# Finding the angular velocity of a cube about different rotation points

1. Apr 27, 2014

### Wavefunction

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

A homogeneous cube of sides $l$ is initially at rest in unstable equilibrium with one edge in
contact with a horizontal plane ($θ = 45$ 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?

2. Relevant equations

$\mathbf{I}_{cm} = \frac{Ml^2}{6}\mathbf{1}$

$\mathbf{J} = \begin{pmatrix}\frac{2}{3}&\frac{-1}{4}&\frac{-1}{4}\\\frac{-1}{4}&\frac{2}{3}&\frac{-1}{4}\\\frac{-1}{4}&\frac{-1}{4}&\frac{2}{3}\end{pmatrix}$

$U_{gravity} = \frac{mg\sqrt{2}l}{2}$

3. 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: $U_{gravity}=T_{translational}+T_{rotational}$

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: $U_{gravity}=T_{rotational}$

Solution:
Part a)

$\frac{mg\sqrt{2}l}{2}=\frac{mV^2}{2}+\frac{1}{2}\frac{ml^2\omega^2}{6}$

$V=\vec{\omega}\times\vec{r}_{cm} \rightarrow V^2 = |\vec{\omega}\times\vec{r}_{cm}|^2= \omega^2(\frac{\sqrt{2}l}{2})^2\sin^2{\beta}$
Since $\beta = \frac{\pi}{2} \Rightarrow V^2 = \omega^2\frac{l^2}{2}$

Then: $\frac{mg\sqrt{2}l}{2}=\frac{m\omega^2\frac{l^2}{2}}{2}+\frac{1}{2}\frac{ml^2\omega^2}{6}$

$= \frac{g\sqrt{2}}{2}=[\frac{\frac{l}{2}}{2}+\frac{1}{2}\frac{l}{6}]\omega^2$
$= \frac{g\sqrt{2}}{2}=[\frac{12l}{48}+\frac{4l}{48}]\omega^2$
$= \frac{g\sqrt{2}}{2\frac{l}{3}}=\omega^2$
$= \frac{3\sqrt{2}g}{2l}=\omega^2 = \omega_{a}^2$

Part B:

$\frac{mg\sqrt{2}l}{2}=\frac{1}{2}\frac{ml^2\omega^2}{6}$
$g\sqrt{2} = \frac{l}{6}\omega^2$
$\frac{g\sqrt{2}}{\frac{l}{6}} = \omega^2$
$\frac{6\sqrt{2}g}{l} = \omega^2=\omega_{b}^2$

Finally $\frac{1}{4}\omega_{b}^2=\omega_{a}^2 \Rightarrow \omega_{b}>\omega_{a}$ 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 $\omega$ should be less than the case where slipping is allowed correct? Thanks in advance for your help guys/gals.

Last edited: Apr 28, 2014
2. Apr 27, 2014

### TSny

A couple of things to consider.

What is the change in vertical height of the CM between initial and final positions?

In part B, is it true that the block only has rotational motion about the CM? Isn't the CM "falling"?

3. Apr 27, 2014

### Wavefunction

Oh I see what your saying I need to consider the final potential energy at height $\frac{l}{2}$. Well if it doesn't just have rotational motion in part B then the center of mass must translate vertically, correct?

4. Apr 28, 2014

### Wavefunction

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

A homogeneous cube of sides $l$ is initially at rest in unstable equilibrium with one edge in
contact with a horizontal plane ($θ = 45$ 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?

2. Relevant equations

$\mathbf{I}_{cm} = \frac{Ml^2}{6}\mathbf{1}$

$\mathbf{J} = \begin{pmatrix}\frac{2}{3}&\frac{-1}{4}&\frac{-1}{4}\\\frac{-1}{4}&\frac{2}{3}&\frac{-1}{4}\\\frac{-1}{4}&\frac{-1}{4}&\frac{2}{3}\end{pmatrix}$

$U_{gravity}^{i} = \frac{mg\sqrt{2}l}{2}$

$U_{gravity}^{f} = \frac{mgl}{2}$

3. 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: $U_{gravity}^{i}=T_{translational}+T_{rotational}+U_{gravity}^{f}$

For part b) since the edge is now allowed to slide this implies that the rotational motion will be a rotation about the the center of mass, and the the center of mass will translate vertically downward. So the equation for energy will look like: $U_{gravity}^{i}=T_{translational}+T_{rotational}+U_{gravity}^{f}$

Solution:
Part a)

$\frac{mg\sqrt{2}l}{2}=\frac{mV^2}{2}+\frac{1}{2}\frac{ml^2\omega^2}{6} + \frac{mgl}{2}$

$V=\vec{\omega}\times\vec{r}_{cm} \rightarrow V^2 = |\vec{\omega}\times\vec{r}_{cm}|^2= \omega^2(\frac{\sqrt{2}l}{2})^2\sin^2{\beta}$
Since $\beta = \frac{\pi}{2} \Rightarrow V^2 = \omega^2\frac{l^2}{2}$

Then: $\frac{mg(\sqrt{2}-1)l}{2}=\frac{m\omega^2\frac{l^2}{2}}{2}+\frac{1}{2}\frac{ml^2\omega^2}{6}$

