Understanding Centrifuge Acceleration on Mercury

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Homework Help Overview

The discussion revolves around a laboratory centrifuge on Earth that produces an acceleration of 3.70 g at its outer end. Participants are attempting to determine the acceleration at a point halfway to the end of the centrifuge, using the relationship between linear velocity, angular velocity, and radius.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are exploring the relationship between radius and acceleration, questioning how changes in radius affect linear velocity and acceleration. There is a discussion about the formulas involved, particularly a = V^2/R and its implications when the radius is altered.

Discussion Status

Participants are actively engaging with the problem, sharing their understanding of the relationships between linear and angular velocity. Some have offered insights into how halving the radius affects both linear velocity and acceleration, while others express confusion and seek clarification on these concepts.

Contextual Notes

There is an indication that participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. The original poster and others express uncertainty about the calculations and the underlying principles.

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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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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?
 
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.
 
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.
 
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?
 
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.
 
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 !
 

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