Mechanics: problem of falling stick by I. E. Irodov

[boson boss]

I'm examining the book shelf for interesting problems. Here's the first
I came upon - it is suposed to be easy and I picked it because I just
don't get it :-))

"Homogenuos stick in horizontal position is falling from height h and
elastically hits a massive plate with one end. Find the velocity of the
center of stick immediately after the impact."

[dx]

Is the answer \frac{R}{2} (\frac{2mgh}{I})^{1/2})?

[jumanicus]

The stick of length l and mass m is falling horizontally and hits the
edge of the plate with one end. We can use conservation of angular
momentum at the point of impact:
Initial angular momentum of the rod is Li = m.sqrt(2gh).l/2, since the
rod is not rotating.
Final angular momentum is Lf = (ml^2) / 3 . 2(v / l).
This is because, (ml^2) / 3 is the momentum of inertia (around the
point of impact) and 2(v / l) is its angular speed (around the point of
impact) after impact.
Equating these two equations we get:
v = (3.sqrt (2gh)) / 4.

[Stephen Riley]

In message <1146873859.459962.251990@g10g2000cwb.googlegroups. com>,
boson boss <junkerade@gmail.com> writes
>I'm examining the book shelf for interesting problems. Here's the first
>I came upon - it is suposed to be easy and I picked it because I just
>don't get it :-))
>
>"Homogenuos stick in horizontal position is falling from height h and
>elastically hits a massive plate with one end. Find the velocity of the
>center of stick immediately after the impact."

It's been a while since I did played about with these, fun thought they
are, hence beware of the following solution! Mostly its just jiggling
around with algebra.

Using conservation of energy and angular momentum:

u = Velocity of stick centre of mass before impact
v = Velocity of stick centre of mass after impact
w = angular velocity of centre stick about impact point after impact
I = Inertia tensor
r = impact distance from stick centre

Energy
[1] Energy before is the kinetic energy of the stick : 1/2mu^2
[2] Energy after is the kinetic energy of the stick + angular energy:
1/2mv^2 + 1/2Iw^2
It's an elastic collision, so energy conserved:
1/2mu^2 = 1/2mv^2 + 1/2Iw^2
Simplifying:
[3] v^2 = (mu^2-Iw^2)/m

Momentum:
[4] Angular momentum about impact point before collision:
mur + 0
[5] Angular momentum about impact point after collision:
mvr + Iw
Angular momentum conserved, so:
mur = mvr + Iw
Simplifying:
[6] w = (mur - mvr ) / I

Inserting [6] into [3] and solving for V:
v^2=u^2-mr^2(u-v)^2/I

[7] Giving v = -u(I-mr^2)/(I+mr2)

So if r=0 (middle of stick hits plate) v = -u (stick reflects off plate)
whereas if r were infinite (hits end of a looooong light stick) v = u
(COM velocity of infinitely long light stick is unaffected by a
collision at an end!). Seems reasonable.

Plugging in rotational inertia of a thin rod: 1/12ml^2 into [7]

v = u/2

(Assumes the rod rotates about its centre).

So the centre of stick continues on at half the original speed, and I
didn't need the height :)

--
Stephen Riley

[Igor Khavkine]

On 2006-05-06, boson boss <junkerade@gmail.com> wrote:
> I'm examining the book shelf for interesting problems. Here's the first
> I came upon - it is suposed to be easy and I picked it because I just
> don't get it :-))
>
> "Homogenuos stick in horizontal position is falling from height h and
> elastically hits a massive plate with one end. Find the velocity of the
> center of stick immediately after the impact."

It's strange that the problem doesn't specify how much of the stick
comes into contact with the (I'm assuming infinitely) massive plate. If
the stick lands completely on the place then, by symmetry, it will have
to bounce up just like a particle would. If the end of the stick barely
misses the plate, then its velocity is not altered. Clearly, there is a
dependence, unless I'm misunderstading the problem. But given the
ambiguity we are all free to consider the most interesting variation of
the problem.

In this formulation, here's the geometry just before the collision:


stick
*********************
| .--------------
v | | plate v = sqrt(2gh) <= mgh = mv^2/2
V |
|
|<--------->|<------->|
s t s + t = L.

As the stick is falling, there is no difference whether it's considered
as a solid body or just as a bunch of mass segments that happen to be
falling synchronously. So the collision can be treated for each mass
segment (a * in the diagram) can be treated individually. For a point
mass, as it falls onto a hard horizontal surface with velocity v, after
an elastic bounce, its velocity is just -v. The same happens to each of
the rod's mass segments. Hence the velocity field along the rod, just
after the collision, is

^ ^ ^
| | | -v
| | |
*********************
| | | .--------------
v | | | |
V V V |
|
|

Knowing the velocity distribution is enough to calculate everything
else. For instance the velocity of the center of mass u is obtained by
dividing the total momentum by the total mass of the stick. Since the
stick is uniform, the masses of the segments that miss the plate, strike
the plate, and the total mass are respectively proportional to the
lengths s, t, and L. Hence, the answer is

s-t
u = --- v, where u = v if s = L and t = 0
L u = 0 if s = L/2 and t = L/2
u = -v if s = 0 and t = L .

