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Linear motion converted into rotatory motion problem

  1. Oct 21, 2012 #1
    1. The problem statement, all variables and given/known data
    In a ballistic pendulum an object of mass m is fired with an initial speed v_0 at a pendulum bob. The bob has a mass M, which is suspended by a rod of length L and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement theta.
    Find an expression for v_0, the initial speed of the fired object.
    Express your answer in terms of some or all of the variables m, M, L, and theta and the acceleration due to gravity, g.
    a figure is attached for better understanding!!
    please help! :(


    2. Relevant equations

    v= rω, s=rθ, ω=θ/t

    3. The attempt at a solution

    i have tried using the above examples but i dont know what to remove time since we are not asked to write in terms of time
     

    Attached Files:

  2. jcsd
  3. Oct 21, 2012 #2

    lewando

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    Approach the problem from an energy and momentum point of view. Find final (maximum) PE, KE after the collision, KE before the collision.
     
    Last edited: Oct 21, 2012
  4. Oct 21, 2012 #3
    that was helpful! but while finding the potential energy, how will we find the height?
     
  5. Oct 21, 2012 #4

    Doc Al

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    Staff: Mentor

    Use a bit of trig.
     
  6. Oct 21, 2012 #5
    i got the answer v = (2g(m+M)(L-Lcostheta)/m)^1/2
    what do u think?
     
  7. Oct 21, 2012 #6

    Doc Al

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    Rather than just give your final answer, describe how you arrived at your answer.
     
  8. Oct 21, 2012 #7
    ok so i got the height as L-Lcostheta

    the kinetic energy of the block is 0 and the kinetic energy of both of them at the heighest point is 0. so
    1/2mv(initial)^2 = (m+M)g(L-Lcostheta)
    then i just rearranged to find v

    i am still a little doubtful about the height though
     
  9. Oct 21, 2012 #8

    Doc Al

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    The height is just fine. But your assumption that energy is conserved during the collision is not.
     
  10. Oct 21, 2012 #9
    umm but then how will we find the equations?
    K.E before the collision is 1/2mvinitial^2
    after the collision = 1/2(m+M)v^2 + mg(L-Lcostheta)
    momentum is conserved so
    mvinitial = (m+M)v
     
  11. Oct 21, 2012 #10

    Doc Al

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    Good. That's what you were missing.

    Treat the motion as having two stages:
    - The collision. Momentum (not energy) is conserved.
    - The rise after the collision. Energy is conserved.
     
  12. Oct 21, 2012 #11
    so momentum is mvinitial = (m+M)v and after rearranging it will be v= mvinitial/(m+M)
    then energy is conserved after collision 1/2(m+M)v^2 = mg(L-Lcosθ)

    then we will substitute the value of v from the conservation of momentum equation so it will be
    1/2 *mvinitial^2/(m+M) = g(L-Lcosθ)
    and then rearranging, vinitial = 2g(m+M)(L-Lcosθ) / m

    correct?
     
  13. Oct 21, 2012 #12

    lewando

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    The PE side of the equation's mass should be (m+M)
     
  14. Oct 21, 2012 #13
    o yes yes! sorry! so
    the energy equation after substituting for v will be

    [ (m+M) m^2 * vinitial ^2 ] /2(m+M) = (m + M)g (L-Lcostheta)

    and then vinitial = [2g(m+M)^2 (L-Lcos theta)] ^1/2 / m
     
  15. Oct 21, 2012 #14

    lewando

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    I didn't get that. Show all steps. At least show what you got for v.
     
  16. Oct 21, 2012 #15
    since momentum is conserved, collision before the moemntum is = after collision
    mvinitial = (m+M)v
    we rearrange for v so v= m vinitial/ (m+M)

    then energy is conserved after the collision so

    1/2 (m+M) v^2 = (m+M)g(L-Lcosθ)

    then we replace v after collision with initial velocity so we substitute for it from the first equation in the second equation

    1/2 (m +M) [m vinitial /(m+M) ] ^2 = (m+M)g (L-Lcosθ)
    we simplify it and (m+M) is cancelled

    so it becomes
    1/2 * (m vinitial)^2 / (m+M) = (m+M) g(L-Lcosθ)

    then rearrange (m vinitial)^2 = 2g(m+M)^2 (L-Lcosθ)
    then take the square root of both the sides and send m to the other side
    vinitial = [ 2g(m+M)^2 (L-Lcosθ) ] ^1/2 / m
    vinitial = (m+M) [ 2g(L-Lcosθ ] ^1/2 /m

    hope this clears up a little
     
  17. Oct 21, 2012 #16

    lewando

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    Clears it up a lot. Good work by you!
     
  18. Oct 21, 2012 #17
    you have no idea how thankful i am of u!!!!
    u have helped me in return of no personal gain!!
    i am literally moved by your act of kindness!!
    thank you so so much!!
    hope all your problems(physics and otherwise :P) work out !!


    :)
     
  19. Oct 21, 2012 #18

    lewando

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    You are welcome. You wish will be shared by me, Doc_Al, and many others.
     
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