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Homework Help: 2 swinging balls (pendulum) problem

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data

    A pendulum of mass m_1 is released at an angle theta_1 and strikes the pendulum of mass m_2, which rises to angle theta_2. The coefficient of restitution e for the two masses is given. Derive an expression for the angle theta_2.

    2. Relevant equations

    KE_1+PE_1=KE_2+PE_2

    e=(v2'-v1')/(v1-v2)


    3. The attempt at a solution

    Solving for v1 at time of impact using conservation of energy I get v1=Sqrt(2gl(1-cos(theta_1)))

    I solve the same equation for v'2=sqrt(2gl(1-cos(theta_2)))

    I then plug in the coefficient of restitution equation and solve for theta_2

    theta_2= arccos(1-((e*Sqrt(2gl(1-cos(theta_1)))+v'1)^2)/2gl

    My issue is determining whether or not this is a complete elastic collision. If so I assume the v'1 would be omitted, correct?
     
  2. jcsd
  3. Mar 30, 2010 #2

    kuruman

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    To determine if the collision is perfectly elastic, you need to look at the coefficient of restitution. If e =1, then the collision is perfectly elastic. If e = 0, the collision is perfectly inelastic (the masses stick together). If 0 < e < 1, the collision is inelastic. Unless a number for e is given, that's all you can say.

    Why do you say that v'1 = 0 for an elastic collision? This would be the case only if m1=m2.
     
  4. Mar 30, 2010 #3
    The problem is purely symbolic and since its not specified otherwise I assume that m1=m2. If that was the case then v'1 would be zero correct? I leave the coefficient of restitution in since the problem states thats it's a known.

    Besides those questions you had, does it look like I am on the right track or answered correctly?
     
  5. Mar 30, 2010 #4

    kuruman

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    If it is not specified otherwise, then you must assume the most general case, and that is that the two masses are not the same.
    Correct, but like I said above you may not assume that the masses are equal in which case v'1 is not zero. You need to find what it is. I suggest that you use the momentum conservation equation to do so.
    That is good.
    Address the issues that I raised, and you will be on the right track.
     
  6. Mar 30, 2010 #5
    I have taken your suggestion and used the conservation of momentum equation to solve for v'1 and the resultant equation I get is:

    v'1=v1-(m2)/(m1)*v'2

    plugging this into the coefficient of restitution equation along with the other solved variables leaves me with:

    e=(Sqrt(2gl(1-cos(theta_2))-Sqrt(2gl(1-cos(theta_1))-(m2/m1)*Sqrt(2gl(1-cos(theta_2)))/(Sqrt(2gl(1-cos(theta_1)))

    Before I begin the "painful" exercise of simplifying and solving this, is this looking a bit better?
     
  7. Mar 31, 2010 #6

    kuruman

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    Much better. The solution is not all that painful. Note that you have a common factor of sqrt(2gl) in both numerator and denominator. Once you get rid of that, solve for sqrt(1-cos(θ2)) in terms of sqrt(1-cos(θ1)) and you are almost done. :wink:
     
  8. Mar 31, 2010 #7
    Ok lets see how this looks for a final answer

    Theta2 = arccos(1-((e+1)2*m22*(1-cos(theta1)))/(m2-m1)2)

    The only issue I see with this answer is that if m1 and m2 are identical, we then have a 0 in the denominator. I have worked it out twice and gotten the same answer both times, so I think this is right.
     
  9. Mar 31, 2010 #8

    kuruman

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    After looking at your expression for e more carefully, I see that you have a wrong sign in front of the mass ratio. If
    e = |v'2 - v'1|/v1
    and
    v'1 = v1 - v'2*m2/m1
    then
    e = |v'2 - (v1 - v'2*m2/m1)|/v1
    and
    e = |v'2*(1+m2/m1) - v1|/v1

    You can take it from here.
     
  10. Mar 31, 2010 #9
    Aha! I forgot to distribute that minus sign!

    Here is what I get now and no longer have a 0 in denominator.

    Theta2 = arccos(1-((e+1)2*(m2)2*(1-cos(theta1)))/(m2+m1)2)

    Look better?
     
  11. Mar 31, 2010 #10

    kuruman

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    I haven't done the algebra, but it looks OK. Why don't you test your expression? What should θ2 be when m1=m2 and you have a perfectly elastic collision, i.e. e = 1? Does your expression give you what you expect?
     
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