# Minimum energy for turning over a cube [closed]

This is a strange question from the physics Olympiade I wish to get some explanation for:

Using a certain type of wood, cubes are sawn with different edge lengths $R$. When they are on the ground, they are turned over one of the edges of the face touching the ground, so they rotate 90 degrees in the end. What is the relationship between the minimum energy $E$ required for this action and $R$?

• A. $E$ is proportional with $R$.
• B. $E$ is proportional with $R^2$.
• C. $E$ is proportional with $R^3$.
• D. $E$ is proportional with $R^4$.

The answers is D, but why?

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## closed as off-topic by David Z♦Sep 25 '13 at 21:29

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I disagree that it's a "strange question". It looks pretty straightforward. – Mark Eichenlaub Sep 25 '13 at 19:43

The energy is

$$E = h*g*m$$

with $h$ the height you raise the center of mass, $g$ gravitational acceleration, and $m$ the mass.

How does each individual term scale with $R$? When you figure that out, multiplying gives you the total scaling.

(Hint: both $h$ and $m$ depend on $R$)

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haha i was going to say that the answer is too quick :P, esp if the OP is doing the physics olympiad... he should be thinking about this on his own and providing his reasoning. but i see you've updated your answer to make it less 'spoon feeding' – nervxxx Sep 25 '13 at 19:46
Lets see, h = R/2, m = R^3 * c, g = some value, so the final thing will be E = something * R ^ 4, well, pretty straightforward. But I didn't think of this because: does it only costs you energy when you raise the centre of mass? How about the rotation you apply, the translation of the centre of mass to the right? Won't that have effect on the proportionality between R and E? – user209347 Sep 25 '13 at 19:53