- #1
thecommexokid
- 70
- 2
Homework Statement
A homogeneous cube, each edge of which has a length [itex]\ell[/itex], 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.
Homework Equations
The moment of inertia for a cube rotating about an edge is [itex]I=\frac{2}{3}m\ell^2[/itex].
Gravitational potential energy is [itex]mgh[/itex].
Angular kinetic energy is [itex]\frac12 I\omega^2[/itex].
Linear kinetic energy is [itex]\frac12 mv^2[/itex].
The Attempt at a Solution
Balanced on edge, the cube has initial potential energy [itex]U_i=mg\frac{\ell}{\sqrt{2}}[/itex].
At the moment before it hits the ground, it has final potential energy [itex]U_f=mg\frac{\ell}{2}[/itex].
At the moment before it hits the ground, it has final angular kinetic energy [itex]\kappa_f=\frac{1}{2}I\omega_f^2[/itex].
For part (a), where the edge stays fixed, that's everything. [itex]U_i=U_f+\kappa_f[/itex], solve for [itex]\omega_f[/itex], 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 [itex]\frac{\ell}{\sqrt{2}}-\frac{\ell}{2}[/itex]. The COM's velocity at this point is given by [itex]v_f^2=v_i^2+2ad=0+2g\ell(\frac{1}{\sqrt2}-\frac{1}{2})[/itex] and we conclude that [itex]v_f=\sqrt{g\ell(\sqrt2-1)}[/itex].
Therefore the final linear kinetic energy of the block would be [itex]K_f=\frac{1}{2}mv_f^2=\frac{1}{2}mg\ell(\sqrt2-1)[/itex].
Then I would think I would proceed to say [itex]U_i=U_f+\kappa_f+K_f[/itex] and solve for [itex]\omega_f[/itex]. But if I try that, I find that, as I have calculated them, [itex]U_f+K_f=U_i[/itex] all by themselves, leaving 0 energy left for [itex]\kappa_f[/itex].
I suspect that my error is in calculating [itex]K_f[/itex]: 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: