Conservation of Angular Momentum Problem with Rod and Bullet

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SUMMARY

The discussion revolves around a conservation of angular momentum problem involving a uniform thin rod and a bullet. The rod, with mass (M_R) and length (L), rotates about a vertical axis after being struck by a bullet of mass (M_B) at an angle (A) degrees. The user initially calculated the rotational inertia incorrectly by not accounting for the bullet's distance from the axis, leading to an incorrect velocity calculation for the bullet just before impact. The correct rotational inertia should be calculated as (1/12)M_R*L^2 + M_B*(L/2)^2.

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DoTell
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Alright, this is my first experiment with this forum. Hopefully, I do this right!

Homework Statement


I'm going to give the problem in terms of variables, not the numbers I'm given, so that people can't give me a direct answer.
A uniform thin rod of length (L)m and mass (M_R)kg can rotate in a horizontal plane about a vertical axis through its center. The rod is at rest when a bullet of mass (M_B)kg traveling in the horizontal plane of the rod is fired into one end of the rod. As viewed from above, the direction of the bullet's velocity makes an angle (A) degrees with the rod. If the bullet lodges in the rod and the angular velocity of the rod is (w) radians per second immediately after the collision, what is the magnitude of the bullet's velocity, (V_B)m/s just before impact?
Here's an image that is similar if it helps (with A=60 degrees): http://s3.amazonaws.com/answer-board-image/ec17a0c0-ec15-42e9-879e-d40b70321ef1.jpeg

Homework Equations


Angular Momentum is conserved (Initial L=Final L)
Rotational inertia of a rod about its center: (1/12)MR^2
L=Iw=m(R x V)

The Attempt at a Solution


First, I found the rotational inertia of the entire system: I=(1/12)M_R*L^2+M_B*L^2
Now I know the inertia of the system and can plug into conservation of momentum.

M_B*L*V_B*sin(A)=Iw
...and I can solve for V, which is (Iw)/(M_B*L)=(((1/12)M_R*L^2+M_B*L^2)w)/(M_B*L*sin(A))
When I input this solution into WebAssign, it's incorrect. Can someone pinpoint my error? Thanks in advance!

Also, anyone know about this error on PF? "You specified a tag that was too long. A tag can only be 20 characters." I had to remove all of the fancy tags to get this to post.
 
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DoTell said:
First, I found the rotational inertia of the entire system: I=(1/12)M_R*L^2+M_B*L^2
Now I know the inertia of the system and can plug into conservation of momentum.
If L is the length of the rod, what's the embedded bullet's distance from the axis?

Also, anyone know about this error on PF? "You specified a tag that was too long. A tag can only be 20 characters." I had to remove all of the fancy tags to get this to post.
Not sure what you mean. Were you trying to use Latex?
 
Ah yes, that's a problem! I must use the radius for the bullet part of the total inertia. So the rotational inertia should be (1/12)M_R*L^2+M_B*(L/2)^2
But why is it still the wrong answer? If that's the only error, I must be making a computation mistake.

And about the tag error, I don't know, I just tried making subscripts and superscripts but it gave me that message as an error at the top of the page when I tried to preview it.
 
DoTell said:
Ah yes, that's a problem! I must use the radius for the bullet part of the total inertia. So the rotational inertia should be (1/12)M_R*L^2+M_B*(L/2)^2
But why is it still the wrong answer? If that's the only error, I must be making a computation mistake.
The same issue applies to calculating the angular momentum of the bullet.
 
Thank you very much!
 

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