n=4.476 xx 10^6(revolution)/(min)`

Explanation:

centripetal acceleration is
`0.35 xx 10^6 g= 0.35 xx 10^6 xx 9,8 m/s^2 = 3,43 xx 10^6 m/s^2`
and you can calculate it throught:
`a_c= v^2/r`
`v^2= a_c /r =(3,43 xx 10^6 m/s^2) /(0.025 m) = 137 xx 10^6 m^2/s^2`
`v= 11,7 xx 10^3 m/s`
betwenn tangential velocity and angular velocity there is the relationship
`v= omega xx r`
`omega=v/r=(11,7 xx 10^3 m/s)/(0.025 m)= 468 xx 10^3 (rad)/s`
but
`1 (revolution)/ (min)= (2pi (rad))/(60 s)`
`1(rad)/s = 1/(2 pi)(revolution) /((1min)/(60s))`
`n=(60/(2pi)) xx 468 xx 10^3= 4.476 xx 10^6(revolution)/(min)`