# Cube tipping on horizontal surface

1. May 24, 2014

### raving_lunatic

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

A cube of mass m and side 2a is held on one of its edges on a horizontal surface. It is released from this position and allowed to tip. Find an expression for the angular velocity of the cube as its face strikes the surface of the table in the following cases:

a) the surface is sufficiently rough to stop the edge from moving
b) the edge of the cube can slide freely on the table

2. Relevant equations

G = r x F
G = dJ/dt (G torque, J angular momentum)

I = 2/3 m a2 - moment of inertia for cube about its center of mass

I = 2/3 m a 2 + 2ma2 - moment of inertia for a cube about one of its edges (using the parallel axis theorem)

1/2 I w2 = rotational kinetic energy

3. The attempt at a solution

Okay. So, first I tried to work out the torque of the weight about the rotation axis on one of the cube's edges, which I determined to be mga tan θ in the direction of increasing θ, where θ is defined as the angle between the vertical and the line between the center of mass, and the edge about which the cube tips. The reaction force acts through this point, so it exerts no torque about this axis. I wasn't able to integrate the equation for torque to find angular momentum (as θ is a function of time) so I tried a conservation of energy approach, reasoning that the center of mass moved by (√2-1)a and equating the loss in gravitational potential energy to the rotational kinetic energy. This gave me an answer for the case where the motion is purely rotational (which I think is the first case), but I wasn't sure how to deal with the case when the block is sliding freely. Possibly a term should be added to compensate for the translational kinetic energy, but I can't find the relation between this and the angular velocity? Any help with this problem would be greatly appreciated

2. May 24, 2014

### BvU

Hello Looney,

At first I was a little surprised by your statement that in the first case "the motion is purely rotational", but I think you are correct there, because your rotation is around an edge and you take that into account.

So the hint for the other case jumped on me: where is the axis of rotation ?

(Not an expert, but intrigued by this 'simple' elegant problem!)

3. May 24, 2014

### raving_lunatic

I'm guessing the axis of rotation is the one that passes through the center of mass, parallel to the surface? (There's only two that seem to make sense to me.)

Okay - so we have a combination of translational motion of the CM and rotational motion about the CM in the case where the edge of the block is free to slide, is that correct? So we need to take into account the velocity of the system's center of mass as well as the rotation about it? I can see how that would be possible (we can get it in terms of the angle of rotation, at least, defining a suitable origin...)

4. May 24, 2014

### BvU

Again, at first I had a different idea: there's a gravitational force acting at the CM axis and a normal force at the supporting edge. So my axis was halfway, but that leads to nonsense (I don't think the CM can be moved sideways). Your approach looks a lot better. And indeed, CM dropping straight down imparts some kinetic energy. But due to lower moment of inertia the final answer might well come out the same. Or more. Or less .
Let me (and other listeners) know what you find!