- #1
Wavefunction
- 99
- 4
Homework Statement
A homogeneous cube of sides [itex]l[/itex] is initially at rest in unstable equilibrium with one edge in
contact with a horizontal plane ([itex]θ = 45[/itex] degrees initially). The cube is given a small angular
displacement and allowed to fall. What is the angular velocity of the cube when one face
contacts the plane if:
a) the edge in contact with the plane cannot slide?
b) the plane is frictionless so the edge can slide?
Homework Equations
[itex]\mathbf{I}_{cm} = \frac{Ml^2}{6}\mathbf{1} [/itex]
[itex]\mathbf{J} = \begin{pmatrix}\frac{2}{3}&\frac{-1}{4}&\frac{-1}{4}\\\frac{-1}{4}&\frac{2}{3}&\frac{-1}{4}\\\frac{-1}{4}&\frac{-1}{4}&\frac{2}{3}\end{pmatrix} [/itex]
[itex] U_{gravity} = \frac{mg\sqrt{2}l}{2} [/itex]
The Attempt at a Solution
Setup: For part a) since the edge is not allowed to slide this implies that there will be translational velocity of the center of mass in addition to the angular velocity of the cube so the equation for energy will look like: [itex]U_{gravity}=T_{translational}+T_{rotational} [/itex]
For part b) since the edge is now allowed to slide this implies that the only motion will be a rotation about the the center of mass. so the equation for energy will look like: [itex] U_{gravity}=T_{rotational} [/itex]
Solution:
Part a)
[itex] \frac{mg\sqrt{2}l}{2}=\frac{mV^2}{2}+\frac{1}{2}\frac{ml^2\omega^2}{6} [/itex]
[itex] V=\vec{\omega}\times\vec{r}_{cm} \rightarrow V^2 = |\vec{\omega}\times\vec{r}_{cm}|^2= \omega^2(\frac{\sqrt{2}l}{2})^2\sin^2{\beta}[/itex]
Since [itex] \beta = \frac{\pi}{2} \Rightarrow V^2 = \omega^2\frac{l^2}{2} [/itex]
Then: [itex] \frac{mg\sqrt{2}l}{2}=\frac{m\omega^2\frac{l^2}{2}}{2}+\frac{1}{2}\frac{ml^2\omega^2}{6} [/itex]
[itex] = \frac{g\sqrt{2}}{2}=[\frac{\frac{l}{2}}{2}+\frac{1}{2}\frac{l}{6}]\omega^2 [/itex]
[itex] = \frac{g\sqrt{2}}{2}=[\frac{12l}{48}+\frac{4l}{48}]\omega^2 [/itex]
[itex] = \frac{g\sqrt{2}}{2\frac{l}{3}}=\omega^2 [/itex]
[itex] = \frac{3\sqrt{2}g}{2l}=\omega^2 = \omega_{a}^2 [/itex]
Part B:
[itex] \frac{mg\sqrt{2}l}{2}=\frac{1}{2}\frac{ml^2\omega^2}{6} [/itex]
[itex] g\sqrt{2} = \frac{l}{6}\omega^2 [/itex]
[itex] \frac{g\sqrt{2}}{\frac{l}{6}} = \omega^2 [/itex]
[itex] \frac{6\sqrt{2}g}{l} = \omega^2=\omega_{b}^2 [/itex]
Finally [itex]\frac{1}{4}\omega_{b}^2=\omega_{a}^2 \Rightarrow \omega_{b}>\omega_{a} [/itex] Which makes since because when slipping is not allowed the cube is rotating about an axis that is not its center of mass so it's moment of inertia should be greater and for the same given initial energy [itex] \omega [/itex] should be less than the case where slipping is allowed correct? Thanks in advance for your help guys/gals.
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