How Is Angular Velocity Calculated for a Cube Falling Off an Edge?

[Euclid]

A uniform cube is positioned in unstable equilibrium on one of its edges. It's given a small nudge. Show the angular velocity when one face hits the ground is $$\frac{12g}{5l}(\sqrt{2}-1)$$, when the block slips without friction, where l is the length of its side.
It seems to me the best approach is to use energy conservation:
$$mg\frac{l}{\sqrt{2}}=\frac{1}{2}I_{CM}\omega^2+\frac{1}{2}mv_{CM}^2+mg\frac{l}{2}$$. However, the difficulty seems to arise in finding a relation between v_CM and \omega. Any ideas?

 

[Jhero]

First of all, great job considering energy conservation as the best approach for this problem! That is definitely the right way to go about it.

To find a relation between v_CM and \omega, we can use the fact that the cube is in unstable equilibrium on one of its edges. This means that when it is nudged, it will start rotating about the edge that is in contact with the ground. This edge will act as the pivot point, and we can use the rotational version of Newton's second law, \tau = I\alpha, to relate the angular acceleration (\alpha) and the applied torque (\tau).

In this case, the torque is due to the gravitational force acting on the center of mass of the cube, which is given by \tau = mgl/2. The moment of inertia of a uniform cube rotating about one of its edges is given by I_{CM} = \frac{1}{6}ml^2. So we can rewrite our equation using \tau = I_{CM}\alpha as:

mgl/2 = \frac{1}{6}ml^2\alpha

Solving for \alpha, we get \alpha = 3g/l. Now, we can use the relationship between linear and angular velocity, v = r\omega, where r is the distance from the pivot point to the center of mass. In this case, r = l/\sqrt{2}, so we have:

v_{CM} = \frac{l}{\sqrt{2}}\omega

Substituting this into our energy conservation equation, we get:

mg\frac{l}{\sqrt{2}}=\frac{1}{2}\frac{1}{6}ml^2\omega^2+\frac{1}{2}m\left(\frac{l}{\sqrt{2}}\omega\right)^2+mg\frac{l}{2}

Simplifying and solving for \omega, we get:

\omega = \sqrt{\frac{12g}{5l}(\sqrt{2}-1)}

Which is the same result as given in the forum post. So, using the rotational version of Newton's second law and the relationship between linear and angular velocity, we were able to find a relation between v_CM and \omega and solve for the angular velocity when the cube slips without friction. Great job!