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I’ve been trying to mechanically model a cube tumbling down a straight slope and hit some difficulties early on. So as a starting point I want to look at the kinematics of a cube allowed to gravity-rotate freely around a pivot point on a horizontal plane and work out what happens when the lower side of the cube hits the plane.

enter image description here

In the diagram above, the ‘Before:’ part shows a cube balanced on its lower side. Ignore all friction (except the friction force that prevents the lower side from sliding on the plane) and drag. When $\theta >0$, the tangential force $F_t$ provides torque around $O$ and the cube starts rotating, until $\theta=\frac{\pi}{4}$ and the lower right side hits the horizontal plane.

I’ve been playing with some square and rectangular objects by allowing them to gravity-rotate around their lower side and noticed that the objects always continue to travel horizontally for a bit, coming to rest after $\Delta x$ (in the ‘After:’ part of the above diagram) of horizontal translation. Obviously this is caused by energy/momentum conservation but I can’t work out how.

Going back to the ‘Before:’ part, $F_t$ provides a couple $T=F_tD$ with $D$ half the diagonal length of the cube.

It’s also clear that $F_t=mg\sin\theta$, so $T=D mg\sin\theta$ and the equation of rotational motion becomes:

$D mg\sin\theta=I\dot{\omega}$.

(There is no other motion possible because the other component of $mg$, i.e. $mg\cos\theta$, is countered by an equal reaction force in the pivot point)

Re-written we get:

$D mg \sin \theta=I \frac{d \omega}{d\theta} \frac{d \theta}{dt}$.

Or: $D mg\sin\theta d\theta=I\omega d\omega$.

Integrating with boundaries $t=0, \theta=0, \omega=0$ to $\theta=\frac{\pi}{4}$ we get:

$\frac{I\omega^2}{2}=\frac{2-\sqrt{2}}{2}D mg$.

So now we know kinetic energy and (obviously too) angular momentum at $\theta=\frac{\pi}{4}$.

Now where above I wrote “the objects always continue to travel horizontally for a bit”, I was a bit economical with my observations. In reality, when the lower plane of the cube hits the horizontal plane a ‘thudding’ noise is heard, the object seems to rebound a bit but does always end up a little to the right of its original position. Obviously the impact is somewhat inelastic (so some energy is lost to friction), some energy is converted to translation kinetic energy and that is then lost to translational friction. I just can’t see how rotational kinetic energy is converted to translational energy on this case.

Edit:

Evaluating translational kinetic energy at $\theta=\frac{\pi}{4}$:

From the energy balance $U=E_{K,R}+ E_{K,T}$ between $\theta=0, \theta=\frac{\pi}{4}$, $ E_{K,T}$ can be determined, since as $ E_{K,R}$ is already known.

During the rotation the centre of gravity of the cube is lowered from $D$ to $\frac{L}{2}$ with $L$ the length of a side of the cube and potential energy is converted to kinetic energy.

With Pythagoras:

$L^2=D^2+D^2=2D^2$, so $L=\sqrt{2} D$.

$U=mg(D-\frac{L}{2})=mg(D-\frac{\sqrt{2}}{2}D)$.

Or $U=\frac{2-\sqrt{2}}{2}mgD$.

Since as $E_{K,R}=\frac{2-\sqrt{2}}{2}mgD$.

This suggests that $E_{K,T}=0$!

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As you say, you cannot ignore friction here. If we imagine a frictionless surface, the center of mass of the cube would fall straight down. It would rotate about the center of mass (as the center fell) rather than about the corner of the cube. In that case there would be no horizontal movement of the center.

But because there is friction, there is a net horizontal force on the cube in the direction it falls.

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  • $\begingroup$ But as long as $\theta<\frac{\pi}{4}$ the motion is purely rotational, right? Assuming friction, of course. $\endgroup$ – Gert Sep 11 '15 at 20:11
  • $\begingroup$ The friction is only in one direction, and unopposed by any other horizontal forces. That makes it a net force. A net force will accelerate the (center of mass) of the object. $\endgroup$ – BowlOfRed Sep 11 '15 at 20:15
  • $\begingroup$ ah, all clear now. Blindness is a terrible affliction. $\endgroup$ – Gert Sep 11 '15 at 20:45
  • $\begingroup$ While this is not untrue, it skips over an interesting part: as the cube rotates, I believe the normal force exerted by the corner on the ground decreases - and so does the friction. Normally, if the corner were a pivot point the final motion of the center of mass would be straight down - but this would require "negative friction" and a net force to the left. This suggests to me that the corner will lift off at a certain angle of rotation? Worth calculating? $\endgroup$ – Floris Sep 15 '15 at 6:26

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