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Homework Help: Proportionality Statement

  1. Apr 16, 2007 #1
    Hey everyone, I'm new to the forums and this is my second thread started. This is an excellent resource in my opinion.

    Regardless of the formalities I have a question regarding a proportionality statement. I have here a question asking for a proportionality statement (in words) relating centripetal force and frequency.

    Is a proportionality statement something (alpha) something when written in proper notation and from there I could derive an equation by use of a constant? I just don't really know what to write.

    Thanks for the help!
    - Murdoc
  2. jcsd
  3. Apr 16, 2007 #2
    Say you have a function
    [tex] f(x) = \frac{x^3}{p^2} [/tex]

    A proportionality statement for the function will be that the function is proportional to the cube of x and inversely proportional to the square of p.

    So, the typical way to write centripetal force is
    [tex] F_c = m \frac{v_t^2}{r} [/tex] where v_t is the tangential velocity and r is the radius.

    Can you relate this equation to an equation with angular velocity, and then further relate angular velocity to frequency?
  4. Apr 16, 2007 #3


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    I just have a few points about that

    1. Yes, what you have described is what a proportionality statement *looks like*, although not necessarily what it means. Note also that you have described a mathematical statement of proportionality, whereas your assignment just asked for one in words.

    2. This is just a nitpick, but the symbol for "proportional to" isn't the same as the greek letter alpha:

    [tex] \textrm{alpha} \rightarrow \ \alpha \ \textrm{vs.} \propto \ \leftarrow \textrm{proportional to} [/tex]

    3. Getting back to what it means, I think you probably have an intuitive sense of what it means when we say that something is proportional to something else. Take force and acceleration. When we say that force and acceleration are proportional, we mean that the larger the force, the larger the acceleration, the smaller the force, the smaller the acceleration. As one increases or decreases, so too does the other. If one changes by a certain proportion, then the other changes by the same proportion. If you double the force, you double the acceleration. In other words, from all of this, we can deduce that the statement "the acceleration of an object is directly proportional to the net force acting on it" really boils down to a statement that there is a linear relationship between the two quantities. If x did not vary linearly with y, but quadratically, or in some other way, then x would NOT be proportional to y. It would be proportional, however, to y^2 (or whatever).

    4. What about the so-called "constant of proportionality?" Mathematically, of course, you can't equate a force to an acceleration, since they are two different quantities with two different dimensions. So you need the constant there as a sort of conversion factor. But that doesn't offer much insight. A more intuitive physical interpretation of this "conversion factor" is how much acceleration you get per unit of force, how many m/s^2 you'll get per newton. Therefore, we can conclude that the mass of an object is a measure of how hard it is to change its state of motion (to accelerate it), since the more massive an object, the less acceleration you get for the same force (the less bang for your buck [itex] a \propto \frac{F}{m} [/itex]). In Newtonian mechanics, this is a good operational definition of mass (inertial mass, anyway). I cite this example because it was the first "constant of proportionality" that took on a real physical meaning for me, and illustrated the usefulness of determining the relationships between quantities. I hope that this will help you understand what it is that you need to state about the centripetal force and the frequency.
    Last edited: Apr 16, 2007
  5. Apr 16, 2007 #4
    Alright so from the graphs I have,I come to an equation to represent the function on the graph of centripetal force versus time I get in this case y=8.3x+7.59 for the equation what would I say? I'm suddenly really lost :S
    Last edited: Apr 16, 2007
  6. Apr 16, 2007 #5


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    Huh, what graphs? I thought you were meant to derive a theoretical relationship between centripetal force and frequency using the definitions, as Mindscrape was getting at. You didn't mention any graphs. I think we need the full details of your problem to help you.
  7. Apr 17, 2007 #6
    Is this a different problem? What is y and what is x? I assume from your wording that y represents force and x represents time.

    Are you asking about how the pieces of the equation relate to proportionality?
  8. Apr 17, 2007 #7
    Alright essentially I have these things to do, all relative to information I recorded during a lab.

    This is due wednesday so I'm trying to get it done tonight / morning. I've plotted the graph and it is already in a linear form so I found the slope and the put into slope y intercept form in order to come to a linear equation.

    The equation is y = 8.3x - 7.59 to represent the line on the graph. I'm just realizing now I may be doing something horribly wrong because it shouldn't be in linear form right away but should cubic possibly... but it is what my data has shown.

    Any suggestions?

    P.S. here is a what my graph looks like http://s7.photobucket.com/albums/y266/murdoc8888888/?action=view&current=hmwrks1.jpg
    Last edited by a moderator: Apr 22, 2017
  9. Apr 17, 2007 #8
    Also, I know the function is Centripetal Force in terms of frequency.

    So could I not find Fc in terms of frequency in #5 using my equation of f(x) = 8.3x - 7.59 where 'x' is Hz, because everything is the same except the frequency as my lab? My teacher isn't really teaching this section he is more or less allowing us to figure it out on our own.
    Last edited: Apr 17, 2007
  10. Apr 17, 2007 #9
    I looked at your graph and something's not quite right. If you write out the theorectical equation for Centripeltal force versus frequency, you will see that a plot of Fc (Y axis) versus f ought to curve upwards, while yours curves down. I suspect that frictional forces or the motion of your wrist are adding significant experimental error that is reducing the force for the higher frequencies.
  11. Apr 17, 2007 #10
    That's what I'm thinking so ... I'll just elaborate that in the lab that there was an error somewhere. He'll accept it as long as I properly identify the problem, and how it can be adjusted so I'm not worried to much about it.
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