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Finding the Tension of string supporting a disc

  1. Dec 30, 2009 #1
    I am trying to write a simulation program for a relatively complicated mechanical system. Parts of it will be strings wrapped around discs under tension - with the discs moving and rotating. So, I decided to figure out the basics, and am stuck.

    The situation is thus:
    An unstretchable string is attached to a point on a ceiling. The other end, is attached at some point on the circumference of a disc. The disc is rotated and moved to some position, and then let go. How it is rotated, and moved, would determine the exact point at which the string pulls on the disc. Assuming that I know the velocities (linear and angular), and the positions and angles of the disc and string-effect-on-disc points, how would I determine the tension in the string when the disc is let go?
    Or, simpler - if you glue a string to the edge of a quarter, and attach the other end to the ceiling, move the quarter to the side a bit, and rotate it at some angle - the moment when you let go of the quarter, what is the tension in the string?

    I have figured out / stated a bunch of stuff below..still stuck

    [tex]\ddot{X} = \frac{m\vec{g}+\vec{T}}{m}[/tex]

    [tex]\ddot{\theta}_{cm}I_{cm} = (\vec{R}\times\vec{T})_{k}[/tex]

    from here, I had three unknowns and two equations.
    I tried using conservation of energy, but that got me no where:


    [tex]0=\frac{dE}{dt}=(\frac{dKE_{linear}}{dt}) + (\frac{dKE_{angular}}{dt}) + (\frac{dPE_{gravitational}}{dt})[/tex]

    [tex]0=(m\dot{X}\cdot\ddot{X}) + (I_{cm}\dot{\theta}_{cm}\ddot{\theta}_{cm}) + (mg\dot{X}_{j})[/tex]

    and plugging in the first two formulas into the last...

    [tex]0=(\dot{X}\cdot({m\vec{g}+\vec{T}})) + (\dot{\theta}_{cm}(\vec{R}\times\vec{T})_{k}) + (mg\dot{X}_{j})[/tex]


    [tex]0=(-mg\dot{X}_{j}+\dot{X}\cdot \vec{T}) + (\dot{\theta}_{cm}(\vec{R}\times\vec{T})_{k}) + (mg\dot{X}_{j})[/tex]

    which means

    [tex]0=(\dot{X}\cdot \vec{T}) + (\dot{\theta}_{cm}(\vec{R}\times\vec{T})_{k})[/tex]

    however, this tells me nothing about T, since every T solves the equation
    interestingly, this equation just shows the relationship between angular velocity and linear velocity - neato...but currently useless.

    So after making a nice head-shaped indent in the wall of my room, I decided to just look at it from a different perspective.

    Instead of taking the point of rotation to be the center of the disc, I decided to take it as the point at which the string is acting on the disc (this would be the point of attachment, unless the string is wrapped around, and then it would be a tangential point...but that is irrelevant)

    from the attached picture...

    [tex]\ddot{\theta}_{cm} = \frac{(\vec{R}\times\vec{T})_{k}}{I_{cm}}[/tex]

    converting to scalar notation, where T is the tension in the string

    [tex]\ddot{\theta}_{cm} =\frac{ rTsin({\phi}_{1})}{I_{cm}}[/tex]

    similarly cn = 'connection point on disc or point at which the string is acting on the disc'

    [tex]\ddot{\theta}_{cn} =\frac{ (rmg)sin({\phi}_{2})}{I_{cn}}[/tex]

    Now, this may be a wrong assumption, but it seems valid in my mind for several reasons, but..I assumed that

    [tex]\ddot{\theta}_{cm} = \ddot{\theta}_{cn} [/tex]

    if this is true, then

    [tex]\frac{ rTsin({\phi}_{1})}{I_{cm}} = \frac{ (rmg)sin({\phi}_{2})}{I_{cn}}[/tex]

    which would mean that

    [tex]T = \frac{I_{cm}}{I_{cn}}\frac{sin({\phi}_{2})}{sin({\phi}_{1})}mg[/tex]

    moment of inertia:
    [tex]I_{cm} = \frac{mr^{2}}{2}[/tex]
    parallel axis theorum
    [tex]I_{cn} = I_{cm} + mr^{2}[/tex]

    [tex]I_{cn} = \frac{3mr^{2}}{2}[/tex]


    [tex]\frac{I_{cm}}{I_{cn}} = \frac{1}{3}[/tex]

    which leaves

    [tex]T = \frac{1}{3}\frac{sin({\phi}_{2})}{sin({\phi}_{1})}mg[/tex]

    which seems fine and dandy until you look at what happens when both angles are very very small...u get a 0/0...and both num and den can approach 0 at the same rate (in some particular case)

    which would mean that, in the limit as the angles approach 0, in that particular case ^

    [tex]T = \frac{mg}{3}[/tex]

    even though T should equal mg for all cases when the angles approach 0.

    did I just screw up somewhere?...or is my last limiting argument incorrect?

    And, if I didn't get near the answer...how would you determine the tension?

    if you got to this point..thanks for reading this whole giant post :)
  2. jcsd
  3. Dec 30, 2009 #2
    I think i might be missing something with regards to torques around accelerating axes. In both cases, the axes are accelerating - so maybe there is another set of mechanics formulas to describe that, but I don't intuitively think so.
  4. Dec 30, 2009 #3
    o..and a good point was brought up by my friend - my solutions fails when the initial condition is the pendulum position: pull the disc as far as u can at some angle...then, phi_1 = 0, phi_2 isn't 0, and the tension would be infinity by my formula
    so its wrong...but how and why

    also...what of the normal disc pendulum?
    initial phi_1 =0...so with respect to the center, there would not be any torque...so how does the disc start to rotate....or is it that at first, it doesn't rotate at all..and then, after a dt..there is some dphi1?
  5. Dec 30, 2009 #4
    and...my answer is wrong since it doesn't take into account the velocity of the disc..which, when the disc is at the bottom of the pendulum...could be arbitrarily large, and would lead to a very large tension in the string for the centripetal forces.
  6. Dec 30, 2009 #5
    ahh....i think the problem is the frame of reference...i'm using accelerating frames..and treating them as stationary fulcrums over small intervals of time

    "As covered in kinematics the absolute angular velocity of a solid object is the angular velocity in the local coordinates plus the angular velocity of the frame of reference."
    from http://www.euclideanspace.com/physi...otation/rotationfor/index.htm#angularvelocity
  7. Dec 30, 2009 #6
    a good thing to point out as well..even though it's almost new year's, and i doubt anyone will read this anytime soon, if at all....

    is that, in the case there the string is vertical, and tangential to the side of the disc. T=mg/3 is the correct answer!
    So, perhaps i'm not wrong, and only my limiting question is wrong (or something like the function not having a limit since the limit may be different depending on how you approach it!)
  8. Jan 3, 2010 #7
    well, I found the solution for the case where the string isn't wrapped around at all, and the point on the disc that the string is attached too is moving only perpendicularly to the string. (if it isn't moving perpendicularly, then the disc will start moving in such a way as to give the string slack (0 tension), or, is about to bounce back (since the string doesn't stretch)...and that is another solution.

    If anyone is interested...reply, or else there is no point for me to spend the time writing stuff here.
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