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Homework Help: Circular Motion

  1. May 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the the theoretical velocities and accelerations at various points around a circle.
    Catch; the only information we have is the radius length of 5.2m

    2. Relevant equations

    Ok, so i got pretty far with this;
    At the top of the circle. T = 0
    T = mv(sq) / r - mg
    mg = mv(sq)/r
    m cancel
    v = [tex]\sqrt{rg}[/tex]

    Well, change in GPE = change in KE
    mgh = .5m (v) sq
    m cancel
    v = [tex]\sqrt{2gh}[/tex]
    where h = 2r

    change in GPE = change in KE
    mgh = .5m (v) sq
    m cancel
    v = [tex]\sqrt{2gh}[/tex]

    g = 9.81
    T = 0 at top

    a centripetal = v(sq) / r

    3. The attempt at a solution

    Vmin = [tex]\sqrt{rg}[/tex]
    =7.14 m/s

    Vmax = [tex]\sqrt{2gh}[/tex]
    = 14.28 m/s

    Vmid = [tex]\sqrt{2gh}[/tex]
    where h = r
    =10.1 m/s

    a min = v(sq) / r
    = 9.8 m/s

    a max = v(sq) / r
    = 39.22 m/s

    a mid = v(sq) / r
    = 19.6 m/s

    So is all of this working right?
    I assumed that v at the top is 0 with all of the working and equations. If it wasn't 0, but was continuing after a revoltion. Would i just add vmin to each of the velocities?
    Thanks for any help.
  2. jcsd
  3. May 29, 2008 #2


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    Hi Noir,

    It looks to me like there is a lot more to the problem statement than just the two sentences you gave. If so, could you post the exact problem?
  4. May 29, 2008 #3
    Thanks for your reply, in regard to the question. I pretty much paraphrased to try and squash it down.

    "Describe the physcis associated with the motion of the object as it moves around the circle. Include Centripetal forces, energy changes etc. Predict the forces the bob would 'experience', estimate the velocities at various points during its course and relate all of this to what you have learnt this term".

    She told us in class that motion is that of an amusement park ride - So i guess the bob could be seen as a platform with people on it? I have no idea, that sounds logical to me.

    The only solid information is that the radius of the circle is 5.2m.

    She gave us a few 'hints'. She told us to research pendulum's, disregard mass (ie, cancel it out) and 'tops of circles'.

    Someone in class asked for a typical problem solving question we'd get in our exam in a couple weeks so she came up with this. I hope this helps!
  5. May 29, 2008 #4


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    So we assume this is an object on a string, going in a vertical circle? I'll assume that in my remarks.

    Here you have assumed that the mass is going as slow as possible if it's going in a circular path, and you found the speed at the top, so this part looks good.

    Like you said in your comments at the bottom of this post, this assumed that the speed was zero at the top, which is not true. But you just can't add the top speed to your answer here; instead, the change in kinetic energy is

    \Delta({\rm KE})= \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

    and then solve for Vmax (which is [itex]v_f[/itex]). You would follow the same kind of procedure to find Vmid.

    (By the way, your energy statement is not quite right. It needs to be the change in GPE + change in KE equals zero; this works out because the change in GPE is -mgh since the object is moving downwards.)

    Once you make that change, I believe that would give you the two new velocities you are looking for. What do you get?

    Once you have those, what are the accelerations at the three points?
  6. May 30, 2008 #5
    Thanks for your reply. Using the equation

    [tex]\Delta({\rm KE})= \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2[/tex]

    [tex]v_f = \sqrt{2g \Delta h + v_i^2}[/tex]
    [tex]v_f = \sqrt{2*9.81*10.4 + 7.14^2}[/tex]
    [tex]v_f = 15.96 m/s [/tex]

    [tex]v_m = \sqrt{2g \Delta h + v_f^2}[/tex] Where [tex]v_m[/tex] is middle velocity - is that right?
    [tex]v_m = \sqrt{2*9.81*5.2 + 7.14^2}[/tex]
    [tex]v_m = 12.37 m/s [/tex]

    Now i have all three velocites, i can find the accelerations;
    [tex]a_c = \frac{v^2}{r}[/tex]

    [tex]a_i = 9.81 m/s^2 [/tex]
    [tex]a_m = 49 m/s^2 [/tex]
    [tex]a_f = 29.5 m/s^2 [/tex]
    are the max velocites at those points.

