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Conservation of Energy of a rolling sphere

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data

    A sphere rolling with an initial velocity of 30 ft/s starts up a plane inclined at an angle of 30o with the horizontal as shown. How far will it roll up the plane before it rolls back down?

    2. Relevant equations

    [itex] T_1+V_1=T_2+V_2 [/itex]

    3. The attempt at a solution
    We are doing rigid bodies so I started with

    [itex] T_1+V_1=T_2+V_2 [/itex] Where [itex]V_1=T_2=0[/itex] So I have

    [itex] T_1=V_2 [/itex]

    [itex].5 m v^2 = m g h [/itex]

    [itex] h=x\sin(30) [/itex]

    Which gives me [itex] .5(30)^2 = (32.2)(x \sin(30)) [/itex]

    [itex] x=27.95 [/itex]


    And that is not one of the answers, I assume inertia is supposed to be used somewhere but I have no idea where to plug it in because no radius of the circle is give.. Any help would be very much appreciated\



    EDIT:
    I tried using Inertia like this:

    [itex] T_1 = V_2 [/itex]

    [itex] T_1 = .5 I \omega^2 + .5mv^2 \quad, \qquad \omega = v/r [/itex]

    [itex] I=.5 m r^2 [/itex]

    [itex] T_1 = .5((.5 m r^2)(\frac{v}{r})^2) + .5 m v^2 [/itex]

    [itex] T_1 = \frac{1}{4} m v^2 + \frac{1}{2} m v^2 = \frac{3}{4}mv^2 [/itex]

    [itex] \frac{3}{4}mv^2 = m g x \sin(30) [/itex]

    [itex] x = \frac{3v^2}{4 g \sin(30)} \qquad, x=41.9255 [/itex]

    Still not an answer but closer than I was.. Any suggestions?
     

    Attached Files:

    Last edited: Apr 13, 2012
  2. jcsd
  3. Apr 13, 2012 #2
    It's a sphere! Assuming this to be a case of rolling without slipping, the the total energy in the sphere is [itex]\frac{7}{10}[/itex]mv[itex]^{2}[/itex]

    Here's how:

    Total energy
    = Translational Kinetic energy + Rotational kinetic energy
    = [itex]\frac{1}{2}[/itex]mv[itex]^{2}[/itex] + [itex]\frac{1}{2}[/itex]I[itex]\omega[/itex][itex]^{2}[/itex]

    (m = mass
    v = velocity
    I= moment of iniertia about the centre
    [itex]\omega[/itex]=angular velocity about the centre
    r=radius of the sphere
    )

    =[itex]\frac{1}{2}[/itex]mv[itex]^{2}[/itex] + [itex]\frac{1}{2}[/itex]([itex]\frac{2}{5}[/itex]mr[itex]^{2}[/itex])[itex]\omega[/itex][itex]^{2}[/itex]


    (Because I = [itex]\frac{2}{5}[/itex]mr[itex]^{2}[/itex], for a sphere)

    =[itex]\frac{1}{2}[/itex]mv[itex]^{2}[/itex] + [itex]\frac{1}{2}[/itex]([itex]\frac{2}{5}[/itex]mr[itex]^{2}[/itex])([itex]\frac{v}{r}[/itex])[itex]^{2}[/itex]


    (Because v = [itex]\omega[/itex]r, for rolling without slipping)

    Solving this gives [itex]\frac{7}{10}[/itex]mv[itex]^{2}[/itex].

    Try putting that and you should get an answer.
     
  4. Apr 13, 2012 #3
    Ugh, I'm an idiot haha. Thank you! I got it now!
     
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