Rod, Putty, and an Inelastic Collision

[e(ho0n3]

Yet another problem I'm breaking my head with: A thin rod of mass M and length L is suspended vertically from a frictionless pivot at its upper end. A mass m of putty traveling horizontally with a speed v strikes the rod at its center of mass and sticks there. How high does the bottom of the rod swing?

I can't apply conservation of momentum because of gravity. The collision is inelastic so kinectic energy isn't conserved. This just leaves me with using torques and potential energy. I figured I could calculate the work done by the torque due to gravity and then equate this with the change in potential energy but then my answer will be independent of v which suggests that my reasoning is bogus. What other things can I look at?

[Doc Al]

Treat the collision as happening very quickly: angular momentum is conserved. Then apply conservation of energy.

[e(ho0n3]

Treat the collision as happening very quickly: angular momentum is conserved. Then apply conservation of energy.
Ah! I see what you mean. Let u be the velocity of m + M after the collision. By conservation of momentum u = mv/(M + m) right.Then I use conservation of energy and get I\omega^2 = gh(M+m) where I is the combined moment of inertia of M and m, h is the maximum attainable height of the center of mass and \omega = 2u/l. I get as an answerh = \frac{m^2v^2(3m+4M)}{3g(M+m)^3}.Since this is the height of the center of mass which is located a distance L/2 from either end of the rod, the end of the rod would have swung a distance 2h. Apparently this is wrong since the answer in my book is\frac{3m^2v^2}{g(3m+4M)(m+M)}.I checked my algebra so the problem must be with my equations. I'm guessing u is wrong, but why?

[Doc Al]

Ah! I see what you mean. Let u be the velocity of m + M after the collision. By conservation of momentum u = mv/(M + m) right.
No. Angular momentum is conserved, but not linear momentum. The frictionless pivot will not exert a torque on the rod, but it will certainly exert a force.

[e(ho0n3]

No. Angular momentum is conserved, but not linear momentum. The frictionless pivot will not exert a torque on the rod, but it will certainly exert a force.
OK then. So by conservation of angular momentum,Lmv/2 = I_M\omega + Lmu/2where \omega = u/L. I still don't get the right answer though.

[Doc Al]

OK then. So by conservation of angular momentum,Lmv/2 = I_M\omega + Lmu/2where \omega = u/L. I still don't get the right answer though.
Forget about "u", I think that's messing you up. Conservation of angular momentum will give you:
Lmv/2 = I_{(rod + putty)} \omega
Figure out the rotational inertia of the "rod plus putty" about the pivot, then find \omega. Then you can apply conservation of energy to see how high the rod swings.

[arildno]

An "alternative" to Doc Al's argument of conservation of angular momentum about the pivot for the system rod+putty before and after the collision, is impact theory.
Clearly, all velocities remain horizontal before and after the collision phase, and we may assume that wharever impulses acts on the two objects, these impulses are also horizontal.

Let the initial velocity of the putty be v_{p,0}\vec{i}
The linear impulse equation for the putty reads:
-I\vec{i}=m(v_{p,f}-v_{p,0})\vec{i}(1)
For the rod, we know it will continue to have zero velocity at the pivot point, i.e, the rod will rotate about the pivot.
Hence, instead of using the linear impulse equation for the rod, we use the angular impulse equation about the pivot (eliminating the effect of the impulse from the pivot on the rod):
-\frac{L}{2}\vec{j}\times(I\vec{i})=\mathcal{I}_{W} \omega_{f}\vec{k}(2)

We must also require that the contact point (normal) velocities equal during the collision:
v_{p,f}=\frac{L}{2}\omega_{f} (3)

For an elastic collision, there will be a secondary, rebound impulse associated with the reversal of elastic deformation, which will induce a velocity difference between the two objects (and hence, their departing from each other). Since we have an inelastic collision between the rod and putty, however, no such rebound impulse will occur.

Putting the information from (1) and (3) into (2), we gain:
\frac{L}{2}mv_{p,0}\vec{k}=(\mathcal{I}_{W}+m(\fra c{L}{2})^{2})\omega_{f}\vec{k}

But this is equivalent to conservation of angular momentum about the pivot..

[e(ho0n3]

Forget about "u", I think that's messing you up. Conservation of angular momentum will give you:
Lmv/2 = I_{(rod + putty)} \omega
Figure out the rotational inertia of the "rod plus putty" about the pivot, then find \omega. Then you can apply conservation of energy to see how high the rod swings.
OK, I get the correct answer. Question: Why is it that I_m\omega \neq Lmu/2. It's as if \vec{r} \times \vec{p} and I\vec{\omega} are not equal for the putty in this case. Why is this?

[Doc Al]

OK, I get the correct answer. Question: Why is it that I_m\omega \neq Lmu/2. It's as if \vec{r} \times \vec{p} and I\vec{\omega} are not equal for the putty in this case. Why is this?
I think you're just defining "u" as the speed of the tip, for some strange reason. (You wrote \omega = u/L.) If "u" was supposed to represent the speed of the putty post collision, then you should have written \omega = 2u/L. If so, then I_m\omega = m(L/2)^2 (2u/L) = Lmu/2.

[e(ho0n3]

I think you're just defining "u" as the speed of the tip, for some strange reason. (You wrote \omega = u/L.) If "u" was supposed to represent the speed of the putty post collision, then you should have written \omega = 2u/L.
The strange reason being than I'm an idiot. If there is one thing I've learned from doing all these problems is that I'm dumber than I thought I was.

Anyways, thanks a lot.