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Problem proving tension in string

  1. Sep 25, 2004 #1
    Ok so i have a homework problem again :) This time it's a simple pendulum swing and the question is as follows:

    "Prove that the tension in the string of a simple pendulum when the string makes an angle [tex]\theta[/tex] with the vertical is approximately [tex]mg(1 + \theta^2_0 - (3/2)\theta^2[/tex] where m is the mass of the bob and [tex]\theta_0[/tex] is the value of [tex]\theta[/tex] at the extremes of the motion."

    Now our teacher gave us a bit of help, saying that we should use a taylor polynom to solve this, as well as the formula:

    [tex]x(t) = x_0 * cos(\omega * t)[/tex]

    Where [tex]x_0[/tex] is x (or s if you prefer) at the extremes of it's motion.

    So i basicly want to find the formula for the total force on the string right? So i got the resultating forces:

    [tex]F_{res} = mg*cos(\theta) + F_{string} = ma = m(v^2/r)[/tex]

    And hence i isolate the force on the string

    [tex]m(v^2/r) - mg*cos(\theta) = F_{string}[/tex]

    I'm not entirely sure what to do next. I'm not sure if i should be making the Taylor Polynom out of

    [tex]x(t) = x_0 * cos(\omega * t)[/tex]

    Can i simplify that down to only cos(theta) ? That would be easy enough and could perhaps get that to fit the formula given in the question (i'm thinking maybe if i could do that i could replace v^2 in the above example with 2Ax and use the taylor polynom insted of x in the point theta_0 and then substitute cos(theta) with the taylor polynom of theta). Someone also tried to hint to me i should use the preservation of energy, but i'm not seeing where that should enter the picture.

    If someone could perhaps nudge me in the right direction that would be great :eek: (now i just gotta hope all that latex came through as intended).
    Last edited: Sep 25, 2004
  2. jcsd
  3. Sep 25, 2004 #2

    Doc Al

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    Staff: Mentor

    Tip #1: Fix your signs. The radial component of the weight opposes the tension in the string.
    Tip #2: Use conservation of energy to eliminate [itex]v^2[/itex]. Hint: find an expression for the KE in terms of [itex]cos\theta_0[/itex] and [itex]cos\theta[/itex].

    Tip #3: Once you've eliminated [itex]v^2[/itex], replace all cos terms with the first two terms in its Taylor expansion.
  4. Sep 25, 2004 #3


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    Homework Helper

    Hint: You can find [itex]\frac {d \theta}{dt}[/itex] by integrating the equation of motion
    [tex]\frac {d^2 \theta}{dt^2} = - \frac {g}{L} \sin \theta[/tex]
    to find
    [tex]\frac {d \theta}{dt} = \sqrt {\frac {2g}{L} (\cos \theta - \cos \theta_0)}[/tex]
    [itex]L \times d\theta / dt[/itex] is the tangential velocity you need to calculate the centripetal force. You should end up with
    [tex] T = 3 mg \cos \theta - 2mg \cos \theta_0[/tex]
    which is exact, by the way, and you can now do a taylor expansion to get the approximation!
  5. Sep 25, 2004 #4
    The concepts in this problem aren't too bad but the math may be a little intimidating. What you are trying to find in this problem is [tex]T = T(\theta)[/tex], the tension in the string as a function of the angle. You already correctly wrote down what this is:

    [tex]T(\theta) = m\frac{v(\theta)^2}{l} - mg*cos\theta[/tex]

    where [tex]v(\theta)[/tex] is the tangential velocity of the mass as a function of angle. At this point you don't know what [tex]v(\theta)[/tex] is. This is where conservation of energy is handy. Since only conservative forces (i.e. gravity) do work on the mass, energy is conserved. I will pick [tex]U = 0[/tex] where the string is connected to the ceiling. I will take

    [tex]E_{initial} = -mglcos\theta_{0}[/tex]

    where the mass is released from rest and

    [tex]E_{final} = -mglcos\theta + \frac{1}{2}mv(\theta)^2[/tex]

