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Rotational kinetic energy of a dumbell

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data
    A dumbell made of two point masses separated by a massless rod of fixed length has centre of mass velocity [tex]V[/tex] and rotates about its CM with angular velocity [tex]\omega[/tex]. Two point masses [tex]m_1[/tex] and [tex]m_2[/tex] are located at [tex]r_1[/tex] and [tex]r_2[/tex] repsectivelt wrt the CM. Let [tex]T[/tex] be the kinetic energy of the dumbell in the lab frame and [tex]T'_{rot}[/tex] be the kinetic energ of rotation about the CM. Show that [tex]T = \frac{1}{2}MV^2+T'_{rot} [/tex]and derive and expression for [tex]T'_{rot}[/tex]


    2. Relevant equations



    3. The attempt at a solution

    In the lab frame, the velocity of the point masses is [tex]v'_1=V+v_1[/tex] and [tex]v'_2=V+v_2[/tex] where the little v refers to velocity wrt the CM. So Kinetic energy is given by

    [tex]\frac{1}{2}m_1v'_1\cdotp v'_1 + \frac{1}{2}m_2v'_2\cdotp v'_2[/tex]
    [tex]=\frac{1}{2}m_1(V^2+2V\cdotp v_1+v_1^2) + \frac{1}{2}m_2(V^2+2V\cdotp v_2+v_2^2)[/tex]
    let [tex]M = m_1+m_2[/tex]
    [tex]=\frac{1}{2}MV^2 + V\cdotp (m_1v_1+m_2v_2) + \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2[/tex]

    The last two terms can be reduces to the expression for the rotational kinetic energy, which means that I have the extra term [tex]V\cdotp(m_1v_1+m_2v_2)[/tex] in my answer. Now, either I did something wrong, or this is identically 0 for some reason, but I can't work out which. Can someone help me?
     
    Last edited: Nov 8, 2009
  2. jcsd
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