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Acrobat grabbing monkey

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data
    A circus acrobat of mass ##M## leaps straight up with initial velocity ##v_0## from a trampoline. As he rises up, he takes a trained monkey of mass ##m## off a perch at height ##h##, above the trampoline. What is the maximum height attained by the pair?

    2. Relevant equations
    center of mass

    3. The attempt at a solution

    I put that ##y_1(t)## is the trajectory of the acrobat, ##y_2(t)## the trajectory of the monkey, and ##t_e## grabbing time. I look at the system acrobat-monkey before and after grabbing.

    Before grabbing: Between ##0\le t\le t_e## :
    1. ##y_1(t) = tv_0 - \frac{gt^2}{2}##
    2. ##y_2(t) = h##
    so the vertical component of the center of mass is
    ## R_y^{(b)}(t) = \frac{1}{m+M} (my_2+My_1) = \frac{1}{m+M} (mh+Mt(v_0 - \frac{gt}{2}) )##


    After grabbing: When ##t_e \le t##, the center of mass is under external force ##-(m+M)g\vec j##, and its motion is given by:
    ## R_y^{(a)}(t) = h + (t-t_e) v_1 - \frac{g}{2} (t-t_e)^2 ##

    At maximum height, we will have ## \dot R_y^{(a)}(t_{\max}) = 0 ## so that ##t_{\max}- t_e = \frac{v_1}{g}##.
    The maximum height will be ## R_y^{(a)}(t_{\max}) = h + \frac{v_1^2}{2g}##

    Now we must find ##v_1##.
    I assume continuity of the center of mass speed at time ##t_e## ( I'm not sure about that! ), and get that ## v_1 = \frac{M}{m+M}(v_0 - g t_e) ## where the parenthesized term is the speed of the acrobat at grabbing time.

    To conclude I must get rid of ##t_e## which is not a constant of the problem. How should I do that please?
     
  2. jcsd
  3. Dec 10, 2014 #2

    BvU

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    When grabbing, y1(te) = h too.

    However, there might be a much simpler way to deal with this exercise...
     
  4. Dec 10, 2014 #3
    Hi,
    So you solve the equation ## \frac{g}{2} t_e^2 -v_0 t_e + h = 0 ## ? You probably keep the first root since the pair will fall back to height ##h##, therefore ## t_e = \frac{v_0 - \sqrt{v_0^2 - 2 hg}}{g}##

    and maximum height is

    ##(1 - {(\frac{M}{m+M})}^2) h + {(\frac{M}{m+M})}^2 \frac{v_0^2}{2g} ##

    Correct?
     
    Last edited: Dec 10, 2014
  5. Dec 10, 2014 #4

    BvU

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    I get that too.

    And my idea of a simpler way doesn't apply: the grabbing process is an inelastic collision, so there is no conservation for kinetic + potential energy.
    There is momentum conservation (your continuity of c.o.m. speed) that in fact demonstrates this non-conservation.
    Nice exercise!
     
  6. Dec 10, 2014 #5
    Thanks BvU !
    I am a starter in mechanics so I have no knowledge of what is energy for now.
    I am interested by your explanation about continuity of c.o.m speed which I assumed but not quite well understood.
    You justified it by momentum conservation which I understand it prooves continuity. But why is momentum conserved around time ##t_e## ?
     
  7. Dec 10, 2014 #6

    BvU

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    Yes. There is no external force doing work on the system man+monkey during the grabbing. With ## F = {dp\over dt} ## that means p is the same before and after (it's not constant, but there's no sudden change. But you have me in doubt now ... )
     
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