# A cube balanced on edge falls.

1. May 15, 2013

### thecommexokid

1. The problem statement, all variables and given/known data

A homogeneous cube, each edge of which has a length $\ell$, is initially in a position of unstable equilibrium with one edge in contact with a horizontal plane. The cube is then given a small displacement and allowed to fall. Find the angular velocity of the cube when one face strikes the plane if: (a) the edge cannot slip on the plane. (b) sliding can occur without friction.

2. Relevant equations

The moment of inertia for a cube rotating about an edge is $I=\frac{2}{3}m\ell^2$.

Gravitational potential energy is $mgh$.

Angular kinetic energy is $\frac12 I\omega^2$.

Linear kinetic energy is $\frac12 mv^2$.

3. The attempt at a solution

Balanced on edge, the cube has initial potential energy $U_i=mg\frac{\ell}{\sqrt{2}}$.

At the moment before it hits the ground, it has final potential energy $U_f=mg\frac{\ell}{2}$.

At the moment before it hits the ground, it has final angular kinetic energy $\kappa_f=\frac{1}{2}I\omega_f^2$.

For part (a), where the edge stays fixed, that's everything. $U_i=U_f+\kappa_f$, solve for $\omega_f$, and we're done.

My question is about part (b). My approach was to say that the center of mass of the block will accelerate linearly downward in free-fall, so by the time the block is about to hit the ground, the COM will have fallen a distance $\frac{\ell}{\sqrt{2}}-\frac{\ell}{2}$. The COM's velocity at this point is given by $v_f^2=v_i^2+2ad=0+2g\ell(\frac{1}{\sqrt2}-\frac{1}{2})$ and we conclude that $v_f=\sqrt{g\ell(\sqrt2-1)}$.

Therefore the final linear kinetic energy of the block would be $K_f=\frac{1}{2}mv_f^2=\frac{1}{2}mg\ell(\sqrt2-1)$.

Then I would think I would proceed to say $U_i=U_f+\kappa_f+K_f$ and solve for $\omega_f$. But if I try that, I find that, as I have calculated them, $U_f+K_f=U_i$ all by themselves, leaving 0 energy left for $\kappa_f$.

I suspect that my error is in calculating $K_f$: since the block is in contact with the ground, it seems a little fishy to say that the COM undergoes free-fall acceleration. But I don't see what other forces would be involved. Any guidance?

Last edited: May 15, 2013
2. May 15, 2013

### TSny

Does the COM have free-fall acceleration g? How many vertical forces act on the block?

3. May 15, 2013

### thecommexokid

Thanks for the reply. As I mentioned at the end of my post, I recognize that the block is in contact with the ground. But I can't figure out what the ground's force on the block would be. "Normal" forces are so named because they act in a direction normal (i.e. perpendicular) to the surface of contact, but in this situation there is no surface of contact, only a line segment of contact. So I can't even come up with a particularly good argument for what direction the force exerted by the ground ought to be in. (I mean, the COM of the block does not accelerate left or right, so obviously the ground contact force must be straight upward, but I can't figure out how I would determine that without looking at the resultant motion.)

My one thought was that maybe I could establish an axis that passes through both the COM and the point of contact with the ground, and resolve the block's weight vector $\mathbf{w}$ into a component $w_{||}$ along that axis and a component $w_{\bot}$ perpendicular to that axis. Then maybe the force exerted on the block by the ground would be the negative of $w_{||}$??? But that would have a left-right component, so that seems wrong.

4. May 16, 2013

### TSny

With no friction, the ground can exert only a vertically upward (normal) force on the edge of the block that is in contact with the ground. You are right that the COM will move vertically downward as the block slips. But the normal force affects the acceleration of the COM, so you can't assume the COM will have free-fall acceleration.

You're also on the right track with conservation of energy. You know how to get the loss in potential energy. Can you find a way to express the final total kinetic energy in terms of the unknown final angular velocity?

5. May 16, 2013

### thecommexokid

Okay, here's my attempt to do that:

At the moment before the cube lands on the ground, it is rotating about the edge in contact with the ground with angular velocity $\omega_f$. Therefore the COM is moving at a linear velocity of $v_f=\omega_f r=\omega_f \frac{\ell}{\sqrt2}$. So the block has final linear kinetic energy $K_f=\frac12 m v_f^2=\frac14 m\omega_f^2\ell^2$.

(1) Is that a valid approach? (2) Is that the approach you had in mind?

6. May 16, 2013

### thecommexokid

Well, doing it out using this approach, I wind up with a ωf that's 76% of the answer I got for part (a) where the pivot-edge was held fixed. This seems like a very plausible answer numerically. So that's a good sign.

7. May 16, 2013

### TSny

This is not quite right.

The block is not undergoing pure rotation about the edge in contact with the ground at the moment the face strikes the ground. The edge is still sliding horizontally at that moment. Pure rotation about the edge would not give the COM a vertical velocity.

One approach is to find the axis about which the block is in pure rotation at the moment the block hits the ground. You can find the location of that axis by using the fact that the COM is moving vertically while the edge in contact with the ground is moving horizontally. See instantaneous center of rotation .

8. May 16, 2013

### D H

Staff Emeritus
Another approach is to model the cube as simultaneously translating and rotating about the center of mass. Kinetic energy is ½mv2 + ½Iω2, where I is now the cube's moment of inertia about an axis through the center of mass. That an edge is always in contact with the surface means that the center of mass velocity can be expressed in terms of the angular velocity. An advantage of this approach is that the moment of inertia of a cube rotating about the its center of mass is well known.

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