$= \frac{g(\sqrt{2}-1)}{2}=[\frac{\frac{l}{2}}{2}+\frac{1}{2}\frac{l}{6}]\omega^2$
$= \frac{g(\sqrt{2}-1)}{2}=[\frac{12l}{48}+\frac{4l}{48}]\omega^2$
$= \frac{g(\sqrt{2}-1)}{2\frac{l}{3}}=\omega^2$
$= \frac{3(\sqrt{2}-1)g}{2l}=\omega^2 = \omega_{a}^2$

Part B:

$\frac{mg\sqrt{2}l}{2}=\frac{mV^2}{2}+\frac{1}{2}\frac{ml^2\omega^2}{6} + \frac{mgl}{2}$
$g(\sqrt{2}-1) =\frac{V^2}{l}+ \frac{l}{6}\omega^2$ Use $V=\omega l$?
$g(\sqrt{2}-1) =\omega^2 l+ \frac{l}{6}\omega^2$
$g(\sqrt{2}-1) =[ l+ \frac{l}{6}]\omega^2$
$\frac{g(\sqrt{2}-1)}{\frac{7l}{6}} = \omega^2$
$\frac{6g(\sqrt{2}-1)}{7l} = \omega^2=\omega_{b}^2$

Finally $\frac{7}{4}\omega_{b}^2=\omega_{a}^2 \Rightarrow \omega_{b}<\omega_{a}$

I don't think this is right either because I think that $\omega_{b}>\omega_{a}$ I put $V=\omega l$? Because I couldn't think of anything else to do, but I know that the position vector from the fixed point to the center of mass is the zero vector since the fixed point is the center of mass so $\vec{\omega}\times\vec{r}_{cm} = \vec{0}$. So I'm stuck on this part.

5. Apr 28, 2014

### TSny

Part A looks correct to me.

For part B, you need to come up with some reasoning for getting the relation between the center of mass speed Vc and the angular speed ω at the instant the block collides with the table. I don't believe Vc = ωl is correct. You are correct that the center of mass only has a vertical component of velocity during the fall.

6. Apr 28, 2014

### Wavefunction

Okay I had an idea after I had a chance to sleep on the problem: in part b) The center of mass in part b) is not fixed; however, the sliding edge is vertically fixed despite the fact that it is translating horizontally so If I use $\vec{\omega}\times\vec{r} =v$ where $\vec{r}$ is the vector pointing from a point along the edge of the cube to the center of mass I'll get:

$\frac{mg\sqrt{2}l}{2}=\frac{mV^2}{2}+\frac{1}{2}\frac{ml^2\omega^2}{6} + \frac{mgl}{2}$
$g(\sqrt{2}-1) =\frac{V^2}{l}+ \frac{l}{6}\omega^2$ Use $V=\frac{\sqrt{2}\omega l}{2}$
$g(\sqrt{2}-1) =\frac{l}{2}\omega^2+ \frac{l}{6}\omega^2$
$g(\sqrt{2}-1) =[ \frac{6l}{12}+ \frac{2l}{12}]\omega^2$
$\frac{g(\sqrt{2}-1)}{\frac{8l}{12}} = \omega^2$
$\frac{3g(\sqrt{2}-1)}{2l} = \omega^2=\omega_{b}^2$

Now If I compare $\omega_{a}^2$ to $\omega_{b}^2$ I find that $\frac{7}{4}\omega_{a}^2 = \omega_{b}^2 \Rightarrow \omega_{b}>\omega_{a}$

7. Apr 28, 2014

### TSny

You've changed your answer to part A. It's no longer correct. If you want to use the moment of inertia about the fixed axis of rotation in part A, then the total kinetic energy is $\frac{1}{2} I_{\rm axis}\, \omega^2$. You would not add an additional $\frac{1}{2}mV_c^2$.

For part B I don't see how you are getting $V = \frac{\sqrt{2}\omega l}{2}$. As you say, the edge that is slipping on the table can have only a horizontal velocity. Suppose you consider the cube just as it hits the table. The picture shows the cube at this instant, where the red dot indicates the edge that remained in contact with the table during the fall. The blue dot shows the midpoint of the left face of the cube. Its velocity may be thought of as the velocity of the blue dot relative to the CM plus the velocity of the CM. The velocity relative to the CM is due to rotation about the CM. These velocities are indicated on the picture.

Since the red dot has zero vertical component of velocity, what does that tell you about the vertical component of velocity of the blue dot?

#### Attached Files:

• ###### Falling Cube 2.png
File size:
1.7 KB
Views:
162
8. Apr 29, 2014

### Wavefunction

Well if the red dot has zero vertical velocity then the blue dot must also have zero vertical velocity which means that the $v_c = \omega r$ where in this case $r=\frac{l}{2}$ right?

Last edited: Apr 29, 2014
9. Apr 29, 2014

### Wavefunction

Okay so how does that look now?

10. Apr 29, 2014

### TSny

Yes, that all looks good. (You meant 12/5 > 3/2.)