# Homework Help: Rod swinging and hitting a ball

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1. Dec 4, 2017

### haruspex

The problem is that for the ball to transition immediately to rolling there must be a horizontal frictional impulse at the ground during the collision. This invalidates your angular momentum equation.
To fix this, suppose the impulse at the collision point is J. In terms of that, you can now write the angular momentum equation for the rod about its hinge and that for the ball about its pivot (the point of contact with the ground). J can then be eliminated between the two equations.
Indeed, we can generalise the problem to two objects with MoIs I1, I2, about their hinge points, some initial angular velocities about those hinges, and respective final angular velocities.
Let the line of action of the collision impulse be r1, r2 respectively from the hinges. (I.e the perpendicular distances to the impulse vector.)
It turns out that the usual linear velocity relationship still arises. If, at the point of impact, the first body's velocity goes from u11r1 to u'1, etc. and KE is conserved then u'2-u'1=u1-u2.

2. Dec 5, 2017

### erfz

Hm that's interesting.
I just realized a few hours ago that I can write $I_{rm}\omega_i = I_{rm}\omega_f + Mv'L$ where $v'$ is the translational velocity of the ball right after the collision and before it starts rotating. This is before the angular impulse has an effect.
Of course, the ball has to start rolling after this. If we assume that the energy lost to sliding is negligible (because the distance it slides is $\approx 0$), I can simply write $\frac{1}{2}Mv'^2 = \frac{1}{2}Mv^2 + \frac{1}{2}I_b\omega^2$. $v$ is the new translational velocity after the ball has started rolling.
If I treat the problem like this, using the coefficient of restitution $e = 1$ allows me to obtain the energy equation with the angular momentum expression I've written.
Effectively, I guess this is the same as what you've suggested.

3. Dec 5, 2017

### haruspex

I don't think that we can treat the impact and the transition to rolling as separate events. In effect, that would be saying that the ball initially sits on a frictionless surface and only reaches the (extremely) frictional surface later. That could alter the proportion of energy that ends up in the two objects. In the simultaneous treatment, the frictional impulse from the ground increases the collision impulse, which in turn pushes back the rod more.

4. Dec 6, 2017

### erfz

In that case, let me see if I've got the equations correct.
I'm not entirely sure what you mean here.
How I imagine it is:
For the rod: $I_{rm}\omega_i - NLt = I_{rm}\omega_f$ (about hinge)
For the ball: $-fRt = - I_b v/R$ (about ball's COM)
where $f$ is the force of friction; $N$ is the normal force between the rod and ball; $t$ is the collision time.
And then we know $Ma = N - f$ and use that to solve the equations, I think? But that still leaves time to be solved.

I'm not entirely sure how the J's cancel and what J would be here.
What equations would you set up?

5. Dec 9, 2017

### erfz

@haruspex
Sorry for bumping; haven't heard from you in a while.
What equations would you use here?

6. Dec 9, 2017

### haruspex

As I posted, it will be easier to take moments for the ball about its point of contact with the ground. That eliminates friction, while producing a second equation with Nt in it.
The time should not matter because N and t always appear as the combination Nt. This is what I was calling J, the impulse between the objects.

7. Dec 10, 2017

### erfz

So..
For the rod: $I_{rm}\omega_i - NLt = I_{rm}\omega_f$
For the ball: $-NRt = -MRv - I_bv/R$?
(I am not 100% sure on the signs of the angular impulse. How does it depend on my conventions?)

Combining these, I get that $I_{rm} \omega_i = I_{rm} \omega_f + MLv + I_bvL/R^2$. Is there an intuitive explanation here, assuming this is correct?

Last edited by a moderator: Dec 10, 2017
8. Dec 10, 2017

### haruspex

It looks right. If, when combined with work conservation, it leads to the usual "relative velocity after = -relative velocity before" then that's as close as I can see to an intuitive explanation.

9. Dec 10, 2017

### erfz

Yes, this expression does allow me to reconstruct work conservation using $e = 1$.

10. Dec 12, 2017

### erfz

@haruspex
I'm thinking now, would it be possible to set your axis on the table that the ball sits on and treat the rod and ball as a single system?
This would eliminate frictional torque, I think.
Do you see anything wrong with that?

Like so:
$$m \omega_i L/2 (R + L/2) - I_r \omega_i = m \omega_f L/2 (R + L/2) - I_r \omega_f + MvR + I_b v/R,$$ where I have used the linear velocities of the rod's COM, $I_r$ is the rod's MOI about its COM, $I_b$ is about the ball's COM.
I don't seem to get the same numbers with this expression; something is wrong here mathematically or conceptually.

Last edited by a moderator: Dec 12, 2017
11. Dec 12, 2017

### haruspex

But then you would have torque from the unknown reaction at the rod's hinge.
You can take anywhere as your reference points for angular momentum, but if that means dragging in the reaction force from the hinge or the friction from the table then you will need to bring in a linear momentum equation as well in order to eliminate the unknown. The great benefit of using the rod's axis for its equation and the point of contact of the ball with the table for its equation is that we never get those impulses in the equations.

12. Dec 12, 2017

### erfz

Ah, shoot. I forgot about that completely. Thank you very much!