Conservation of Angular Momentum Problem with Rod and Bullet

In summary, a problem is given involving a uniform thin rod of length (L)m and mass (M_R)kg that can rotate in a horizontal plane about a vertical axis through its center. A bullet of mass (M_B)kg is fired into one end of the rod at an angle (A) degrees with the rod, and the bullet lodges in the rod. The angular velocity of the rod immediately after the collision is (w) radians per second. The task is to find the magnitude of the bullet's velocity, (V_B)m/s just before impact. The rotational inertia of the system is calculated using the formulas (1/12)M_R*L^2+M_B*L^2 and (1/12)
  • #1
DoTell
6
0
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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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
Thank you very much!
 

Related to Conservation of Angular Momentum Problem with Rod and Bullet

1. What is the conservation of angular momentum?

The conservation of angular momentum is a fundamental principle in physics that states the total angular momentum of a system remains constant in the absence of external torques.

2. How does the conservation of angular momentum apply to the problem of a rod and bullet?

In the problem of a rod and bullet, the conservation of angular momentum means that the initial angular momentum of the system (which includes the rod and bullet) must equal the final angular momentum after the bullet is fired. This is because there are no external torques acting on the system.

3. Can the conservation of angular momentum be applied to all types of systems?

Yes, the conservation of angular momentum can be applied to all types of systems as long as there are no external torques acting on the system.

4. How can the conservation of angular momentum be used to solve the problem of a rod and bullet?

By applying the conservation of angular momentum, we can set the initial angular momentum of the system (before the bullet is fired) equal to the final angular momentum (after the bullet is fired). This allows us to solve for the final angular velocity of the system, which will tell us how fast the rod is spinning after the bullet is fired.

5. What are the limitations of using the conservation of angular momentum to solve problems?

The conservation of angular momentum assumes that there are no external torques acting on the system. In real-world situations, this may not always be the case and other forces or torques may need to be taken into account. Additionally, the conservation of angular momentum only applies to systems that are isolated and do not exchange angular momentum with other objects or systems.

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