Angular Momentum of Rod Question

In summary, the problem involves a uniform rod and a bullet colliding in a horizontal plane. The bullet lodges into the rod, and the resulting angular velocity is given. The correct approach is to use conservation of angular momentum, taking into account the angular momentum of the bullet before collision. The mistake made in the initial attempt was trying to compare translational and rotational momentum. The correct equation for angular momentum of the bullet before collision is L_b=m(\frac{L}{2})^2\frac{V_b\sin(\theta)}{\frac{L}{2}}.
  • #1
americanforest
223
0
http://loncapa2.physics.sc.edu/res/sc/gblanpied/courses/usclib/hrw/hrwpictures/12-42.jpg

A uniform thin rod of length 0.40 m and mass 3.0 kg can rotate in a horizontal plane about a vertical axis through its center. The rod is a rest when a 3.0-g bullet 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 velocity makes an angle of 60° with the rod (see the figure). If the bullet lodges in the rod and the angular velocity of the rod is 12.0 rad/s immediately after the collision, what is the magnitude of the bullet's velocity just before impact?

What I've done:

As I read this I thought of three ideas:
1. Conservation of Angular Momentum, [tex]P_b=L_r[/tex]
2. Inelastic Collision
3. Parrallel Axis Theorem (with the Mass of the Bullet at the end of the rod after it hits it.

First thing I tried was quite simple. I simply set [tex]P_b=L_r[/tex] but the velocity that I found was incorrect. Suspecting that I was neglecting using the angle, I then tried [tex]P_b\sin(\theta)=L_r[/tex]. This, however, also left me with an incorrect result. As I don't see anything wrong in my methods for finding velocity I think I may have made a mistake in calculating the Moment of Inertia of the rod in it's final state which I have as:

[tex]I_f=\frac{(m_r)(L^2)}{12}+\frac{(m_b)(L^2)}{4}[/tex]

and thus:

[tex]V_b=\frac{[\frac{(m_r)(L^2)}{12}+\frac{(m_b)(L^2)}{4}]\omega}{(m_b)\sin(\theta)}[/tex]

Where is the mistake here? I assume it is legal to convert from translational momentum to rotational momentum like this?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
americanforest said:
http://loncapa2.physics.sc.edu/res/sc/gblanpied/courses/usclib/hrw/hrwpictures/12-42.jpg


1. Conservation of Angular Momentum, [tex]P_b=L_r[/tex]

Incorrect. Use the angular momentum of the bullet instead of its translational momentum.

ehild
 
Last edited by a moderator:
  • #3
OK, so is this correct:

[tex]L_b=m(\frac{L}{2})^2\frac{V_b\sin(\theta)}{\frac{L}{2}}=L_d[/tex]

So, you cannot convert straight from Translational to Rotational Momentum? Why not?

Also, is the angular momentum equation for the rod+bullet correct from my first post?
 
Last edited:
  • #4
americanforest said:
So, you cannot convert straight from Translational to Rotational Momentum? Why not?

You can not compare angular momentum with translational momentum. They are different quantities. The conservation of angular momentum means that the angular momentum of a system does not change when no external torgue is applied. The initial angular momentum is that of the bullet, as the rod is in rest. How do you calculate the angular momentum of the bullet before collision?

ehild
 
  • #5
ehild said:
You can not compare angular momentum with translational momentum. They are different quantities. The conservation of angular momentum means that the angular momentum of a system does not change when no external torgue is applied. The initial angular momentum is that of the bullet, as the rod is in rest. How do you calculate the angular momentum of the bullet before collision?

ehild

The angular momentum of the bullet relative to the centre of mass of the rod just before the collision is

[tex]L_b=m(\frac{L}{2})^2\frac{V_b\sin(\theta)}{\frac{L}{2}}[/tex]

I've got the answer, thanks for the help.
 

What is Angular Momentum of a Rod?

Angular momentum of a rod is a measure of the rotational motion of the rod around a fixed axis. It is the product of the rod's moment of inertia and its angular velocity.

How is Angular Momentum of a Rod Calculated?

The formula for calculating the angular momentum of a rod is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

What Factors Affect the Angular Momentum of a Rod?

The angular momentum of a rod is affected by its moment of inertia, angular velocity, and the position of the axis of rotation. The longer the rod, the higher its moment of inertia and the greater its angular momentum.

How is Angular Momentum Conserved in a Rod?

In a closed system, the total angular momentum of a rod remains constant. This means that if the rod's angular velocity changes, its moment of inertia will also change in order to keep the angular momentum constant.

What is the Importance of Angular Momentum in Physics?

Angular momentum is an important concept in physics as it helps us understand rotational motion and conservation of energy. It is also used in various fields such as astronomy, engineering, and mechanics.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
894
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
23
Views
915
  • Introductory Physics Homework Help
2
Replies
62
Views
9K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
210
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
2K
Back
Top