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Ball hitting a hanging rod

  1. Aug 11, 2015 #1
    1. The problem statement, all variables and given/known data
    A rod of length L hangs on a nail. a ball of mass m hits and sticks to it. the rod rotates to angle θ.
    What is the velocity of the ball and what's the loss of energy.

    2. Relevant equations
    Kinetic energy of a rigid body: ##E=\frac{1}{2}I\omega^2##
    Angular momentum: ##I\omega##

    3. The attempt at a solution
    Location of C.O.M:
    $$x_{c.m.}=\frac{mL+M\frac{L}{2}}{m+M}=\frac{L}{m+M}\left( m+\frac{M}{2} \right)$$
    The potential energy at the inclined position, the final position:
    $$E_f=(m+M)g\cdot y_{c.m.}=(m+M)\frac{L}{m+M}\left( m+\frac{M}{2} \right)g\cos\alpha=Lg\left( m+\frac{M}{2} \right)\cos\alpha$$
    Ef equals the kinetic energy after the hit:
    $$E_f=\frac{1}{2}I\omega^2\rightarrow Lg\left( m+\frac{M}{2} \right)\cos\alpha=(mL^2+\frac{1}{3}ML^2)\omega^2\Rightarrow \omega^2=\frac{g\left( m+\frac{M}{2} \right)\cos\alpha}{L\left( m+\frac{1}{3}M \right)}$$
    Conservation of angular momentum:
    $$mvL=I\omega \Rightarrow mvL=\left( mL^2+\frac{1}{3} ML^2 \right) \omega$$
    $$\Rightarrow v=\frac{ L\left( m+\frac{1}{3}M \right) \omega }{m}=\sqrt{Lg\left( m+\frac{1}{3}M \right)\left(m+\frac{M}{2} \right) }$$
    $$\Delta E=Lg\left( m+\frac{M}{2} \right)\cos\alpha-\frac{1}{2}mv^2=Lg\left( m+\frac{M}{2} \right) \left[ \cos\alpha-\frac{1}{2m}\left( m+\frac{1}{3}M\right) \right]$$
    Is it true?
     

    Attached Files:

  2. jcsd
  3. Aug 11, 2015 #2

    haruspex

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    What baseline are you taking for the PE? As theta (alpha) increases, does the PE increase or decrease?
     
  4. Aug 11, 2015 #3
    $$E_f=(m+M)g\cdot y_{c.m.}=(m+M)\frac{L}{m+M}\left( m+\frac{M}{2} \right)g(1-\cos\alpha)=Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha)$$
    $$E_f=\frac{1}{2}I\omega^2\rightarrow Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha)=(mL^2+\frac{1}{3}ML^2)\omega^2\Rightarrow \omega^2=\frac{g\left( m+\frac{M}{2} \right)(1-\cos\alpha)}{L\left( m+\frac{1}{3}M \right)}$$
    $$v=\frac{ L\left( m+\frac{1}{3}M \right) \omega }{m}=\sqrt{Lg\left( m+\frac{1}{3}M \right)\left(m+\frac{M}{2} \right)(1-\cos\alpha) }$$
    $$\Delta E=Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha)-\frac{1}{2}mv^2=Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha) \left[1-\frac{m}{2}\left( m+\frac{M}{3}\right) \right]$$
     
  5. Aug 12, 2015 #4

    haruspex

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    The first two lines look fine.
    In the third you seem to have lost a 1/m. (Check the dimensions.)
    In the middle expression of the last line you seem to have lost an entire factor, but maybe that was just a transcription error.
    The final expression is dimensionally inconsistent (inside the square brackets you have a constant term minus a mass2 term).
     
  6. Aug 12, 2015 #5
    $$v=\frac{ L\left( m+\frac{1}{3}M \right) \omega }{m}=\frac{1}{m}\sqrt{Lg\left( m+\frac{1}{3}M \right)\left(m+\frac{M}{2} \right)(1-\cos\alpha) }$$
    I don't think i missed a factor:
    $$\Delta E=E_f-\frac{1}{2}mv^2$$
    $$\Delta E=Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha)-\frac{1}{2}mv^2=Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha) \left[1-\frac{1}{2m}\left( m+\frac{M}{3}\right) \right]$$
    And the units are correct, at least
     
  7. Aug 12, 2015 #6

    ehild

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    Haven't you lost an 1/2?
     
  8. Aug 12, 2015 #7
    $$E_f=\frac{1}{2}I\omega^2\rightarrow Lg\left( m+\frac{M}{2} \right)(1-\cos\alpha)=\frac{1}{2}(mL^2+\frac{1}{3}ML^2)\omega^2\Rightarrow \omega^2=\frac{2g\left( m+\frac{M}{2} \right)(1-\cos\alpha)}{L\left( m+\frac{1}{3}M \right)}$$
    And i understand that the rest is good
     
  9. Aug 12, 2015 #8

    ehild

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    With the factor 2, it is correct.
     
  10. Aug 12, 2015 #9
    Thank you ehild and haruspex (factor of 2...)
     
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