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Physical Pendulum - Rod - Period and Distance From Center of Mass

  1. Apr 16, 2016 #1
    1. The problem statement, all variables and given/known data
    We have a rod (length L, mass m) suspended at a point whose distance from the center of mass is a.
    1) prove that (generally) there exist two values of a (a1, a2) for which the pendulum oscillates with the same period.
    2) derive and explain: [tex] T = 2\pi\sqrt{\frac{a_1+ a_2}{g}}[/tex]
    3) with what value of a do we get the smallest period?​

    2. Relevant equations
    [tex] E = 1/2 I w^2 [/tex]
    [tex]J= J^* + ma^2[/tex]
    [tex]J= 1/12 mL^2 + ma^2 [/tex]
    [tex] T = 2\pi\sqrt{\frac{J}{mga}}[/tex]
    [tex] T = 2\pi\sqrt{\frac{L}{g}}[/tex]

    3. The attempt at a solution
    1) I thought that two periods being the same would mean T1 = T2
    [tex] 2\pi\sqrt{\frac{J_1}{mga}}= 2\pi\sqrt{\frac{J_2}{mga}}[/tex]
    [tex]J^* + ma_1^2 =J^* + ma_2^2[/tex]
    [tex]a_1^2 =a_2^2[/tex]
    [tex]a_1 = +/- a_2[/tex]

    and since a1 = a2 would be the same location a1 = - a2 which would mean that the only way to get the same period for two different values of a, would be to have the a be the same but in different directions from the center of mass.

    2) I noticed that [tex] T = 2\pi\sqrt{\frac{a_1+ a_2}{g}}[/tex] looks a lot like [tex] T = 2\pi\sqrt{\frac{L}{g}}[/tex]
    So I thought that a1 + a2 = L
    and that that would mean that the rod would be suspended on one end and the mass would be concentrated on the other end of the rod? I guess it sounds silly but I just couldn't figure out anything else.

    3) [tex] \lim_{x\rightarrow 0} T = 2\pi\sqrt{\frac{J}{mga}} = 2\pi\sqrt{\frac{1/12 L^2 + a^2}{ga}} [/tex]
    since [tex] 2\pi\sqrt{\frac{1}{g}} [/tex] is a constant I'll derive the rest because a limit is just a special type of derivation (I think?)
    [tex] \frac{1/2 L^2 + a^2}{a}[/tex] derived is:

    [tex] La + a^2 + {\frac {L^2}{2}}[/tex]

    and from that I can see that the smaller the a the smaller the period of the oscillation so we would get the shortest period if the pendulum was hung from its center of mass.


    I was told my solution is wrong. I don't know which parts, I don't know why and I don't know how to correct it. If you could help me out somehow I'd be really grateful.​
     
    Last edited: Apr 16, 2016
  2. jcsd
  3. Apr 16, 2016 #2

    haruspex

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    where r represents....?
     
  4. Apr 16, 2016 #3
    That would be the distance between the point of rotation and the center of mass. Sorry, I got it from a different problem and messed it up. It should be a.
    It affects the first equation, but the way I reasoned the error wouldn't make a difference.

    I'm going to go edit it so it fits this problem.
     
  5. Apr 16, 2016 #4

    haruspex

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    but there is no 'a', there is a1 and a2, so it does make a difference.
     
  6. Apr 16, 2016 #5
    Oh, okay so in the first part of my problem I can't just disregard the mga anymore because one is mga1 and the other mga2. That will completely change the problem, you're right!
    Thank you :) I'm going to try to fix it.

