How fast in RPM must a centrifuge rotate if a particle 9.00 cm from the axis of rotation is to experience an acceleration of 115,000 g's?
Please explain your answer...
2006-10-04 15:46:08 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Gravity is accepted as 9.81 m/s^2
If Kevin_R was a student of mine in a Physics OR Engineering class, I would take a point or two from him if he continues insisting that gravity equals 10 m/s^2. In a previous question, I could live with estimating gravity - but we live in the real world and estimates no longer cut it.
Rather than give you the correct answer, replace 10 with 9.81 and you will find the right answer. The explanation is the centripetal force balances out gravity when one does a force balance on gravity (which points striaght down) and centripetal force (which points in).
2006-10-05 06:26:30 · answer #1 · answered by Anonymous · 0⤊ 0⤋
1 g of acceleration is approximately 10 m/2^s. So we are looking for an acceleration of 1,115,000 m/s^2
for centripetal acceleration a = v^2/r. where v is the velocity and r is the radius. The velocity come from the rotating centrifuge. Therefore v = w*r. where w is the rotation rate of the centrifuge in radians per second and r is the radius (same r as above). since there are 2*pi radians in a single rotation, w = 2*pi*(RPS) where RPS = revolitions per second. Since there are 60 RPM for 1 RPS the above equation becomes w = 2*pi*RPM/60.
Plugging in we get v = 2*pi*RPM*r/60
And since a = v^2/r, a = (2*pi*RPM*r/60)^2/r
This simplifies to a = r*(2*pi*RPM/60)^2
Plugging in the numbers (after converting 9cm to 0.09m)
1,115,000 = 0.09*(2*pi*RPM/60)^2
Solve for RPM, I get 33611 Revolutions per Minute
Karpenisi is correct that 9.81 m/s^2 ought to be used. The problem indicates that 3 significant figures is used for this problem, therefore the constants that I use need to be at least to that number of significant digits. But I do not believe that using 10 or 9.81 is the point of this exercise. Both numbers are approximations.
Karpenisi's explanation is incorrect. In this problem 'g' is used for a constant representing the magnitude of gravitational acceleration, not the vector. The centripetal force is not 'balancing out' gravity in this problem. An 'inward' force could never balance a 'downward' force. The vectors will never cancel.
2006-10-04 15:59:00 · answer #2 · answered by Kevin R 2 · 1⤊ 0⤋
You will find that Physics comes down to F=ma (Force = mass x acceleration)most of the time.
You have the force part in g's. You have the radius. Think about the relationship between acceleration and velocity and force and you will be able to find the missing piece...rotational velocity.
2006-10-04 16:06:57 · answer #3 · answered by Perry L 5 · 0⤊ 0⤋
F = mv^2/r = m 115000g
m cancels
you know g, and r, calculate v. convert v into angular freq.
2006-10-04 15:56:14 · answer #4 · answered by Anonymous · 1⤊ 0⤋
do you own homework
2006-10-04 15:47:33 · answer #5 · answered by SeJon 1 · 0⤊ 4⤋
DO YOUR HOME WORK BOY
2006-10-04 15:47:20 · answer #6 · answered by gary a 2 · 0⤊ 4⤋