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2 swinging balls (pendulum) problem

  • Thread starter awiest82
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Homework Statement



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.

Homework Equations



KE_1+PE_1=KE_2+PE_2

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


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?
 

Answers and Replies

  • #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.
 
  • #3
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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?
 
  • #4
kuruman
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The problem is purely symbolic and since its not specified otherwise I assume that m1=m2.
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.
If that was the case then v'1 would be zero correct?
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.
I leave the coefficient of restitution in since the problem states thats it's a known.
That is good.
Besides those questions you had, does it look like I am on the right track or answered correctly?
Address the issues that I raised, and you will be on the right track.
 
  • #5
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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?
 
  • #6
kuruman
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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?
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:
 
  • #7
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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.
 
  • #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.
 
  • #9
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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?
 
  • #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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