Angular speed of rod shot by bullet

[songoku]

Homework Statement

A rod with mass $M$ and length $L$ meter is pivoted at its center and put in vertical position, then a bullet of mass $m$ with initial speed $u$ is directed at a point $x$ meter from the top of the rod ($x$ < $0.5~L$) and the bullet will penetrates through the rod with final speed $v$. Considering no energy lost and ignoring mass of rod lost when the bullet penetrates, find the angular speed of the rod

Relevant Equations

Conservation of linear momentum

Conservation of angular momentum

Circular motion

1) Applying conservation of linear momentum:

$$m.u = M.V + m.v$$

where $V$ is final linear speed of the rod

$$V=\frac{m.u-m.v}{M}$$2) Applying formula of circular motion:
$$V=\omega . r$$
$$\omega = \frac{\left(\frac{mu-mv}{M} \right)}{\frac{1}{2}L-x}$$

Is this correct?And can this be solved by using conservation of angular momentum? This is what I tried:

Applying conservation of angular momentum:
$$m.u.(\frac{1}{2}L-x)=m.v.(\frac{1}{2}L-x)+I \omega$$
$$(\frac{1}{2}L-x)(m.u-m.v)=\frac{1}{12} ML^2 \omega$$
$$\omega = \frac{(\frac{1}{2}L-x)(m.u-m.v)}{\frac{1}{12} ML^2}$$I am really not sure about my working. Thanks

 

[kuruman]

Linear momentum is not conserved because the pivot exerts an external force. Angular momentum about the pivot is conserved assuming that the bullet goes through very fast.

 

[Delta2]

Something tells me you have to use conservation of energy because the statements says no energy lost.

 

[wrobel]

Angular momentum of the whole system is conserved. Only that is needed.
Condition x≤L/2 is strange; condition about the energy is strange as well
$$xmu=J\omega+xmv$$

 

[kuruman]

Something tells me you have to use conservation of energy because the statements says no energy lost.

I agree that that's what the problem says but the issue is that if we conserve energy, the answer is different from what one gets from angular momentum conservation. Angular momentum is conserved because the torque due to gravity about the pivot can be ignored if the bullet goes through fast enough.

The energy conservation route gives a result that is independent of the distance $x$ from the pivot where the bullet strikes. So why is that distance given? Also, unless the energy transfer from bullet to rod is mediated by conservative forces, I don't see how energy can be conserved. Bullets going through rods usually do not exert conservative forces.

I favor ignoring the energy conservation clause and proceeding with angular momentum conservation.

 

[songoku]

Angular momentum of the whole system is conserved. Only that is needed.
Condition x≤L/2 is strange; condition about the energy is strange as well
$$xmu=J\omega+xmv$$

Shouldn't the distance measured from the pivot, so $\frac{1}{2}L - x$ since $x$ is measured from top of rod?

Thanks

 

[haruspex]

I agree that that's what the problem says but the issue is that if we conserve energy, the answer is different from what one gets from angular momentum conservation. Angular momentum is conserved because the torque due to gravity about the pivot can be ignored if the bullet goes through fast enough.

The energy conservation route gives a result that is independent of the distance $x$ from the pivot where the bullet strikes. So why is that distance given? Also, unless the energy transfer from bullet to rod is mediated by conservative forces, I don't see how energy can be conserved. Bullets going through rods usually do not exert conservative forces.

I favor ignoring the energy conservation clause and proceeding with angular momentum conservation.

Yes, the problem is overspecified. We are given u, v, m, M, x and L. From these we can calculate the energy lost.
Conversely, we can take energy as conserved and calculate v.

 

[kuruman]

Shouldn't the distance measured from the pivot, so $\frac{1}{2}L - x$ since $x$ is measured from top of rod?

Thanks

Yes, it should.

 

[songoku]

Yes, the problem is overspecified. We are given u, v, m, M, x and L. From these we can calculate the energy lost.
Conversely, we can take energy as conserved and calculate v.

Will the energy lost be: $\frac1 2 mu^2 - \frac 1 2 mv^2 - \frac 1 2 J \omega^2$ where $J$ is the moment of inertia of the rod?

Thanks

 

[kuruman]

Will the energy lost be: $\frac1 2 mu^2 - \frac 1 2 mv^2 - \frac 1 2 J \omega^2$ where $J$ is the moment of inertia of the rod?

Thanks

It would be that with $\omega$ being what you get using angular momentum conservation.

 

[wrobel]

I wrote the formula for the following configuration

 

[songoku]

Thank very much for all the help and explanation kuruman, Delta2, wrobel, haruspex