Centripetal acceleration centrifuge

AI Thread Summary
A centrifuge operates by rotating a container at high speeds, causing denser materials to settle at the bottom. The centripetal acceleration of the sample is 5.92 x 10^3 times gravity, and the radius of rotation is 6.37 cm. The calculation for velocity involves using the formula Ac = V^2/r, leading to a velocity of approximately 60.79 m/s. To convert the period to revolutions per minute (RPM), the relationship between period and frequency must be correctly applied. The final conversion from period to RPM requires careful attention to units to ensure accurate results.
pookisantoki
Messages
44
Reaction score
0
A centrifuge is a device in which a small container of material is rotated at a high speed on a circular path. Such a device is used in medical laboratories, for instance, to cause the more dense red blood cells to settle through the less dense blood serum and collect at the bottom of the container. Suppose the centripetal acceleration of the sample is 5.92 x 10^3 times as large as the acceleration due to gravity. How many revolutions per minute is the sample making, if it is located at a radius of 6.37 cm from the axis of rotation?

Acceleration due to gravity is 9.80 and I converted 6.37cm to .0637m.
So this is how I solved it: Ac=V^2/r so set it up where V^2=Ac*r=sqrt(9.80*5920)*.0637= 60.79160
Then V=(2pi*r)/T so T=(2pi*r)/V: (2pi*.0637)/60.79160=.006584

Is this right??
 
Physics news on Phys.org
I didn't double check your numbers, but you solved for period and the question asks for rpm, just remember to do the appropriate conversion.
 
I'm not sure on how to go from period to RPM
 
Well period is seconds per revolution, while rpm is revolutions per minute. Recall 1/T=F, the frequency, which has units revolutions per second.
 
So would it be .006584*60=.39504??
 
No... check your units and make sure they come out alright. What you have is this:
.0065 \frac{\textrm{seconds}}{\textrm{revolution}} \times 60\frac{\textrm{seconds}}{\textrm{minute}} = .395 \frac{\textrm{seconds}^2}{\textrm{revolution minute}}

Which is obviously not right. Remember where you want to go, which is revolutions/minute.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top