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Potential Energy of Masses on a Pivot

  1. Aug 22, 2009 #1
    A straight rod of negligable mass is mounted on a frictionless pivot (see attached diagram)Blocks having masses [tex]m_1,m_2[/tex] are attched to the rod at distances [tex]l_1,l_2[/tex]. (a) Write an expression for teh gravitational potential energy of the blocks-Earth system as a function of the angle [tex]\theta[/tex] made with the horizontal. (b) Find the angle of minimum potential energy. (c) Show that if [tex]m_2l_2=m_1l_2[/tex] then the system is in equilibrium regardless of the angle [tex]\theta[/tex]

    I find that the potential energy function is:

    [tex]u(\theta)=(m_2l_2-m_1l_1)g_E\sin(\theta)[/tex]

    [tex]\frac{du}{d\theta}=(m_2l_2-m_1l_1)g_E\cos(\theta)[/tex]

    Setting the derivative equal to zero gives [tex]m_2l_2=m1_l_1[/tex].


    This seems to address part c. Setting this equal to zero doesn't give an angle (part b). How do I solve part (b)?
     
  2. jcsd
  3. Aug 22, 2009 #2
    I almost forgot the diagram. Attached.
     

    Attached Files:

  4. Aug 23, 2009 #3

    ideasrule

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    Use your intuition. If you hold a rod by the middle and attach a mass on one side, at what angle does the rod come to rest?
     
  5. Aug 23, 2009 #4

    kuruman

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    And what happens to your potential energy if this condition is satisfied?
     
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