    Thank you for all your help, and in regard to the issue with the energy statement. This is just frame of reference, no? Easily fixed with a -mgh, correct?
  7. May 30, 2008 #6


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    Two notes on these: first, I think when you calculated these centripetal forces, you switched the middle and lower values; a_m should be 29.5 and a_f should be 49.

    However, those are the centripetal accelerations, and it looks to me like you want to find the total acceleration. So a_m would be more than just the centripetal part. What do you get?

    No, that's not quite true. You can add whatever you want to a single value of the potential, but then you have to do the same to all the other values. So the change in gravitational potential energy will always be a negative number if the object is moving downwards.

    For example, you want the change in energy for the particle moving from the top to the bottom. If we set the height=0 at the bottom, we get:

    \Delta ({\rm PE}) = mgh_f-mgh_i = m g (0) - m g (+10.4) = - m g (10.4)

    If we set the height=0 at the top:

    \Delta ({\rm PE}) = mgh_f-mgh_i = m g (-10.4) - m g (0) = - m g (10.4)

    or height=0 at the middle:

    \Delta ({\rm PE}) = mgh_f-mgh_i = m g (-5.2) - m g (5.2) = - m g (10.4)

    So although the values of the potential energy at a specific point can be made into whatever you want, the change in the potential energy is related to the work done by the force, and so will have a definite value.
  8. May 30, 2008 #7
    Thank you for clearing that up - It was a mistake with the two accelerations, i had them mixed around. With the -mgh, i know it has to be negative, but doesn't that give a nonreal answer when subbed into the equation?
    [tex]v_m = \sqrt{v_f^2 - 2g \Delta h }[/tex]
    Just a thought. With
    [tex]a_f = 49 m/s[/tex]

    or 5G's. Would this be total or is another g added due to gravity? I'm sorry if this sounds stupid but my head isn't working to well this time of morning.
    Last edited: May 30, 2008
  9. May 31, 2008 #8


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    I don't think so; I think you get:

    v_m = \sqrt{14.28^2 - 2 (9.8) (5.2)}

    because v_f is the speed at the bottom and v_m is the speed at the middle height. Since 14.28^2 is greater than (2 * 9.8 * 5.2) the result will be real.

    The values of a_i=9.81 m/s^2 and a_f=49 m/s^2 (total acceleration at top and bottom) look right to me.

    I think the problem is with a_m (the acceleration at the middle height). The centripetal acceleration there is 29.5 m/s^2; but like your post indicates there is a tangential component of the acceleration due to gravity. So to find the total acceleration at the middle, there is a value of g 'added', but since the directions of the centripetal and tangential components are perpendicular, we can't just add them together like normal numbers.
  10. May 31, 2008 #9
    Thanks for all your help, but one more thing.
    You got:
    [tex]v_m = \sqrt{14.28^2 - 2 (9.8) (5.2)}[/tex].
    Does the equation before it's in term of Vm look like this?
    [tex]2g \Delta h = \frac{1}{2}mv_m^2 - \frac{1}{2}mv_f^2[/tex]
    Where the it is [tex]v_m - v_f[/tex] because of the -mgh?
    Thanks once again :)
  11. May 31, 2008 #10


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    (You have a two instead of an m on the left side of that equation.) The starting point would be:

    \Delta (KE) + \Delta(PE) =0

    Expanding, and using v_m as the initial point:

    \frac{1}{2}m v_f^2 - \frac{1}{2} m v_m^2 + m g (\Delta h) =0

    So this agrees with the equation in your post (once you move the kinetic energy terms to the right side and fix up the two):

    m g (\Delta h) = \frac{1}{2} m v_m^2 -\frac{1}{2}m v_f^2

    Solving for [itex]v_m[/itex] gives:

    v_m = \sqrt{ v_f^2 + 2 g (\Delta h)

    Then plug in the numbers v_f = 14.28 and [itex]\Delta h=(-5.2)[/itex].
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