# Ball hitting a rod

1. Feb 6, 2009

1. The problem statement, all variables and given/known data
A rod of length L and mass M is suspended vertically from a pivot (and is stationary). A ball traveling horizontally at speed v and mass m sticks to the rod at a distance L/2 from the pivot. What is the minimum speed v required so that the rod will undergo a full rotation around its pivot (as opposed to just being a pendulum).

2. Relevant equations

L=Iw
L=rmv

3. The attempt at a solution
I havent had much luck with this chapter on rotation.
I said that initial momentum = final momentum
so
$$L_{0}=L$$
$$L_{0}=\frac{1}{2}mv$$
$$L=(\frac{1}{3}ML^2+\frac{1}{4}mL^2)\omega$$
but equating the 2 is nonsense to me because you got omega and v to mess things up. I would imagine the result to be a function of m, M and L but no idea how that pops out, given that those are pretty much the only equations given in the entire chapter (gyro precession equation is the other equation)

Thanks for any assistance

2. Feb 6, 2009

### Carid

PotentialEnergyOfRodAndBallAtTopOfCycle = KineticEnergyOfTheBall

3. Feb 6, 2009

sounds simple enough
but i thought i read somewhere that in cases like that (inelastic collision i think they call them)
you cant use the conservation of energy

4. Feb 6, 2009

### Staff: Mentor

Attack the problem in two steps:
(1) The collision, in which energy is not conserved but angular momentum is.
(2) The rising of the rod+ball after the collision, where energy is conserved. (How much energy must it have at the bottom to make it over the top?)

5. Feb 6, 2009

### Staff: Mentor

Correct this. (You left out the L.)
That's fine, but it's only part of the solution. (It's not nonsense at all.)

6. Feb 6, 2009

### Carid

doc al

Thanks for the correction

But doesn't the expression

PotentialEnergyOfRodAndBallAtTopOfCycle = KineticEnergyOfTheBall

actually give us the minimum value of v i.e. when the interaction is at the limit of elastic and inelastic?

7. Feb 6, 2009

### Staff: Mentor

I don't understand that last phrase. But that expression doesn't seem to apply in any case, whether the collision is elastic or inelastic.

8. Feb 6, 2009

### Carid

Doc Al

Doesn't the expression

PotentialEnergyOfRodAndBallAtTopOfCycle = KineticEnergyOfTheBall

actually give us the minimum value of v?

After all if the ball transfers less energy than this the rod won't get to the top.

9. Feb 6, 2009

### Staff: Mentor

No.
I'm puzzled at your statements since we know that some of its KE is dissipated as internal energy.

It sounds like you are saying something like: It requires a certain amount of energy to get the rod+ball over the top. If the ball had that amount of energy, and transferred it all to the system, then that would be the minimum speed required. But that's a big if that we know isn't true. I don't see how that helps solve this problem.

10. Feb 6, 2009

### Carid

Sorry if this gets a bit boring. I feel I'm being a bit obtuse, but here goes...

Yes, the bit about the ball sticking to the rod is a codeword for inelastic collision and consequent dissipation of energy.

So v must be more than what would be required is all energy was transmitted. But how much more? Maybe in an extreme case only a tiny delta that we could safely ignore. Maybe with a great big heavy ball and a very light rod...

If you feel you've more urgent matters to attend to that reply to this I will understand!

11. Feb 6, 2009

### Staff: Mentor

Why not just solve the problem exactly and then worry about making approximations for special cases?

12. Feb 6, 2009

ok
so i can say angular momentum is conserved at the collision
$$\frac{L}{2}mv=I\omega$$
and after the collision energy is conserved
(not sure but assuming i have to use centre of mass here,hence the m+M term)
$$\frac{1}{2}I\omega^2=(m+M)gL$$
from these 2 equations i can eliminate omega to solve for v
but are the above 2 equations correct?

13. Feb 7, 2009

### rl.bhat

At the top the rod has rotational energy, center of mass has linear kinetic energy and potential energy. Equate all these to the initial kinetic energy at the bottom.

14. Feb 7, 2009

### Staff: Mentor

Perfectly correct. (And yes, when you calculate the change in gravitational PE all you have to worry about is the change in the position of the center of mass--which raises by a distance L.)

The stick can be considered in pure rotation about one end, so there's no need to separate rotational and linear kinetic energy. Furthermore, since we are interested in the minimum speed, the kinetic energy at the top is zero.

15. Feb 7, 2009

thanks for your help Doc Al