- #1
Hernaner28
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Homework Statement
Hi. Just look at the picture, I will explain it:
It says that a bullet impacts the rod as you can see (in the center of mass forming a 30 degree angle with the horizontal and keeps inside the rod) with a speed of: 25m/s. The bullet has a mass 0.1Kg, the rod 0.9Kg and the length of the rod is 0.45m.
I'm asked to find the maximum angular displacement that has the rod respect to the vertical. (the bullet keeps in the rod).
Homework Equations
The Attempt at a Solution
What I did first was to realize that the angular momentum is conserved just when the bullet impacts the rod, after that what is conserved is the energy so I wrote:
[tex] \displaystyle {{L}_{i}}=r\wedge mv[/tex]
[tex] \displaystyle {{L}_{i}}=\frac{L}{2}mv\sin 120{}^\text{o}[/tex]
And the final angular momentum (when the bullet gets inside the rod):
[tex] \displaystyle {{L}_{f}}=I\omega [/tex]
[tex] \displaystyle {{L}_{f}}=\left( \frac{M{{L}^{2}}}{3}+m\frac{{{L}^{2}}}{4} \right)\omega [/tex]
So,
[tex] \displaystyle \frac{L}{2}mv\sin 120{}^\text{o}=\left( \frac{M{{L}^{2}}}{3}+m\frac{{{L}^{2}}}{4} \right)\omega [/tex]
And soving for w I get that the initial angular speed is 8.13 rad/s. Now I use conservation of energy:
[tex] \displaystyle \frac{1}{2}I{{\omega }^{2}}=\left( m+M \right)gh[/tex]
where h is the maximum height of the center of mass. Solving for h I get h=0,22m.
But now, how can I get the angle? The correct answer is 79,5 degrees.
Thanks!