Angular Momentum of a uniform rod

[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

 

[ideasrule]

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?

 

[denverdoc]

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.

 

[Shifty]

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

 

[Shifty]

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

 

[ideasrule]

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.

 

[ideasrule]

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

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.

 

[Shifty]

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.

 

[Shifty]

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 :).

 

[cdotter]

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

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?