Understanding Centrifuge Acceleration on Mercury

In summary, the laboratory centrifuge on Earth has an angular velocity of n rpm and produces an acceleration of 3.70 g at its outer end. To calculate the acceleration at a point halfway out, one must use the equation a = (V^2)/R and consider the relationship between linear velocity and angular velocity. By halving the radius, the linear velocity is also halved, resulting in a quadrupled acceleration.
  • #1
cdlegendary
15
0

Homework Statement



A laboratory centrifuge on Earth makes n rpm (rev/min) and produces an acceleration of 3.70 g at its outer end.

a.) What is the acceleration g's at a point halfway out to the end?

Homework Equations



a = (V^2)/R

The Attempt at a Solution



I really don't know what to do. I thought that if you divide the radius by 2, the acceleration would double, according to the equation. That doesn't seem to be the case since the answer 7.40 is incorrect. Any help would be greatly appreciated.
 
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  • #2
cdlegendary said:

Homework Statement



A laboratory centrifuge on Earth makes n rpm (rev/min) and produces an acceleration of 3.70 g at its outer end.

a.) What is the acceleration g's at a point halfway out to the end?

Homework Equations



a = (V^2)/R

The Attempt at a Solution



I really don't know what to do. I thought that if you divide the radius by 2, the acceleration would double, according to the equation. That doesn't seem to be the case since the answer 7.40 is incorrect. Any help would be greatly appreciated.

Do you remember how to calculate linear velocity from angular velocity? If radius changes, but angular velocity is constant, does linear velocity change?
 
  • #3
Uhhh no I don't know how to calculate the linear velocity. I'm guessing the linear velocity changes if the radius changes eh? Only formula I'm familiar with is the a = V^2/R, and the other variation.
 
  • #4
cdlegendary said:
Uhhh no I don't know how to calculate the linear velocity. I'm guessing the linear velocity changes if the radius changes eh? Only formula I'm familiar with is the a = V^2/R, and the other variation.

w * r = v

because w = radians / second

What is archlength? radians * radius (for example, 2pi radians (360 degrees) * r = circumference , the entire arch length)
so we get archlength/ second, which is a velocity along the rim of a circle.

The fact that v changes with r makes sense too: if you were going 1 revolution(360 degrees) a second, and you were right next to the pivot, you'd travel less distance per second. If you were 70 thousand miles away from the pivot going around every second, you'd be traveling VERY fast.
 
  • #5
xcvxcvvc said:
w * r = v

because w = radians / second

What is archlength? radians * radius (for example, 2pi radians (360 degrees) * r = circumference , the entire arch length)
so we get archlength/ second, which is a velocity along the rim of a circle.

The fact that v changes with r makes sense too: if you were going 1 revolution(360 degrees) a second, and you were right next to the pivot, you'd travel less distance per second. If you were 70 thousand miles away from the pivot going around every second, you'd be traveling VERY fast.

Ah okay, thanks. So that would mean since the radius is halved, the linear velocity would be halved as well? If that's the case, then I would assume that the acceleration is quadrupled since both velocity and radius are being halved?
 
  • #6
cdlegendary said:
Ah okay, thanks. So that would mean since the radius is halved, the linear velocity would be halved as well? If that's the case, then I would assume that the acceleration is quadrupled since both velocity and radius are being halved?

If a = V^2/ r = w^2 r^2 / r = W^2 r
then changing r by a factor of 1/2 would half the acceleration.
 
  • #7
xcvxcvvc said:
If a = V^2/ r = w^2 r^2 / r = W^2 r
then changing r by a factor of 1/2 would half the acceleration.

Haha wow I got confused there for a sec. I see how it works now, thanks a lot !
 

1. What is a centrifuge and how does it work?

A centrifuge is a laboratory instrument used to separate components of a mixture based on their density. It works by spinning the mixture at a high speed, causing the denser components to move towards the bottom and the lighter components to move towards the top.

2. Can a centrifuge be used on Mercury?

Yes, a centrifuge can be used on Mercury. However, the gravitational force on Mercury is only about 38% of that on Earth, so the centrifugal force generated may not be as strong.

3. How would a centrifuge be affected by Mercury's extreme temperatures?

The extreme temperatures on Mercury can potentially affect the materials used to construct the centrifuge. The instrument would need to be made from materials that can withstand the high temperatures, such as specialized metals or ceramics.

4. What are the potential applications of using a centrifuge on Mercury?

The potential applications of using a centrifuge on Mercury could include studying the composition and density of different materials found on the planet, separating and purifying samples for analysis, and possibly even extracting valuable resources.

5. Are there any challenges or limitations to using a centrifuge on Mercury?

There are several challenges and limitations to using a centrifuge on Mercury, including the extreme temperatures, the low gravity, and the potential for materials to degrade or malfunction due to the harsh conditions. Specialized equipment and materials would need to be developed to overcome these challenges.

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