Calculating the Frequency of a Centrifuge in Hertz and Revolutions per Minute

In summary: I cannot see your numbers or your equations, and I cannot see where you are going wrong if you are going wrong.In summary, using the given values of a research apparatus undergoing centripetal acceleration with a magnitude of 3.3x10^6m/s^2 and a radius of 8.4cm, the frequency can be calculated to be 300.7746 Hz or 18,046.47 RPM using various equations such as ac=4*pi^2*r/f^2 and ac=v^2/r.
  • #1
Balsam
226
8

Homework Statement


A research apparatus called a centrifuge undergoes a centripetal acceleration with a magnitude of 3.3x10^6m/s^2. The centrifuge has a radius of 8.4cm. Calculate the frequency of the centrifuge in hertz and in revolutions per minute.

Ac=3.3x10^6,/s^2 r=0.084m

Homework Equations


ac=4pi^2rf^2

The Attempt at a Solution


I tried solving this by isolating for f and plugging in all of my given values, but the answer I got was 683963.585Hz while the correct answer is 10 000Hz. Also, how do you convert Hz to revolutions per minute- do you just divide by 60?
 
Physics news on Phys.org
  • #2
Acceleration = v^2/r. Tangential velocity is the square root of (acceleration X radius), the tangential velocity is 1,665 m / s. The frequency is the tangential velocity divided by the circumference of rotation, or (1,665 m / s) / (5.27 m), or 315.46 cycles per second (Hertz). That gives a frequency of 18,927 cycles per minute.

Let's assume that the correct answer is 10,000 Hz. Then the tangential velocity is 52,779 m / s. The acceleration would be 3.3 *10^9 m / s^2.

ac=4pi^2rf^2 should be ac=4pi^2r/f^2. And by the way, the f is not frequency but time for a single rotation. Check the dimensional analysis.
 
Last edited:
  • #3
Balsam said:
A research apparatus called a centrifuge undergoes a centripetal acceleration with a magnitude of 3.3x10^6m/s^2. The centrifuge has a radius of 8.4cm. Calculate the frequency of the centrifuge in hertz and in revolutions per minute.

Ac=3.3x10^6,/s^2 r=0.084m

Homework Equations


ac=4pi^2rf^2

Balsam said:
683963.585Hz

check your calculation;

Balsam said:
Also, how do you convert Hz to revolutions per minute- do you just divide by 60?

to find out rev. per minute from rev per sec one should not divide by 60
 
  • #4
Speaking of checking calculations (picture me with a red face), the tangential velocity using a = (v^2 / r) is 1,587.5 m/s. This gives a rate of 300.8 Hz or 18,046.5 RPM.

Using a = (4 * pi^2 * r / t^2) results in a time of 0.00332 seconds per revolution. This agrees at 300.8 Hz, and the tangential velocity agrees at 1,587.5 m/s.
 
  • #5
Balsam said:
I tried solving this by isolating for f and plugging in all of my given values, but the answer I got was 683963.585Hz while the correct answer is 10 000Hz. Also, how do you convert Hz to revolutions per minute- do you just divide by 60?

OldYat47 said:
Speaking of checking calculations (picture me with a red face), the tangential velocity using a = (v^2 / r) is 1,587.5 m/s. This gives a rate of 300.8 Hz or 18,046.5 RPM.

Using a = (4 * pi^2 * r / t^2) results in a time of 0.00332 seconds per revolution. This agrees at 300.8 Hz, and the tangential velocity agrees at 1,587.5 m/s.

well i am getting about 1000 Hz and if converted to per minute it will be 60 thousand rev per minute but where did i do wrong?
 