    Equating these two energies allows us to solve for [tex]v(\theta)[/tex]. Thus,

    [tex]v(\theta) = \sqrt{\frac{2g}{l}(cos\theta - cos\theta_{0})}[/tex]

    Plugging this back into the equation for the string tension gives

    [tex]T = mg(3cos\theta - 2cos\theta_{0})[/tex]

    At this point use the small angle approxmation where

    [tex]cos\theta \approx 1 - \frac{1}{2}\theta^2[/tex]

    That should get you the right answer.
    Last edited: Sep 25, 2004
  6. Sep 25, 2004 #5
    Ok thanks guys, it's getting quite late here so i think i'll wait until tomorrow to look it all over and hopefully get the right answer :) Will report in with my findings to see if i got the right idea.
  7. Sep 26, 2004 #6
    Ok so i just went through it and i seem to get it now.

    I have

    [tex]m(v^2/l) + mg*cos(\theta) = F_{string}[/tex]

    Ok, here i had a bit of a problem, i couldn't find a sensible way to represent h in E=mgh, i'd imagine that it was the height from which the bob is raised from it's lowest position right? So i thought it'd be [tex] h=l-(cos(\theta)*l)[/tex], but that just gives me a lot of confusion :) I saw here above that you used mglcos(theta) to do it, and if i use that i can get the right answer. But i'd imagine that mglcos(theta) would give me l-h, but not h. If anyone could explain how that formula = h, i'd be golden :) Anyway, here's the rest of my calculations assuming that lcos(theta) = h

    [tex]E(\theta_0) = -mglcos(\theta_0)[/tex]

    Where the angle is at the extreme of it's motion where v=0, and

    [tex]E(\theta) = -mglcos(\theta) + 1/2*mv^2[/tex]

    At some other arbitrary point in the range of motion. Now the next step, i'm not quite sure you you guys managed to get 2g/l, but what i did to get what i wanted was:

    [tex]E(\theta) = -mglcos(\theta) + 1/2*mv^2 = -mglcos(\theta_0)[/tex]
    [tex]mglcos(\theta)-mglcos(theta_0) = 1/2*mv^2 [/tex]
    [tex]2glcos(\theta)-2glcos(\theta_0) = v^2 [/tex]
    [tex] 2g(cos(\theta) - cos(\theta_0)) = v^2/l[/tex]

    So now i have a formula for v^2/l, so i pop that into the formula i had before and get

    [tex]F_{string} = m*(2g(cos(\theta) - cos(\theta_0)) + mgcos(\theta) [/tex]

    [tex]2mgcos(\theta) - 2mgcos(\theta_0) + mgcos(\theta) = F_{string} [/tex]
    [tex]mg(3mgcos(\theta)-2mgcos(\theta_0)) = F_{string} [/tex]
    so now if i prop the taylor approximation for theta=0 into this i get:
    [tex] mg(3 - 3 * \theta^2/2 -2 + \theta^2_0/2) = F_{string} [/tex]
    [tex]mg(1 + \theta^2_0 - 3/2*\theta^2) = F_{string}[/tex]

    phew that was a lot of texing, but assuming [tex]mglcos(\theta)[/tex] is right, and i'm just not seeing it o_O, this should be good :)
    Last edited: Sep 26, 2004
  8. Sep 26, 2004 #7

    Doc Al

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    Staff: Mentor

    It shouldn't. But it doesn't matter where you measure the potential energy from, as long as you do it consistently. Using the lowest point as your zero, the initial PE is [itex]mg(l - l cos\theta_0)[/itex]. So the KE at angle [itex]\theta[/itex] is [tex]1/2 mv^2 = mg(l - l cos\theta_0) - mg(l - l cos\theta)[/tex] , which simplifies nicely.

    Nothing wrong with this. Note your use of the minus sign. You are effectively taking the pivot point as your zero of PE. No problem!
  9. Sep 26, 2004 #8
    Right you are, they do go out if i measure l-cos(theta)*l. I guess i should've followed that one through, i just looked at it and it looked like i'd have a lot of variables i couldn't get out of the way. But ok it all looks good now and i think i understand it all good now. Thanks a bunch guys.
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