Optimal shape of a hammer to kill an earthworm?

Question

Suppose I have an earthworm of a particular length, say 10cm, on a plain surface like a table and I want to use a hammer to kill the worm. The hammer must be such that the cross section of the hammer covers the worm from end to end, no matter the shape of the worm (stretched, bent, kinked...). I am allowed to position the hammer any way I want. The conditions precisely are:

* The hammer (its cross section) is able to cover the worm fully, no matter how the worm is stretched, bent, kinked on a given plane surface.
* The cross section (area) of the hammer should be minimal so that the hammer can have minimal weight and manufacturing cost.
* I am allowed to position the hammer any way I want once I see the shape of the worm. While I strike, the worm is not moving.

I suspect the shape of the hammer is somewhat like an ellipse with a length of 10cm in one direction (to cover the case in which the worm is fully stretched) and somewhat narrower in the other direction. But how narrow can it be? And is it really an ellipse or something more elaborate? How do you approach a problem like this?

If I havent made the task clear, contact me for details andrehartmann (at) hotmail.com.

☢☯Chaτнυra☯☢ · Accepted Answer

I think you're
correct. It should
be an ellipse. Also it
must be 10cm long
on the y axis. For the x axis
minimum length
we can imagine a
situation when the
worm stands L
shaped, 5cm each side. So the minimum
length for the x
axis must be
5cos45 = 5/√2
=(5√2)/2 cm I'm pretty sure
that this thing
would cover the
whole worm.

Sharath · Answer

now the maximum length of worm is 10cm.
Now if on end of the worm is fixed to a point, what is the area which the worm can move around?
This is the area of cross-section of the hammer. 
Just imagine a line segment with one end fixed to a point. The area it sweeps encloses a circle. It can move around the point 360 degrees. So the maximum area it sweeps will be the area of the circle. And the radius of the circle will be the length of the worm that is 10cm.
So the minimum area of cross section of the hammer = pi*r^2 = pi*100 = 100pi ~ 314 cm^2

? · Answer

Electron orbitals and their form are actually wave services that satisfy the Schrodinger equation. The shape is a elaborate function expression of the wave operate, and the quantum numbers (n, l, m) that come out of the operate occur because of the boundary conditions placed on the electron around the atom and other electrons. I consider that best the strategy to the hydrogen atom is known exactly, for other atoms the interplay of electrons could be very intricate so the orbitals are best solved numerically or approximately. In view that for electrons the Pauli exclusion principle applies every pair of electrons wishes one other set of quantum numbers and this makes the shape of the s and p sub-shells of electrons.

∅ · Answer

well, i always wondered what i would use calculus for...now i know  :D

personally, i would let the poor worm go. i always preferred algebra anywayz  :S

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Optimal shape of a hammer to kill an earthworm?

4 Answers