  • #6
You may be right, who knows? In any case, here are my numbers and equations.
ac=4pi^2rf^2, above is not correct, should be ac = 4*pi^2*r/f^2. Note that f is not the frequency but the time to complete one revolution. See concurrent dimensional analysis below.
f = [(4*pi^2*r)/ac]^0.5, [(4*9.8696*0.84)/(3*10^6)]^0.5 (meters)/(meters/seconds squared) = (seconds squared)
f=(1.1054*10^-5)^0.5 (seconds squared)^0.5
f=3.325*10^-3 seconds
The frequency is the inverse of that number, or 300.7746 (revolutions (dimensionless) / second).
The tangential velocity is 300.7746 (revolutions per second * meters per revolution), or
(300.7746 * 2 * pi * 0.84) = 1,587.45 (meters/second)

Now let's use ac = v^2/r
v=(ac*r)^0.5, (3*10^6*0.84)^0.5 [(meters/seconds squared)*(meters)]^0.5=(meters squared/seconds squared)^0.5=(meters/second)
v= 1,587.45 (meters/second)

Show your numbers and equations and let's see where the problems lie.
 
  • #7
OldYat47 said:
You may be right, who knows? In any case, here are my numbers and equations.
ac=4pi^2rf^2, above is not correct, should be ac = 4*pi^2*r/f^2. Note that f is not the frequency but the time to complete one revolution. See concurrent dimensional analysis below.
f = [(4*pi^2*r)/ac]^0.5, [(4*9.8696*0.84)/(3*10^6)]^0.5 (meters)/(meters/seconds squared) = (seconds squared)
f=(1.1054*10^-5)^0.5 (seconds squared)^0.5
f=3.325*10^-3 seconds
The frequency is the inverse of that number, or 300.7746 revolutions (dimensionless) / second.
The tangential velocity is 300.7746 revolutions per second * meters per revolution, or
(300.7746 * 2 * pi * 0.84) = 1,587.45 (meters/second)

Now let's use ac = v^2/r
v=(ac*r)^0.5, (3*10^6*0.84)^0.5 [(meters/seconds squared)*(meters)]^0.5=(meters squared/seconds squared)^0.5=(meters/second)
v= 1,587.45 (meters/second)

Show your numbers and equations and let's see where the problems lie.

i did it more simply -therefore i might have made errors-
acceleration = w^2 . R where w is angular velocity as w= 2.pi/T where T is time taken to complete one revolution,
in one second no. of rev. n=1 /T so,

w
= 2.pi.n where n is the no. of revolution per sec

acceleration(a) = 4. Pi^2.n^2 . R therefore n= Sqrt ( a / 4.pi^2 .R) ; n= sqrt [ 3.3 . 10^6) /( 4.(3.14)^2 . 0.084) ]
putting in numbers its approx n =1000 this is equivalent to frequency in Hz ,
so in 1 minute i.e. 60 seconds =1000.60 = 60,000 rev. per minute.

however these numbers are very high
 
Last edited:
  • #8
I cannot see what the problem is other than the insistence that n=1,000. Look at your own equation. When I calculate the value of that equation I get 300.7746 Hz, or 18,046.47 RPM. Using any of the other available equations for these numbers the result is the same.

That's all I've got.
 

1. What is the purpose of a centrifuge?

A centrifuge is a laboratory instrument that spins samples at high speeds in order to separate components based on their density. It is commonly used in scientific research and medical testing to separate substances such as cells, proteins, and DNA.

2. How does the frequency of a centrifuge affect its performance?

The frequency, or speed, of a centrifuge is a crucial factor in its performance. Higher frequencies result in faster separation times, but can also cause damage to delicate samples. Lower frequencies are gentler on samples but may take longer to achieve separation.

3. What is the recommended frequency for a centrifuge?

The recommended frequency for a centrifuge depends on the type of sample being processed and the desired separation results. Generally, a range of 1000-5000 rpm (revolutions per minute) is suitable for most applications.

4. How do I calculate the frequency of a centrifuge?

To calculate the frequency of a centrifuge, you will need to know the speed of the rotor (in rpm) and the radius of the centrifuge chamber. The formula for calculating frequency is: frequency = (rotor speed x 2π) / (radius x 60). This will give you the frequency in revolutions per second (Hz).

5. Can the frequency of a centrifuge be adjusted?

Yes, the frequency of a centrifuge can usually be adjusted by the user. Most centrifuges have a control panel or knob that allows you to set the desired frequency. It is important to refer to the manufacturer's instructions for proper usage and safety precautions when adjusting the frequency of a centrifuge.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
856
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
9K
  • Introductory Physics Homework Help
Replies
9
Views
2K
Back
Top