Constants with variables

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Homework Statement


if a value has a variables that cancel out can it be considered a constant?

looking to make an equation out of proportionality statements


Homework Equations


this is the equation that we are supposed to get: f=√((F)/(4∏2mr))

the 3 proportionality where f2=kF , f= k(1/m) and f= k(1/r)

The Attempt at a Solution


i have reached here : f2= k(F/m2r2)

k being a constant, can its value be : k= (mr)/(4∏2)

so that the values cancel out giving you the required equation?
and would it still be considered a constant?
 

Answers and Replies

  • #2
Simon Bridge
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Welcome to PF;
It is possible that a combination of variables will be a constant for the situation at hand.
But that is not the case for your problem.

You are supposed to get to: $$f=\sqrt{ \frac{F}{4\pi^2mr}}\qquad\text{...(1)}$$... (check that - it does not look right to me) from: $$\begin{align}f^2\propto F & \qquad\text{...(2)}\\ f\propto \frac{1}{m} & \qquad\text{...(3)}\\ f\propto \frac{1}{r} & \qquad\text{...(4)} \end{align}$$

You get $$f^2=k\frac{F^2}{m^2r^2}$$ and ask:
k being a constant, can its value be : k= (mr)/(4∏2)
... No - it cannot be.

##k## has to be a constant - I don't expect that the product ##mr## is a constant

You have another step to perform before you put in the constant of proportionality.
You want to find an equation for ##f## and what you have so far is an equation for ##f^2##.
What do you have to do to your equation to turn it into an equation for ##f##?

Edit:
I think I see your problem - which is why I suggest you check the first equation.
 
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  • #3
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Thanks for the answer!

The first equation gives the magnitude of the force in uniform circular acceleration it was rearranged from: F=4∏2mrf2 this equation i believe is universally known and accepted so i have my doubts that it is wrong

we got the relationships from test result graphs, by using proportionalities im 90% sure they re correct, the only one that im questioning right now is f2=kF which also equals f=k√F

and i believe that that the equation you rearranged from the proportionalities was incorrect
it becomes :

f2= k*(F/m2r2)

where the F wont be squared
 
  • #4
haruspex
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The first equation gives the magnitude of the force in uniform circular acceleration it was rearranged from: F=4∏2mrf2 this equation i believe is universally known and accepted so i have my doubts that it is wrong

we got the relationships from test result graphs, by using proportionalities im 90% sure they re correct, the only one that im questioning right now is f2=kF which also equals f=k√F
Your proportionalities f∝1/m and f∝1/r cannot be right, as is evident from the actual equation.
and i believe that that the equation you rearranged from the proportionalities was incorrect
it becomes :

f2= k*(F/m2r2)

where the F wont be squared
Yes, I think Simon made a typo.
 
  • #5
Simon Bridge
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and i believe that that the equation you rearranged from the proportionalities was incorrect
it becomes :

f2= k*(F/m2r2)
...oh yeah - got carried away with the "^2"'s - that's a good catch: it is exceedingly rare that I make a typo at all in these forums ;)

Too late to edit the original - I'll correct it here: $$f^2=k\frac{F}{m^2r^2}$$... better?

BTW: I intended just to copy your work out in a clearer way - I did not do any derivation.


Looking at post #2: if (1) is correct an unimpeachable, then (3) and (4) cannot be right.
Should f be squared in both those relations?

If (3) and (4) are relations well supported by your experiment, then your experiment does not support (1). That is an acceptable conclusion.

Not every experiment will support the accepted equations - experiments will conclude by whatever their data supports. How the data got that way in your special case is another issue.
 

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