    Okay so calculating
    [tex]2\pi\sqrt{\frac{J^* + ma_1^2}{mga_1}}= 2\pi\sqrt{\frac{J^* + ma_2^2}{mga_2}} [/tex]
    [tex] {\frac{J^* + ma_1^2}{mga_1}}= {\frac{J^* + ma_2^2}{mga_2}} [/tex]
    J* for a rod is:
    [tex] J^* = {\frac{mL^2}{12}}[/tex]
    [tex]{\frac{L^2}{12 a_1}} + a_1= {\frac{L^2}{12 a_2}} + a_2 [/tex]
    [tex]{\frac{L^2}{12}} ({\frac{1}{a_1}} - {\frac{1}{a_2}})= a_2 - a_1 [/tex]
    we can divide that by the [tex]({\frac{1}{a_1}} - {\frac{1}{a_2}})[/tex]
    and get [tex]{\frac{L^2}{12}} = a_1 a_2[/tex]
    which means that [tex] a_1 = {\frac{L^2}{12 a_2}}[/tex]

    Is that correct? Am I missing a +/- from the removal of the square root? And adding that will mean that it is true weather we go to the right from the center of mass or the left.

    So this tells me that there is some correlation between the two distances but that it doesn't mean that they are the same length away from the center of mass?
     
    Last edited: Apr 16, 2016
  7. Apr 16, 2016 #6
    Assuming my calculations for a1 and a2 are correct now. I can plug that into
    [tex]
    T = 2\pi\sqrt{\frac{a_1+ a_2}{g}}[/tex]
    I have decided that I shouldn't be mixing a simple pendulum [tex]
    T = 2\pi\sqrt{\frac{L}{g}} [/tex]
    into this and instead I did this
    [tex]2\pi\sqrt{\frac{ a_1+ a_2}{g}} = 2\pi\sqrt{\frac{J}{mga_2}}[/tex]
    [tex]2\pi\sqrt{\frac{ {\frac{L^2}{12 a_2}}+ a_2}{g}} = 2\pi\sqrt{\frac{ {\frac{mL^2}{12}} + ma_2^2 }{mga_2}}[/tex]
    after simplifying I got 1=1 sooo that means that it's true... but I can't figure out what that means.

    I mean... a pendulum can't be hanging from two points at once so what IS the a1 + a2? Or does the 1=1 just mean that the whole "formula" that was in the problem statement is just the period calculated while hanging from one of the points?
     
  8. Apr 16, 2016 #7

    haruspex

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    I did not understand part 2 at first because T has not been defined. I suppose it must mean where T is the period for either a1 or a2, and that appears to be what you have shown.
    As for how to interpret it, think of it in terms of an equivalent simple pendulum.
     
  9. Apr 16, 2016 #8
    Oh yes, well we weren't given an explanation as to what it was but I assumed that the T is the period.
    So, what it says is that a simple pendulum that is L2/(12a2) + a2 long would have the same period as our physical pendulum?

    Thank you so much for your help!

    Do you have any idea what kind of mistake I made in the third point of the problem? I read this post and tried to follow the instructions given to the OP but OP never followed up and I guess I misunderstood something.
     
  10. Apr 16, 2016 #9

    haruspex

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    No, how do you get that?
    What is the period of a simple pendulum of length a1+a2?

    For part 3, you found a simple relationship between a1 and a2 in post #5. With that constraint, find the minimum of the period given by the expression in part 2.
     
  11. Apr 17, 2016 #10
    Well my reasoning was that since the period for a simple pendulum looks like this [tex]
    T = 2\pi\sqrt{\frac{L}{g}}
    [/tex]
    and the equation for a physical pendulum hanging from a point a2 away from the center of mass can be written like this: [tex]
    T = 2\pi\sqrt{\frac{a_1+ a_2}{g}}
    [/tex]
    and since the equations are so similar I figured that a1+a2 = L and from that I figured that a pendulum hanging at a distance of a2 to the center of mass would have the same period of an oscillation as a simple pendulum of length L= a1 + a2

    isn't the period of a simple pendulum of the length (a1 + a2) this? : [tex]
    T = 2\pi\sqrt{\frac{a_1+ a_2}{g}}
    [/tex]

    I tried: [tex]
    T = 2\pi\sqrt{\frac{a_1+ a_2}{g}} =0
    [/tex]
    [tex]{\frac{L^2}{12 a_2}} + a_2 = 0[/tex]
    but after "cleaning it up" I would get something impossible
    [tex] a_2 = sqrt {{\frac{- L^2}{12}}}[/tex]

    I tried derivating it and then putting it equal to 0 but then I just couldn't free the a2
    [tex] 2 L a_2 - L^2 + 12 a_2^2 =0 [/tex]
     
  12. Apr 17, 2016 #11

    haruspex

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    Ah, ok, yes. I didn't recognise it in the more complicated version you had in post #8. This way of wording it is much clearer.
    T=0?! You are asked to minimise it, but it won't reach zero.
     
