Collision of Two Swinging Bars, Maximum Angle

[tzonehunter]

Homework Statement

Two uniform bars have masses and lengths of m, l and 2m, 2l, respectively. The bars are pivoted at the common frictionless hinge as shown. The bars are released from the horizontal position in such a way that they collide and stick together at the bottom position. After that, the bars swing upward together. Find the maximum angle θ the bars would make with the vertical after the collision.

Homework Equations

I chose to solve this in 3 steps:
1) Use conservation of energy of C.O.M. of each bar to determine ω of each bar just prior to collision.
2) Use conservation of angular momentum to determine ω of the two bar system immediately after collision.
3) Use conservation of energy to determine final height of the C.O.M. of bar 2, and use that to find θ.

The Attempt at a Solution

Step 1:
Bar 1:
U_{gi} + E_{ki} = U_{gf} + E_{kf}
mgl + 0 = \frac {mgl}{2} + \frac {I_{pivot}ω_1^2}{2}

I_{pivot} = \frac {m L^2}{3}
m = m, L = l

mgl = \frac {mgl}{2} + \frac {ml^2ω_1^2}{6}
\frac {g}{2} = \frac {l}{6}ω_1^2
ω_1 = \sqrt{\frac {3g}{l}}

Bar 2:
U_{gi} + E_{ki} = U_{gf} + E_{kf}
2mgl + 0 = 0 + \frac {I_{pivot}ω_1^2}{2}

I_{pivot} = \frac {8m l^2}{3} (m = 2m, L = 2l)
2mgl = \frac {8ml^2ω_2^2}{6}

2g = \frac {8l}{6}ω_2^2
ω_2 = \sqrt{\frac {3g}{2l}} = \frac{1}{2}\sqrt{\frac {6g}{l}}

 

[tzonehunter]

Step 2: Apply conservation of angular momentum to determine the angular velocity of both bars immediately after collision.
L_{before} = L_{after}
I_1ω_1 - I_2ω_2 = -(I_1 + I_2)ω_{both}

\frac {ml^2}{3}\sqrt{\frac{3g}{l}} - \frac {8ml^2}{3}(\frac{1}{2}\sqrt{\frac{6g}{l}}) = -(\frac {ml^2}{3} + \frac {8ml^2}{3})ω_{both}

\frac {1}{3}\sqrt{\frac{3g}{l}} - \frac {4}{3}\sqrt{\frac{6g}{l}} = -(\frac {1}{3} + \frac {8}{3})ω_{both}

\frac {4\sqrt{6g} - \sqrt{3g}}{3\sqrt{l}} = 3ω_{both}ω_{both} = \frac {\sqrt{g}}{9\sqrt{l}}(4\sqrt{6} - \sqrt{3})

 

[tzonehunter]

Step 3: Apply conservation of energy to the swinging 2 bar system to determine h_2.
This picture shows the change in position of the center of mass of each bar from the collision to the top of the swing. I truncated the 2nd half of each bar to try to simplify the picture. I have drawn both bars separately.

Starting with the 2nd bar:
l cos θ = l - h_2
h_2 = l - l cos θ

Now the first bar:
\frac{l}{2} cos θ = \frac{l}{2} - h_1
h_1 = \frac{l}{2} - \frac{l}{2} cos θ = \frac{1}{2}h_2
Since the C.O.M. of the shorter bar starts at a position of \frac{l}{2} above the reference point,
h_{1 final} = \frac{1}{2}h_2 + \frac{l}{2}Now apply conservation of energy:

U_{gi} + E_{ki} = U_{gf} + 0

mg(\frac{l}{2}) + 2mg(0) + \frac{1}{2}(I_{pivot1} + I_{pivot2})ω_{both}^2 = mgh_{1 final} + 2mgh_2

mg(\frac{l}{2}) + \frac{1}{2}(\frac {ml^2}{3} + \frac {8ml^2}{3})ω_{both}^2 = mg(\frac{h_2}{2} + \frac{l}{2}) + 2mgh_2

\frac{1}{2}(\frac {l^2}{3} + \frac {8l^2}{3})ω_{both}^2 = g\frac{h_2}{2} + 2gh_2

3l^2ω_{both}^2 = gh_2 + 4gh_2 **Equation 1


NOTE: THIS IS WHERE I MADE THE ERROR! I DIDN'T FOIL CORRECTLY! MY 7th GRADE MATH TEACHER WOULD BE FURIOUS.
Take a detour to simplify ω_{both}^2:

ω_{both} = \frac {\sqrt{g}}{9\sqrt{l}}(4\sqrt{6} - \sqrt{3})

ω_{both}^2 = \frac {g}{81l}(16*6 - 2*4\sqrt{3*6} + 3)

ω_{both}^2 = \frac {g}{81l}(99 - 8\sqrt{2*9})

ω_{both}^2 = \frac {g}{81l}(99 - 24\sqrt{2})

ω_{both}^2 = \frac {g}{27l}(33 - 8\sqrt{2})Insert our expression for ω_{both}^2 into Equation 1 above:
3l^2(\frac {g}{27l}(33 - 8\sqrt{2})) = 5gh_2

\frac {l}{9}(33 - 8\sqrt{2}) = 5h_2

h_2 = \frac {l}{45}(33 - 8\sqrt{2})

Insert our expression for h_2 into l cos θ = l - h_2 from the very top:

l cos θ = l - \frac {l}{45}(33 - 8\sqrt{2})

cos θ = \frac {12 + 8\sqrt{2}}{45}

θ = 58.8 degrees Note: This is the correct answer. I discovered my mistake above.

 

[tzonehunter]

So, as noted, I found the mistake I made and did get the correct answer. This was an arduous problem.