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Homework Help: Problem on torsional constant

  1. Aug 2, 2004 #1
    I came to find out through books and actual experiment that the torsional constant of a wire in a torsional pendulum is directly inverse to its length
    I find the torsional constant using the rotational version of hooke's law, t=k*theta.
    I have always thought that a greater length should give a greater torque (thus a higher torsional constant) because it has a higher momentum.
    I tried to search for more specific explainations as to why it is a inverse relation on the internet, but without much help. The books I got in the libraries are either too simple or too advaned for me.
    Any explainations would be greatly appreciated. Thx!
     
  2. jcsd
  3. Aug 4, 2004 #2

    Doc Al

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    Staff: Mentor

    Think of it like this. The torsional constant is defined as the amount of torque needed to get a certain angular twist: [itex]k = \tau/\theta[/itex]. So assume a given wire of length L has a constant k. It requires a torque [itex]\tau[/itex] to produce a twist of [itex]\theta[/itex]. What if I only needed a twist of [itex]\theta/2[/itex]? Would you agree that I only need half the torque? (I presume you would.)

    Now consider a wire of length 2L. You can think of it as being composed of two wires of length L in series. What the torsional constant of this composite wire? If I want a net twist of [itex]\theta[/itex], realize that each half of the wire only gets a twist of [itex]\theta/2[/itex]. Thus the same twist ([itex]\theta[/itex]) requires only half the torque. Thus the net torsional constant of a wire of length 2L is 1/2 the constant of a wire of length L. Make sense?

    Note that this is the same thing that happens with springs put in series. Say I have two springs of spring constant k. If I hook them in series, what's the spring constant of the composite double spring? Figure it out the same way as I did above and you'll find that the double spring has a spring constant of k/2.
     
  4. Aug 8, 2004 #3
    Thx a lot for you help, I got the hang of it now
     
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