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Inelastic Pendulum Collision

  1. Dec 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Two pendulums collide and stick together after being released from rest at angles given. (See diagram) What is the final swing angle of the stuck-together masses ? (I'm assuming my teacher means the angle when they get to the highest point in their swing) In what direction is the swing?

    2. Relevant equations

    Conservation of Energy, Conservation of Momentum

    3. The attempt at a solution

    I used cons. of e. to find both velocities just before impact, plugged that into cons. of momentum to get Φ in terms of θ, then used cons. of e. with initial being just as they hit and final being when they swing to a maximum height and got the height to be 5L(1-cosθ), which seems horribly, horribly wrong. Trying to go from there, I got the angle to be arccos(5cosθ-4), which can't be right either. Should I have approached this another way? I've done ballistic pendulum problems, but not when both are moving.
     

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    Last edited: Dec 30, 2012
  2. jcsd
  3. Dec 30, 2012 #2
    I don't see how you can get Φ in terms of θ using conservation of momentum. With conservation of momentum, the final velocity is unknown if you consider your angles to be unknown.

    I believe you're supposed to take the angles as known and your final result should have both Φ and θ.

    Also, where exactly did you believe the two masses to collide?
     
  4. Dec 30, 2012 #3
    Yeah, I wasn't sure about finding one angle in terms of the other. I tired saying P before was 0, P after (right before they collide) was 2mv-mV, ∴ 2v=V, and substituting the v's for what i got using cons. of e, but that would only work assuming the collided in the middle. And I didn't think they'd just be colliding in the middle, but I guess I was getting kinda reckless. I was thinking about solving for time for each of them, but I was thinking that would get messy when trying to do conservation of energy after.
     
  5. Dec 30, 2012 #4
    Momentum conservation only applies when [itex]\vec{dP}/dt[/itex] = 0. From that entire time between start and collision, ask yourself if that's true!

    Also keep in mind momentum is a vector, and as such, all equations involving momentum conservation should account for vector nature.

    You're definitely right to consider momentum and energy conversation but first try turning your attention to how SHM behaves. What does it depend on? In what cases would the two masses collide in the middle? In what cases don't they collide in the middle?

    EDIT: Unless someone can suggest an easier method I'm overlooking, it appears you'll be in for some heavy trig/algebra.
     
    Last edited: Dec 30, 2012
  6. Dec 31, 2012 #5

    haruspex

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    I think the question ought to tell you to assume the angles are small enough that you can treat motion before collision as SHM, despite that the diagram shows the angles as quite large. Without that it becomes extremely difficult, if not impossible.
     
  7. Dec 31, 2012 #6
    Yeah, I was looking at this problem and was kinda baffled, because what I thought might work didn't and I had no other ideas. I made some force diagrams though, got T-2mg=2m(v^2/L) and a = 2gsinθ for the left side and T-mg=m(V^2/L) and A = gsinΦ for the right side. I tried solving for time from there, but things got large and messy. Maybe I'll just confront him about it, hope he made a mistake, and gives me credit for trying.
     
  8. Dec 31, 2012 #7
    My teacher originally drew the diagram with large angles, and labeled the left θo and the right θu. I was thinking maybe he just didn't close the top of his u (although it has a big gap). If the angle was the same for both this would be much simpler as they collide in the middle, right?
     
  9. Dec 31, 2012 #8

    ehild

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    Try to solve the problem with equal angles. In this case, the bobs meet in the middle.

    Conservation of momentum is not valid here but conservation of angular momentum applies.

    ehild
     
  10. Dec 31, 2012 #9

    haruspex

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    If you assume SHM they will collide in the middle even if the angles are different. Do you see why? If your teacher had intended the angles the same I think s/he would not have written any subscripts.
     
