# Rod Falling under gravity hits straight edge

1. Jun 23, 2017

### decerto

1. The problem statement, all variables and given/known data

http://imgur.com/a/Ssolz

2. Relevant equations

Elastic Collision so $$mgh = \frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2 + \frac{1}{2}I\omega^2$$
$$F=ma$$

3. The attempt at a solution

So the problem is similar to this one which allows me to work out the normal force at the point of contact.

I have
$$ma = mg - N$$
$$dN = I\dot{\omega}$$
$$\dot{\omega} = \frac{a}{d}$$

So

$$\frac{dN}{I} = \frac{mg-N}{md}$$

But to be honest I don't know if this is even useful as there is no guarantee the rod sits on the edge after the collision. Any help would be appreciated.

2. Jun 23, 2017

### Staff: Mentor

The collision is instantaneous, there is no meaningful force. There is a well-defined momentum transfer, however.

You also have conservation of angular momentum.

3. Jun 23, 2017

### decerto

Honestly I am confused over each one of those sentences.

What do you mean by there is no meaningful force?

How is there a momentum transfer (to/from?) the rod if there is no force? If so, how do you calculate it?

Surely the rod starts rotating after the collision and wouldn't that mean it gains angular momentum about it's center of mass?

4. Jun 23, 2017

### Staff: Mentor

You cannot calculate a meaningful force value. The collision happens in "zero time" with "infinite force" - don't take these too literal.
Don't work with forces. Work with momentum and angular momentum.
Sure, and how much it gains is related to the change of the momentum transfer - which is related to its change in its linear velocity. You have to find this relation.

5. Jun 23, 2017

### haruspex

Yes, but you are free to choose any stationary point as axis. If all the external forces/impulses acting on the system under consideration all act through some point P then taking P as axis eliminates their moments.

6. Jun 23, 2017

### decerto

I don't understand how this is possible, there is forces so there are no conservation of momentum but the forces are not meaningful so I can't use newtons laws. I clearly have some gap in my understanding here.

7. Jun 23, 2017

### haruspex

mfb is not saying that the linear and angular momentum of the rod do not change. But the change to each comes from the same external impulse. So if you let that impulse be J you can write two equations involving J then eliminate it.
My suggestion in post #5 is slightly different. By judicious choice of axis you can get an angular momentum equation that does not involve J, so you do not need the linear momentum equation at all.

8. Jun 23, 2017

### TSny

I think these two equations are OK. You can simplify the first equation by invoking the usual assumption that $mg$ may be neglected compared to the strong impulsive force $N$ during the collision.
This equation is not valid. It is not consistent with your first two equations given above. The correct relation between $a$ and $\dot{\omega}$ during the collision can be obtained by eliminating $N$ between your two equations $ma = - N$ and $Nd = I\dot{\omega}$.

This will give you an equation that you can integrate once with respect to time to get a relation between $\omega$, $v_i$, and $v_f$. (This will be equivalent to the relation between $\omega$, $v_i$, and $v_f$ that you would get using concepts of angular momentum.)

9. Jun 23, 2017

### haruspex

As mfb points out, it is only strictly correct if the forces are considered functions of time: ma(t)=mg-N(t).
Integrating over the duration Δt of the collision: $m\Delta v=m\int_0^{\Delta t}a.dt=mg\Delta t-\int_0^{\Delta t}N.dt=mg\Delta t-J$.
As Δt→0 we get mΔv=-J.
Yes, I know you are saying the same thing, but I felt it was not quite clear.

I assume v is defined as positive down but N and J as positive up.

10. Jun 23, 2017

### TSny

Yes, definitely. Eliminating $N$ between decerto's two equations gives a relation between $a(t)$ and $\dot\omega(t)$ which can be integrated over the time of the collision. The problem can then be solved by using this result with the energy equation.

OK. It seems to me that the "impulse approximation" is still being used here in order to claim that $mg \Delta t$ can be neglected compared to $\int_0^{\Delta t}N.dt$ for very small $\Delta t$. In other words, a student might wonder why the integral $\int_0^{\Delta t}N.dt$ doesn't also approach zero as $\Delta t$ approaches zero.

I'm not disagreeing with anything you or mfb have said.

I just wanted decerto to realize that his two dynamical equations involving $N$ can be used along with the energy equation to solve the problem. Like you, I prefer to solve the problem using energy and angular momentum (with a judicious choice of origin).

11. Jun 24, 2017

### decerto

Thank you all for the help. Integrating your dynamical equation just gives me back the conservation of angular momentum, is this correct? Should I end up with a quadratic in the final velocity when I combine it with the energy equation?

12. Jun 24, 2017

### haruspex

Please post the equation you have, to make sure there is no misunderstanding, and state the choice of axis.

13. Jun 24, 2017

### decerto

$$ma=-N$$ and $$Nd = I\dot{\omega}$$ gives $$-mad = I\dot{\omega}$$
Integrating wrt to time I have $$-mv_f d + mv_i d = I \omega$$ which is just the conservation of angular momentum about the point of contact with $$v_i= \sqrt{2gh}$$
Solving for omega
$$\omega^2 = \frac{1}{I^2}\left(m^2d^2v_i^2 +m^2d^2v_f^2 - 2m^2v_iv_fd^2\right)$$
which I inserted into the energy equation from the OP to get a quadratic in v_f

14. Jun 24, 2017

### TSny

Yes, that looks right.

15. Jun 24, 2017

### haruspex

Right, which was what I was referring to in post #7. You could have taken moments about the point of contact and got there in one step; no reference to N and no integration.