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Impact of rolling sphere with vertical rod

  1. Mar 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Uniform sphere of mass ##m## and radius ##r##, rolling without slipping on a horizontal surface at a velocity ##v_0##, hits the end of a rod of length ##l## and mass ##10m##, hanging on a hinge (Figure). Determine how the ball will move and the rod after the impact. What time after, the motion of the ball goes into pure rolling? What is the speed of the roller? An impact is perfectly elastic and instantaneous. The coefficient of friction of the ball on the plane is ##k##. Rolling friction neglected.

    1dd8756cdc1aace34b4affd59b383609.jpg


    2. Relevant equations

    Eqs of conservations:
    Angular momentun: ##mv_0l=\dfrac{10ml^2}{3}\omega-mvl##
    Energy: ##\underbrace{\dfrac{mv_0^2}{2}+\dfrac{2}{5}mr^2\dfrac{v_0^2}{2r^2}}_{\dfrac{7mv_0^2}{10}}=\dfrac{10ml^2}{3}\dfrac{\omega^2}{2}+\dfrac{mv^2}{2}##

    3. The attempt at a solution
    I feel that I made a conceptual error in the conservation laws, but I do not know where.
     
  2. jcsd
  3. Mar 21, 2013 #2
    I think I understand. The system of two equations is inconsistent. But if we assume that the sphere after being hit does not lose its spin moment, then everything works:
    ##
    \begin{equation}
    \left\{
    \begin{array}
    \ mv_0l =\dfrac{10ml^2}{3}\omega &- mvl \\
    \dfrac{mv_0^2}{2} =\dfrac{10ml^2}{3}\dfrac{\omega^2}{2} &+ \dfrac{mv^2}{2}
    \end{array}
    \right.
    \end{equation}\text{ without loss of spin angular momentum}
    ##

    Then, we have
    ##v=\dfrac{7}{13}v_0##, ##\omega=\dfrac{6}{13} \dfrac{v_0}{l}##

    But why the sphere does not stop spinning?
     
  4. Mar 21, 2013 #3

    ehild

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    You omitted the rotational energy of the ball from the energy equation.

    ehild
     
  5. Mar 21, 2013 #4
    Yes, exactly! Thank you.
     
  6. Mar 21, 2013 #5

    ehild

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    The rod hits the ball at the height of its centre. So the rod does not exert torque about the centre of the ball. No torque, no change of angular momentum. (The force of friction is finite, it does not have effect on angular momentum in infinitesimally short time)

    ehild
     
  7. Mar 21, 2013 #6
    Oh, really! Thank you!
    Now I start to calculate the time from bounce to pure roll.
    I have two eqns:
    ##m\dfrac{dv}{dt}=-F_{f}##
    ##\dfrac{2}{5}mr^2\dfrac{d\omega}{dt}=-F_{f}r##
    Here not mess with the signs?
     
  8. Mar 21, 2013 #7
    From integrating above eqns:
    ##\dfrac{2}{5}\omega r-v=const##
    Constant determine from initial conditions, e.g. ##const=\dfrac{2}{5}\dfrac{v_0}{r} r-\dfrac{7}{13}v_0=-\dfrac{9}{65}v_0##
    When will the pure rolling ##v=\omega r##, then ##\dfrac{2}{5}\omega r-\omega r= -\dfrac{9}{65}v_0##
    E.g.
    ##\omega r = \dfrac{3}{13}v_0##

    ##v=\dfrac{3}{13}v_0## - velocity of pure rolling.
    Then
    ##v_{at\; the\; finish\; of\; sliding,\; or\; starting\; pure\; roll}=v_{initial\; velocity\; after\; bounce} +at##
    or
    ##\dfrac{3}{13}v_0-\dfrac{7}{13}v_0=at##
    ##-\dfrac{4}{13}v_0=at##
    ##t=-\dfrac{4}{13}v_0\dfrac{1}{a}##
    From the eqn of center mass motion ##a=-kg##

    Answer ##t=\dfrac{4}{13}\dfrac{v_0}{kg}##

    That does not match with the answer book of problems ##t=\dfrac{40}{91}\dfrac{v_0}{kg}##
     
  9. Mar 21, 2013 #8

    ehild

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    Take care of the signs.
    After collision, the ball travels to the right and spins anticlockwise. When it starts to roll it travels to the right with velocity v and rotates clockwise with ω=v/r. The friction accelerates clockwise rotation.

    ehild
     
  10. Mar 21, 2013 #9
    I was not thinking. Thanks!

    ef0d7deae537cab688ad2e55d7843ede.jpg

    Right calculations:

    From integrating above eqns:
    ##\dfrac{2}{5}\omega r-v=const##
    Constant determine from initial conditions, e.g. ##const=\dfrac{2}{5}\dfrac{v_0}{r} r-\dfrac{7}{13}v_0=-\dfrac{9}{65}v_0##
    When will the pure rolling ##v=-\omega r##, then ##\dfrac{2}{5}\omega r+\omega r= -\dfrac{9}{65}v_0##
    E.g.
    ##\omega r = - \dfrac{9}{91}v_0##

    ##v=\dfrac{9}{91}v_0## - velocity of pure rolling.
    Then
    ##v_{at\; the\; finish\; of\; sliding,\; or\; starting\; pure\; roll}=v_{initial\; velocity\; after\; bounce} +at##
    or
    ##\dfrac{9}{91}v_0-\dfrac{7}{13}v_0=at##
    ##-\dfrac{4}{13}v_0=at##
    ##t=-\dfrac{40}{91}v_0\dfrac{1}{a}##
    From the eqn of center mass motion ##a=-kg##

    Answer ##t=\dfrac{40}{91}\dfrac{v_0}{kg}##
     
    Last edited: Mar 21, 2013
  11. Mar 21, 2013 #10

    ehild

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    Gold Member

    Nice work!

    ehild
     
  12. Mar 21, 2013 #11
    Thanks for the help!
     
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