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.


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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