  13. Apr 17, 2016 #12
    Okay I did more searching online and I think I was making a mistake in my drivation anyway but there is what I came up with:
    [tex] {\frac{dT}{da}} = 0[/tex]
    I'm going to go ahead and leave out the constants because I calculated with them and they all "disappeared" anyway
    [tex] [{\sqrt{{{\frac{L^2}{12 a_2}}+ a_2}}}]' = 0 [/tex]
    [tex]{\frac{{\frac{L^2}{-12 a_2^2}}+1}{{\frac{1}{2}}{\sqrt{{\frac{L^2}{12 a_2}}+a_2}}}}=0[/tex]
    since for a fraction to equal 0, the top needs to be equal to zero that means
    [tex]{\frac{L^2}{-12 a_2^2}+1}=0[/tex]
    [tex]L^2 - 12 a_2^2 = 0[/tex]
    [tex]a_2={\sqrt{{\frac{L^2}{12}}}}[/tex]
    and so that last a2 would be the distance from L for the minimal period?
     
  14. Apr 17, 2016 #13

    haruspex

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    It is certainly unsurprising that the case a1=a2 represents a local extremum, but you need to check whether it is a minimum or maximum.
     
  15. Apr 17, 2016 #14
    To do this would I do the second derivative of [tex]
    L^2 - 12 a_2^2 = 0[/tex]
    -24a2 and then if that is smaller than 0 it's a local maximum and if it's greater than 0 then it is a local minimum? so in this case it would be a local maximum?

    But now that I look at it I could have
    [tex]
    -L^2 -+12 a_2^2 = 0[/tex]
    and it would still be correct and then I'd end up with 24a2 which would be a minimum

    which means my logic is flawed.
     
  16. Apr 17, 2016 #15

    haruspex

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    You can simplify the problem a lot. T will be minimised when the sum of the distances is minimised, we don't need to worry about the square root. So the problem reduces to minimising x+y subject to xy=c2. Draw a graph, or consider that (x+y)2=(x-y)2+4xy.
     
  17. Apr 17, 2016 #16
    I've been trying to figure out what to do with what you've told me but I just don't think I know what you mean or exactly how to calculate it. In math we learned how to find the minimum of a function by making the first derivation equal to zero and then either making a determinant from the second derivations (one with respect to x2 the next xy the one after yx(xy) and the one after that y2) or in some cases we just did a second derivation and if it was smaller than 0 then it was the max and if it was bigger than 0 it was the minimum.
    what is xy= c2? is it a1a2 = ?

    I am really grateful for all your help but I just can't figure out what to do with this last tip :(
     
  18. Apr 17, 2016 #17

    haruspex

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    Given that xy=c2, a constant, and that (x-y)2 cannot be negative, what is the minimum possible value of (x+y)2?
     
  19. Apr 18, 2016 #18
    (x+y)2 >= c2 ? So... xy would be the minimum?
     
  20. Apr 27, 2016 #19

    BvU

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    I notice this thread peterered out here... status ?
    Haru's exercise is straightforward: you want to find an x and a y and a minimum vaue for (x+y)2, so the answer xy is not what you are looking for: the minimum has to be some function of c. In fact Haru has put you on top of the answer in #15 (and by now you know how useful a picture can be ! :smile: )
     
  21. Apr 27, 2016 #20
    I marked it as answered because in the end I took that L2 - 12a22 and I got a2 = +/-sqrt( L2/12)
    and when I put that back into the original problem and got that in both cases it will be >0 and that means that it will be the minimum. In school I was told that it's correct and so I marked it as solved.

    Anyway, I never said thank you, but really, thank you Haruspex! I would've never finished this problem without your help :)
     
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