Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Energy in an angular (coil) spring system

  1. Jul 1, 2011 #1
    1. The problem statement, all variables and given/known data

    A solid disk of mass M and radius R is on a vertical shaft. The shaft is attached to a coil spring which exerts a linear restoring torque of magnitude [tex]C\theta,[/tex] where theta is the angle measured from the static equilibrium position and C is constant. Neglect the mass of the shaft and the spring, and assume the bearings to be frictionless. (Please tell me if you want me to attach a figure as well. That is a lot of work but can be arranged.)

    a. Show that the disc can undergo simple harmonic motion, and find the frequency of the motion.

    b. Suppose that the disc is moving according to [tex]\theta=\theta_{0}\sin(\omega t),[/tex] where omega is the frequency found in part a. At time [tex]t_{1}=\pi / \omega ,[/tex] a ring of sticky putty of mass M and radius R is dropped concentrically on the disk. Find:
    (1) The new frequency of the motion
    (2) The new amplitude of the motion

    2. Relevant equations


    3. The attempt at a solution

    a. The initial moment of inertia is [tex]I_{0}=\frac{MR^{2}}{2}[/tex] and the torque is defined as [tex]\tau = I \ddot{\theta}.[/tex] In this case the torque is [tex]\tau=-C\theta,[/tex] which leads to the equation [tex]\ddot{\theta}=-\frac{2C}{MR^{2}}\theta,[/tex] with the solution [tex]\theta=\theta_{0}\sin (\omega t),[/tex] where [tex]\omega=\sqrt{\frac{2C}{MR^{2}}}.[/tex]

    b. Differentiating the equation leads to [tex]\dot{\theta}=\theta_{0}\omega\cos (\omega t),[/tex] so [tex]\dot{\theta}(\pi / \omega)=-\theta_{0} \omega,[/tex] so the disc has a certain velocity. Calculating the new frequency is easy: [tex]\omega_{new}=\sqrt{\frac{C}{\frac{MR^{2}}{2}+MR^{2}}}=\sqrt{\frac{2C}{3MR^{2}}}[/tex] Now since the angular momentum is conserved [tex]\theta_{0}\omega=\theta_{new}\omega_{new},[/tex] so [tex]\theta_{new}=\theta_{0}\sqrt{3}.[/tex]

    Now can this be correct? The energy is defined as [tex]\frac{I\omega^{2}}{2}.[/tex] So in this case [tex]E_{i}=\frac{I_{0}\omega^{2}}{2}=C[/tex] and [tex]E_{f}=\frac{(I_{0}+I_{ring})\omega_{new}^{2}}{2}=C[/tex] so the energy is conserved. I dont understand this because there is a collision and some energy should be lost! Does this mean that my calculation is incorrect. And if so, where did I go wrong?
  2. jcsd
  3. Jul 1, 2011 #2


    User Avatar

    Staff: Mentor

    Nice work. Consider if the putty is dropped on the disk at the moment when it is at the full travel in one direction (so it is momentarily not moving). There is no "collision" in the direction of the oscillatory motion, just a change in the moment of inertia...
  4. Jul 2, 2011 #3
    Thanks Berkeman! Yes, then there would be no energy loss and the amplitude should be the same. However, after thinking it through I have arrived at some other calculations. Firstly since angular momentum is conserved and the angular (radian) speed (what is it called now again?) in fact is given by omega*theta this should lead to [tex]I_{0}\omega_{0}\theta_{0}=3I_{0}\frac{\omega_{0}}{\sqrt{3}}\theta_{new},[/tex] which leads to [tex]\theta_{new}=\frac{\theta_{0}}{\sqrt{3}}.[/tex]
    And since the physical omega comes into the energy equations this should lead to that the energy is not the spring constant, but the spring constant times the amplitude squared, i. e. [tex]E=\frac{C\theta^{2}}{2}[/tex] and therefore there is a loss of 2/3 of the energy. If I put it on the disk "when it is at the full travel in one direction" the amplitude will not change and there will be no energy loss at all.

    Now if this is wrong I dont know what to do. I cant get any further.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook