How Fast Should a Centrifuge Spin to Achieve Specific Acceleration?

[bfusco]

Homework Statement

How fast (in rpm) must a centrifuge rotate if a particle 7.0 cm from the axis of rotation is to experience an acceleration of 1.2×10^5 g's?

The Attempt at a Solution

well first i made 7 cm .07 m. then i changed acceleration to 1.18x10^6 m/s^2 by multiplying 1.2x10^5 g's by 9.8. at this point i no longer know what to do.

i am having trouble relating linear velocity, centripetal velocity, and angular velocity now. rpm means centripetal velocity, which is equal to 2πr/T, and i believe its centripetal velocity that equals rω. and acceleration = v^2/r nd i don't know where to go from there.

 

[Nessdude14]

The centripetal acceleration for circular motion is, as you said, v²/r. Since v = ωr (ω is angular velocity), the centripetal acceleration can also be written as: acceleration = ω²r.

If you use your acceleration of 1.18*10⁶ m/s² and your given radius, you'll get an answer for ω in rads/sec. The answer they're looking for is in revs/minute. You'll have to do a couple unit conversions on your answer to translate it to the right units. Hope that helps.

 

[PeterO]

Homework Statement

How fast (in rpm) must a centrifuge rotate if a particle 7.0 cm from the axis of rotation is to experience an acceleration of 1.2×10^5 g's?

The Attempt at a Solution

well first i made 7 cm .07 m. then i changed acceleration to 1.18x10^6 m/s^2 by multiplying 1.2x10^5 g's by 9.8. at this point i no longer know what to do.

i am having trouble relating linear velocity, centripetal velocity, and angular velocity now. rpm means centripetal velocity, which is equal to 2πr/T, and i believe its centripetal velocity that equals rω. and acceleration = v^2/r nd i don't know where to go from there.

The centripetal acceleration is not only given by v²/R, but also 4∏²R/T²
where T is the period.

Frequency - revolutions per second - is the inverse of Period, and revolutions per second should be pretty easy to convert to revolutions per minute

 

[bfusco]

ok...so using a=4∏^2r/T^2, i solved for T=√(4∏^2r/a)→√(4∏^2(.07)/117600) (the 117600 is 1.2x10^5 times 9.8). so i got T=.0015. next using the equation ω=v/r, i got ω=4099 rps→x60=245880 rpm. which is also wrong. the answer is 39000rpm. but how

 

[bfusco]

ok...so using a=4∏^2r/T^2, i solved for T=√(4∏^2r/a)→√(4∏^2(.07)/117600) (the 117600 is 1.2x10^5 times 9.8). so i got T=.0015. next using the equation ω=v/r, i got ω=4099 rps→x60=245880 rpm. which is also wrong. the answer is 39000rpm. but how

damn i got it, like stated before the answer of 4099 is in rad/sec, which when converted to rpm gives me 39000. damn, thanks guys

 

[PeterO]

damn i got it, like stated before the answer of 4099 is in rad/sec, which when converted to rpm gives me 39000. damn, thanks guys

The Period is how long it takes to do one revolution - you got 0.0015 sec [rounded off] .

AT that stage I would want to know "How many of them fit into 1 minute (60 Seconds)

so 60 divided by 0.0015 [not rounded off though] and there should be your answer without going via radians and ω !