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Homework Help: Please, still need help with rotating rigid body problem

  1. Nov 15, 2004 #1
    :cry: I am still stuck with this problem. Please, help if you can.
    A large, cylindrical roll of tissue paper of initial radius R lies on a long, horizontal surface with the outside end of the paper nailed to the surface. The roll is given a slight shove (V initial = 0) and commences to unroll. Assume the roll has a uniform density and that mechanical energy is conserved in the process. Determine the speed of the center of mass of the roll when its radius has diminished to r = 1.0 mm, assuming R = 6.0 meters.

    This is what I got so far.

    KE=(1/2)I(Omega)^2

    for the cylinder I is :(1/2)MR^2

    So mgh=(1/2)I(Omega)^2+(1/2)mv^2

    I plugged in I

    mgh=(1/2)(1/2)MR^2(Omega)^2+(1/2)mv^2

    Then I used (omega)=v/r to get here

    mgh=(1/4)MR^2(v/r)^2+(1/2)mv^2 the masses cancel

    gh=(1/4)R^2(v/r)^2+(1/2)v^2

    Now what? How do I get h? Did I miss anything? Thank you in advance!!
     
  2. jcsd
  3. Nov 15, 2004 #2

    arildno

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    1. At the start, the total system has only potential energy, MgR, where M is the total mass of the paper.
    2. At time "t", you should find that the mass of the remaining rolled-up paper, m, is:
    [tex]m(t)=(\frac{r(t)}{R})^{2}M[/tex]
    3. The potential energy of the total system remaining at time "t" is:
    [tex]mgr(t)=Mg\frac{r(t)^{3}}{R^{2}}[/tex]
    4. The kinetic energy of the total system at time "t" is:
    [tex]\frac{1}{2}m(t)v^{2}_{G}(t)+\frac{1}{2}(\frac{1}{2}m(t)r(t)^{2})\omega(t)^{2}=\frac{3}{4}(\frac{r(t)}{R})^{2}Mv^{2}_{G}(t)[/tex]
    by using the rolling condition to relate the angular velocity to the center of mass of the rolled up paper.
    5. Hence, you get:
    [tex]v_{G}(t)=\frac{\sqrt{4g(R^{3}-r^{3}(t))}}{\sqrt{3}r(t)}[/tex]
     
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