  11. Jan 1, 2013 #10
    I attempted this using cons. of e. to find each bob's velocity when it meets in the middle and used that velocity in my equations for conservation of angular momentum. I used cons. of e. again from when they were now combined and moving to their final swing angle, got a height, and tried solving for θ, but I get angles that would not happen (like both bobs start at 0 degrees and end up at 60 degrees).


    I was thinking that too, but I thought maybe he gave it a subscript so when I need to say what the final swing angle is I could give it a different subscript. And even assuming they collide at the middle with different angles, I get the same problem stated above.
     
  12. Jan 1, 2013 #11

    haruspex

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    If the masses start off at the same angle (same mass, right?) there should be no motion after collision.
    Pls post your working.
     
  13. Jan 1, 2013 #12

    ehild

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    I do not understand you. You need the final angle of the bobs stuck together, in terms of the θ0, the initial one. It is one angle, not "angles".
    Show what you did.

    ehild
     
  14. Jan 1, 2013 #13

    ehild

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    The bobs are not of the same mass. (m and 2m).

    ehild
     
  15. Jan 1, 2013 #14

    haruspex

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    Yes, I couldn't easily check the diagram in the middle of writing the post. But I think it's clear the initial angles are different, but SHM is to be assumed, so we can take it that collision occurs at the vertical position.
     
  16. Jan 1, 2013 #15
    Sorry about the confusion. I solved for the final angle symbolically, but the answer didn't look right. When I tried to plug in an angle that the bobs started at, such as 0, I got some final angle that didn't make sense for the starting angle, such as 60.

    I used cons. of e. to get the velocity of each bob just before they hit, resulting in the velocity of each to be √(2gL(1-cosθo)). The angular momentum of the left bob before they hit is 2mL2ω and the right is mL2(-ω) (making counterclockwise positive). Total angular momentum before is mL2ω. Angular momentum after they collide is 3mL2ω2. Set equal and ω2=ω/3. I had then changed that to velocity and tried using cons. of e., but I think I see where that went wrong.

    Instead of changing to velocity initially, I tried using cons. of e. with initial being when they start moving together and final being at the final swing angle, and I got (1/2)Iω22 = 3mgL(1-cosθ2). With I being 3mL2, and substituting v/3L for ω2 (after using ω2=ω/3 and v=ωL), I got arccos((8/9)+cosθ) = θ2 as my final answer after simplifying. But any initial angle under 84 comes back undefined.
     
  17. Jan 1, 2013 #16

    SammyS

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    I'm pretty sure that haruspex is correct in saying the problem assumes Simple Harmonic Motion for the pendulums.

    Using conservation of energy, find the velocity of each pendulum bob just prior to the colliding. Use conservation of momentum to analyze the collision, giving you the velocity of the combined bobs immediately after the collision. Then use conservation of energy to find the final swing angle.
     
  18. Jan 1, 2013 #17
    Alright, so the velocity of the left bob just before collision is v1=√(2gL(1-cosθ1)) and the right is v2=√(2gL(1-cosθ2)). Velocity of the combined bobs is v3=(2v2-v1)/3. Using cons. of e. I get (1/2)3mv32=3mgL(1-cosθ3). Simplifying and plugging in for the values of v, I get θ3=arccos((4/9)(1+cosθ1+(1/4)cosθ2+√((1-cosθ1)(1-cosθ2))).

    This looks good to me in terms of the work, but are there any problems with it? Or any ways to simplify it more?
     
    Last edited: Jan 1, 2013
  19. Jan 1, 2013 #18

    ehild

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    You made a little error. cos(θ0) is also divided by 9. arccos((8+cosθ0)/9) = θ2

    ehild
     
  20. Jan 2, 2013 #19

    haruspex

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    Final eqn looks ok if θ1 corresponds to the mass 2m. In the middle there, it seems to have been the other way around.
     
  21. Jan 2, 2013 #20
    Whoops, thanks for catching that.

    Whoops again, v3=(2v1-v2)/3.

    Thanks to both of you for helping me out on all of this!
     
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