Angular Momentum of a uniform rod

• Shifty
In summary, a uniform rod of length L pivots about a fixed frictionless axis at one end and is initially at rest on a frictionless horizontal surface. A bullet with a mass of one-fourth that of the rod and traveling with speed v strikes the center of the rod, becoming embedded in it. According to the principle of conservation of angular momentum, the angular momentum of the system is conserved. Using this concept, it can be determined that the final angular speed of the rod is 6v/19L. This is calculated by considering the initial and final angular momenta of the system, taking into account the different moments of inertia of the bullet and the rod. The final moment of inertia is found to be 19/
Shifty
A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod.

What is the final angular speed of the rod?

Ok, so the answer is 6v/19L, but I don't know how to get there. I have been using angular momentum concepts, but I don't get that answer it wants.

Been on this question for many hours, thanks in advance

Since there's no external torque, the system's angular momentum is conserved. (The system consists of the bullet and the rod.) So what's the angular momentum of the system before the collision? What is it after, in terms of L and v?

Quick question--what did you get for the combined I after collision? Should be something like mL^2(1/3 +1/16) where m = mass os rod.

ideasrule said:
Since there's no external torque, the system's angular momentum is conserved. (The system consists of the bullet and the rod.) So what's the angular momentum of the system before the collision? What is it after, in terms of L and v?

Before:
$$m_{B}=m_{r}/4$$
$$v=L\omega/2$$
$$\omega=2v/L$$

$$P_{net}=m_{r}L^{2}(0)/3+m_{B}\frac{L}{2}\omega=m_{r}L^{2}v/4$$

After:
$$I_p=I_{rod}+m_BL^2/4=19/48*m_rL^2$$

$$P_{net}=19/48*m_rL^2\omega$$

Solution:
$$m_rL^2v/4=19/48*m_rL^2\omega\Rightarrow\omega=12v/19$$

I got close but I mest up somewhere

denverdoc said:
Quick question--what did you get for the combined I after collision? Should be something like mL^2(1/3 +1/16) where m = mass os rod.

Yeah I did get that Moment of Inertia

Shifty said:
Before:

$$P_{net}=m_{r}L^{2}(0)/3+m_{B}\frac{L}{2}\omega=m_{r}L^{2}v/4$$

Check this step. Your L's should cancel.

Shifty said:
$$P_{net}=19/48*m_rL^2\omega$$

Try calculating the angular moment of the rod and of the bullet separately, then adding them together. Once you correct the previous mistake I pointed out, you'll see why.

ideasrule said:
Try calculating the angular moment of the rod and of the bullet separately, then adding them together. Once you correct the previous mistake I pointed out, you'll see why.

$$P_{net}=m_rv/4=m_rL^2\omega/3+m_b(L/2)^2\omega=m_rL^2/3+m_rL^2\omega/16$$

That gives me:
$$P_{net}=19/48*m_rL^2\omega\Rightarrow\omega=12v/19L^2$$

Not quite there yet.

ideasrule said:
Try calculating the angular moment of the rod and of the bullet separately, then adding them together. Once you correct the previous mistake I pointed out, you'll see why.

Got it:
$$P_{initial}=m_rvL/8$$

Then it works. Thanks, I realized that the angular momentum of the bullets should have been the same, your suggestion worked :).

Last edited:
Shifty said:
$$P_{net}=m_{r}L^{2}(0)/3+m_{B}\frac{L}{2}\omega=m_{r}L^{2}v/4$$

Sorry to bump this thread, but could someone explain to me why the last term is divided by four and not 3? In other words, why is the formula for angular inertia changing from (m*l^2)/3 to (m*l^2)/4?

1. What is angular momentum?

Angular momentum is a physics concept that describes the rotational motion of an object. It is a measure of the tendency of an object to keep rotating around a fixed point.

2. How is angular momentum calculated?

The angular momentum of a uniform rod is calculated by multiplying the moment of inertia (I) of the rod by its angular velocity (ω). The equation is L = Iω.

3. What is the moment of inertia of a uniform rod?

The moment of inertia of a uniform rod is a measure of how difficult it is to change the rotation of the rod. It depends on the mass, length, and distribution of mass in the rod.

4. How does the angular velocity affect the angular momentum of a uniform rod?

The angular velocity is directly proportional to the angular momentum of a uniform rod. This means that as the angular velocity increases, the angular momentum also increases.

5. What are some real-life examples of angular momentum of a uniform rod?

Some examples of angular momentum of a uniform rod are a spinning top, a rotating wheel, and a swinging pendulum. These objects all have a fixed axis of rotation and exhibit rotational motion, thus having angular momentum.

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