Physics Math, Lab Theory Question

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The lab we did today was to determine the relationship between radius, speed, mass, and frequency on the centripetal force in a given system. I am typing up my thesis now, and I am having trouble understanding something.

Fc=(mv^2)/r = m*4*pi^2*r*f^2 Given those two equal equations for centripetal force, I would be inclined to believe that centripetal force is INVERSELY related to centripetal force given the first equation, but at the same time, directly related to centripetal force, given the second equation.

So two questions here: Which one would it be, and why are there two equations providing different proportionalities?
 

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  • #2
berkeman
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The lab we did today was to determine the relationship between radius, speed, mass, and frequency on the centripetal force in a given system. I am typing up my thesis now, and I am having trouble understanding something.

Fc=(mv^2)/r = m*4*pi^2*r*f^2 Given those two equal equations for centripetal force, I would be inclined to believe that centripetal force is INVERSELY related to centripetal force given the first equation, but at the same time, directly related to centripetal force, given the second equation.

So two questions here: Which one would it be, and why are there two equations providing different proportionalities?
v = R * omega

so omega = v/R

Is that what has you confused? Also, your question seems to have some typos in it ("centipital force is inversely related to centripital force", rather that inversely related to ____ ?what?)
 
  • #3
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oops, sorry. centripetal force would be inversely proportional to speed in the mv^2 / r equation, while it is directly proportional to r in m4(pi^2)rf^2 (btw, we have not seen/used v*omega... We are just aware of these equations for Fc)

So yes, the question still stands... Why is the proportionality different between Fc and r in the two different equations?

that is:
m4(pi^2)rf^2 = Fc
and
mv^2 / r = Fc
 
  • #4
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oops, sorry. centripetal force would be inversely proportional to speed in the mv^2 / r equation, while it is directly proportional to r in m4(pi^2)rf^2 (btw, we have not seen/used v*omega... We are just aware of these equations for Fc)

So yes, the question still stands... Why is the proportionality different between Fc and r in the two different equations?

that is:
m4(pi^2)rf^2 = Fc
and
mv^2 / r = Fc
Actually the equation Fc = m*4*PI^2*r*f^2 is derived from the equation Fc = mv^2/r
Here is how the equation is derived:

Fc = mv^2/r

from the above equation we know that v = d/t. however since this is circular motion, we can say d = 2*PI*r. Hence subsituting (2*PI*r)/t for v:

Fc = m((2*PI*r)/t)^2/r

now expand and simpify:

Fc = (m*4*PI^2*r^2)/(t^2*r)

As you can see now that r can be canceled out from the equation. However since r in the numerator is squared there will be an r remaining in the numerator (This part should answer your question, no?):

Fc = m*4*PI^2*r/t^2

Now you will notice that there is t in the denominator. Since this is circular motion, it can be represented by a periodic function, thus the time can be represented as the Period of the motion:

Fc = m*4*PI^2*r/T^2

We know that the T = 1/f, thus f = 1/T, hence we can rewrite the equation as:

Fc = m*4*PI^2*r*f^2

Hopefully you noticed during that the r was cancelled out from the denominator while the equation was derived, it is not so much a question of proportionality here but the method in which the equation was derived which produced the r variable to appear in both the numerator and the denominator.

Hopefully this helps....

Sekhar.B
 
  • #5
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okay. So, in summary, what I think you are saying is that in mv^2, it is possible to derive another 2 radii out of there, and so in reality, in the (mv^2) /r equation, there are really 2 r's in the top... so it IS directly proportional to r, even in the (mv^2) /r equation, but the r's in the numerator are jsut hidden.
 
  • #6
Keep in mind that r, v, and f are not independent values. Yes, r appears in the numerator for one expression for centripetal force and in the denominator for the other. But that's only because of the relationship between r, v, and f. Make sure you can understand and write down the equation that relates these three quantities.

Another thing to consider is the dimensions of F. In one expression, the force looks like the kinetic energy divided by a distance r. In the other expression, it looks like a mass times a frequency squared times a distance r, which is a mass times an acceleration. With practice, both these ways of looking at the units for force should make sense to you.
 

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