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So I was trying to figure out how much work someone does when they do a sittup or crunch. I guess to make things simple, I'm imagining a really really thin rod with some uniform mass lying on the ground. Then you tilt it up to 90 degrees.

I tried googling for this but I'm probably missing the technical term for this problem. Anyways, I know it must be less than $mgh$, because we're not lifting all of the rod's weight to height $h$.

I'm thinking this can probably be solved with calculus but I just can't seem to come up with anything that makes sense. I can imagine the rod being broken up into little $\mathrm{d}h$s and then you sum it all up. Something like

$$ W = \int_{0}^{h} g (m\mathrm{d}h/h) x \mathrm{d}x $$

I also feel like the answer should be $.5 mgh$ for some reason.

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If it is at rest in the end, then the work done goes towards the potential energy only. This is determined by the height of the center of mass.

If the rod height is $h$, then the center of mass is located at $\frac{h}{2}$.

$${\rm Work} = m g \tfrac{h}{2} $$

No calculus needed. If you have to use calculus, consider a small mass segment ${\rm d}m$ located at a height $y$. The rod density is $\rho$ and section area $A$

$${\rm Work} = \int g y {\rm d}m = \int_0^h g y \rho A {\rm d}y$$

if the density is non uniform the total mass is $m=\int_0^h \rho A {\rm d}y$. This can help you prove that

$${\rm Work} = m g y_\mbox{center of mass}$$

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  • $\begingroup$ thanks, but can you solve it in a roundabout way with calculus? Just curious. Let's say that the mass of the rod is not uniform and follows an arbitrary distribution, like the human torso. $\endgroup$ – a25bedc5-3d09-41b8-82fb-ea6c353d75ae Mar 19 '16 at 15:44
  • $\begingroup$ You don't have to. It all has to do with the position of the center of mass. $\endgroup$ – John Alexiou Mar 19 '16 at 20:47
  • $\begingroup$ @Floris, the title says upright $\endgroup$ – John Alexiou Mar 19 '16 at 21:13

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