After the collision, we can no longer neglect the fact that the rod is a
rigid body. But the motion or a rigid body is completely characterized
by its total momentum and the angular momentum about the center of mass.
Both of these quantities can be calculated from the instantaneous
velocity distribution above.

A curious thing about the answer is that the direction of motion of the
rod's center of mass does not change, it remains vertical. At first it
seemed a little counter intuitive. But a couple of minutes experimenting
with dropping a pen against the edge of a table confirmed the above
result. In retrospect it seems obvious because, since the table is
flat and horizontal, it can only transfer an impulse in the vertical
direction, not a horizontal one.

A final note about conservation. I noticed that some other posters tried
to appeal to conservation of linear or angular momentum. While energy is
conserved in this elastic collision, neither linear nor angular momentum
is. Before or after the collision, momentum is not constant due to the
influence of gravity. Before and after the collision the angular
momentum is constant. However, during the collision, neither is
conserved because some of each can be provided or absorbed by the
infinitely massive plate.

That was fun!

Igor

[Stephen Riley]

In message <slrne5qtq7.af5.igor.kh@corum.multiverse.ca>, Igor Khavkine
<igor.kh@gmail.com> writes

>A final note about conservation. I noticed that some other posters tried
>to appeal to conservation of linear or angular momentum. While energy is
>conserved in this elastic collision, neither linear nor angular momentum
>is. Before or after the collision, momentum is not constant due to the
>influence of gravity. Before and after the collision the angular
>momentum is constant. However, during the collision, neither is
>conserved because some of each can be provided or absorbed by the
>infinitely massive plate.
>
>That was fun!

Neat!

>I noticed that some other posters tried
>to appeal to conservation of linear or angular momentum.

Yikes, that's true.

The cases you gave were interesting, specifically where s=t and thus
where all the kinetic energy is converted to rotational.

Specifically :

<--- *
<--- *
<--- * and
<--- * plate
---> * .-------------- .--------------
---> * | plate ****************
---> * | | | | | | |
---> * | | | | | | |
---> * | V V V ^ ^ ^

The first where the stick spins around and hits a 'vertical' side and
flies off to the left. And the second, a massive plate of negligible
thickness, where the stick spins around and hits the 'bottom' of the
plate and continues on if nothing happened, apart from doing a rotation.

Now for what would happen with two sticks of the mass and dimensions
hitting each other half way along... :)

--
Stephen Riley

[Fredrik Bulow]

boson boss <junkerade@gmail.com> writes:

> I'm examining the book shelf for interesting problems. Here's the first
> I came upon - it is suposed to be easy and I picked it because I just
> don't get it :-))
>
> "Homogenuos stick in horizontal position is falling from height h and
> elastically hits a massive plate with one end. Find the velocity of the
> center of stick immediately after the impact."


I submitted a solution to this problem two days ago but for some
reason it hasn't shown up in the newsgroup yet so I'm reposting the
link to my (very neatly typed) solution.

http://users.tpg.com.au/helfred/foo/stick.pdf

Oh... and I assume that the stick barely misses the plate.

/Fredrik

[Fredrik Bulow]

boson boss <junkerade@gmail.com> writes:

> I'm examining the book shelf for interesting problems. Here's the first
> I came upon - it is suposed to be easy and I picked it because I just
> don't get it :-))
>
> "Homogenuos stick in horizontal position is falling from height h and
> elastically hits a massive plate with one end. Find the velocity of the
> center of stick immediately after the impact."

I've solved the problem and took the time to LaTeX the solution. It is
available (for a few weeks only though) at

http://users.tpg.com.au/helfred/foo/stick.pdf

Occasionally I make mistakes so please let me know when you find one.

Fun problem by the way!

/Fredrik

[Oh No]

Thus spake Igor Khavkine <igor.kh@gmail.com>
>A final note about conservation. I noticed that some other posters
>tried to appeal to conservation of linear or angular momentum. While
>energy is conserved in this elastic collision, neither linear nor
>angular momentum is. Before or after the collision, momentum is not
>constant due to the influence of gravity. Before and after the
>collision the angular momentum is constant. However, during the
>collision, neither is conserved because some of each can be provided or
>absorbed by the infinitely massive plate.

From what I could see the other posters used conservation of angular
momentum about the point of impact. This is conserved in the impact. I
think the reason that they got different answers was that Stephen used
the formula for rotational inertia about the centre of the rod, whereas
he should have used the formula for rotational inertia about the end, if
that is where the impact takes place.

Regards

--
Charles Francis
substitute charles for NotI to email

[boson boss]

boson boss wrote:
> "Homogenuos stick in horizontal position is falling from height h and
> elastically hits a massive plate with one end. Find the velocity of the
> center of stick immediately after the impact."

The book gives only the final solution:

v=sqrt(gh/2)

This says that no matter what length of stick between the middle and
end hits the "massive plate" the immediate velocity of the center is
the same. If the whole stick would hit the ground like rigid body it
would bounce up with v=sqrt(2gh). So there are 2 distinct velocities
that can happen after impact. That must be the question: why is that
so?