lets say I take two similar rubber bands. One of them I pull until it almost reaches its breaking point. The other I twist until it almost reached its breaking point. Do both of these rubber bands contain (roughly) the same amount of energy? It seems that the twisted band would contain more, but I am having a hard time justifying this.
[Physics] Does a pulled rubber band contain as much energy as a twisted rubber band
energyenergy storagepotential energy
Related Solutions
With two pegs, there are two strips of rubber working in parallel contributing to a total stiffness $K_{\rm total} = 57.6 \;\rm lbf/in$. So each strip is $K = 28.8\;\rm lbf/in$.
With the three pegs you now have two strips at 26° apart, or 13° from vertical each. The effective spring constant in the vertical direction is thus $K_{eff} =2 K \cos^2(13^\circ) = 54.68\; \rm lbf/in$. One of the $\cos(13^\circ)$ comes from the force projection to vertical and the second from the displacement projection to vertical from along the side of the triangle.
Another way to get the same result if the base is $b$ and the height is $h$ is
$$ K_{eff} = \frac{8 h^2}{b^2+4 h^2} K $$ $$ = \frac{8*52^2}{24^2+4*52^2} 28.8 = 54.68 $$
But also the stretch force value has changed. To get this you need the free length of the band which is not given and cannot be calculated from the given values.
Edit 1 With the values given I came up with the following force on the top pin
$$ F = \frac{288 h \left( \sqrt{b^2+4 h^2}+b-82\right)}{5 \sqrt{b^2+4 h^2}} $$
So with two pins $b=0$, $h=52$ the force is $F=633.6\;\rm lbf$ with stiffness $K_{eff}=\frac{\partial F}{\partial h}=57.6$.
With three pins and $b=24$, $h=52$ the force is $F=1367.6\;\rm lbf$ and stiffness $K_{eff}=\frac{\partial F}{\partial h}=56.02$.
Why ? The force on the pin is equal to two times the tension projected vertically (from a Free Body Diagram on the pin).
$$ F = 2 T \cos\left(\frac{\theta}{2}\right) $$
with $\cos\left(\frac{\theta}{2}\right) = \frac{h}{\sqrt{\left(\frac{b}{2}\right)^2+h^2}}$ and tension $T = K (L-L_0) $. The initial band length is $L_0 = 2\cdot 41 = 82\;\rm in$ and the stiffness of each strip is actually $K = 14.4\;\rm lbf/in$. This comes from $K_{eff}=4\,K$ when $b=0$ and with $K_{eff}=57.6\;\rm lbf/in$.
The length of the rubber band is the circumference of a triangle with base $b$ and height $h$
$$ L = b + 2 \sqrt{\left(\frac{b}{2}\right)^2 + h^2} $$
Glue one end of the rubber band to the bottom of the bottle (on the inside) and the other to the inside of the cap. (It may be easiest to cut the bottle open for this and later tape it back together, but with sufficient dexterity and suitable tools, you could do this through the neck of the bottle.)
Attach a weight to the middle of the rubber band, in such a way that the weight can't rotate around the rubber band without twisting it. Make sure the band is short enough that the weight hangs suspended when the bottle is horizontal.
When the bottle is horizontal, gravity will keep the weight below the rubber band, and will thus prevent the middle of the band from turning. Thus, when you roll the bottle on the floor, the band will twist, storing energy. When you let go, the rubber band unwinds, making the bottle rotate backwards and thus roll back to where it started.
Of course, there's a limit to how much energy you can store this way. If you roll the bottle far enough, the tension in the rubber band will eventually overcome gravity and set the weight spinning (or the band will snap).
Best Answer
The band that is pulled contains much more elastic energy. This is because that band is stretched until breaking everywhere. The band that is twisted, only reaches it breaking point at the outer points of the cross section. The strain varies linearly with the distance from the center of the cross section and is zero at the center.
It may be worthy to note that there are two possibilities when you twist a band, if it is short it will twist around the center of it's cross section, and if it is long it will form a coil around another point. In both cases the strain varies linearly with the distance to the center of rotation.