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Homework Help: Uniform circular motion proportionalities

  1. Apr 4, 2005 #1
    I have a lab with which I am having some major difficulties.

    (a) Whats the relationship between frequency of revolution and
    - the magnitude of the force causing the circular motion(centripetal force)?
    - the radius of the circular path?
    - the mass of the object?
    (b) Sketch three graphs to illustrate your answer to (a)
    (c) Find the proportionalities between frequency of revolution and the variables in (a)
    (d) combine the three results from (c) to obtain the equation for frequency in terms of the tension, the radius, and the mass. check your equation using your data points
    (e) The following relationship gives the magnitude of the net force causing the acceleration of an object in uniform circular motion:
    [tex]\sum F = 4\pi^2mrf^2[/tex]
    Rearrange this equation to isolate the frequency. Compare this result with the equation you derived in (d). Indicate the likely causes for any discrepancies.
    (f) Draw a FBD for the rubber stopper***
    (g) Explain how this investigation illustrates all three of Newton's laws of motion.

    ***the rubber stopper is attatched to a string which goes through a hollow tube which is attached to weight .. you twirl the stopper from the tube

    Heres what I know
    (a) the frequency and radius have an inverse relationship
    .. the other variables im not sure
    (b) i know what the graph looks like for radius:
    .. something like that
    (c) same as above
    (d) I need help with
    (e) I can rearrange the equation to:
    [tex]f=\sqrt{\frac{\sum F}{4\pi^2mr}}[/tex]
    the next part is hard to do without (d)
    (f) I think it has a Tension force going to the left or right, and a Force of gravity, not sure.
    (g) Could maybe figure this out after knowing the other answers.

    If you need my results just ask, but I doubt you do.


    BTW i know its alot but any help would be appreciated.
    Last edited: Apr 4, 2005
  2. jcsd
  3. Apr 4, 2005 #2
    i was able to do (c) using logarithms

    frequency is proportional to root of Fc
    frequency is proportional to Radius^-1/2 .. im not sure about this, but .. is this the same as root 1/ radius?
    frequency is proportional to mass the same way it is to the radius

    (d) I am still iffy about
    heres what i did: since f is proportional to root Fc .. it is equal to k*root(m*g)
    F= k root(1/r)
    F= k root(1/m)

    F=k root(mg/mr)
    therefore k must equal root(1/4pi^2)

    which is why there is a slight difference in (e)

    any confirmation would help
  4. Apr 5, 2005 #3


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    Well hopefully you know that the centripetal force is given by:

    [tex] F = \frac{mv^2}{r} [/tex]

    If not, review why that is. I think that this formula will answer a lot of (a), (b), and (c), once you observe the following:

    what is the relationship between velocity and angular frequency ([itex] \omega [/itex])? Can you rewrite this formula to be in terms of [itex] \omega [/itex]? Once you do, the rest should be straightforward.
  5. Apr 5, 2005 #4
    We haven't studied angular frequency yet and i don't think we are.. i searched it up on the net and its basicly the same formula as velocity but with radius multiplied by the top .. but im not sure how to write it in terms of angular frequency. angular frequency is proportional to velocity divided by radius?
  6. Apr 6, 2005 #5


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    Not to worry, it's a really simple concept:

    In uniform circular motion, [itex] \omega [/itex] is the angular velocity (how fast the rotation or spin is occuring in terms of what angle is swept out per second. So it's usually measured in radians per second. However, you're probably used to hearing angular speed measured in these everyday units: rpms "revolutions per minute", such as when reporting engine rotation rate. I hope that last comment makes the concept pretty clear.

    Note that radians are dimensionless, so although we write the units as rad/s to make it clear it's a rate of change of angular position, in fact the actual units are just 1/s = Hz. So it can be though of as a frequency. (Well of course it can: frequency means "often-ness", so omega just measures how often a full rotation occurs). How is it related to the real frequency f?

    What angle is a full revolution? [itex] 2\pi [/itex]. So what is the angular speed? Divide these [itex] 2\pi [/itex] radians by the time it takes to make a full revolution. But that is just the period (T) of the circular motion! So:

    [tex] \omega = \frac{2\pi}{T} [/tex]

    But what is 1/T, the reciprocal of the period? It is just the frequency of the motion! So:

    [tex] \omega = 2\pi f [/tex]

    Circular motion is the only type of motion in which angular frequency has a definite physical meaning (it is the angular velocity). Nevertheless, you see angular frequency being used to describe other types of periodic motion with frequency f that is not circular, such as harmonic oscillators, or wave motion. Huh? Yeah, I agree is doesn't seem to make sense at first, but all you have to do is multiply the frequency f by 2 pi to get omega, which can be interpreted as the equivalent angular frequency you would get if you had a system undergoing circular motion at f.

    I hope this wasn't too convoluted, I was just trying to be thorough.
  7. Apr 6, 2005 #6


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    I just realised you need some more information. You are correct that:

    [tex] v = \omega r [/tex]

    [tex] \omega = v/r [/tex]

    But I wanted to try and explain why that is so. Hopefully you're somewhat familiar with the radian system of angular measure. One radian is defined as the angle between two radii sweeping out an arc length equal to the radii themselves. This angle is therefore the ratio of the arc length to the radius. In general, the angle theta is the ratio of the arc length s to the radius r:

    [tex] \theta = \frac{s}{r} [/tex]

    Which makes sense if you think about it: for a given radius, the larger the arc length along the circle you move, the greater the angle through which you have rotated. We can rewrite it:

    [tex] s = r \theta [/tex]

    Another way of thinking about it is that for a given theta, the larger r is, the larger the arc length you will sweep out.

    Now, say you're measuring the change in the angular position [itex] \Delta \theta [/itex] in a time [itex] \Delta t [/tex]

    How far along the circle does the object travel in this time interval? Use the above formula:

    [tex] \Delta s = r \Delta \theta [/tex]

    So, assuming that the motion is uniform, the velocity is just the distance travelled in that time i.e:

    [tex] v = \frac{\Delta s}{\Delta t} = r \frac{\Delta \theta}{\Delta t} [/tex]

    But [itex] \frac{\Delta \theta}{\Delta t} [/itex] is just the angular velocity, ie how fast you rotate through a certian angle (angle rotated through per unit time).


    [tex] v = \frac{\Delta s}{\Delta t} = r \frac{\Delta \theta}{\Delta t} = r \omega [/tex]
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