Finding the angular velocity of a cube about different rotation points

[Wavefunction]

Homework Statement

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?

Homework Equations

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

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

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

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.

 

[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"?

 

[Wavefunction]

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"?

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?

 

[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?

Homework Statement

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?

Homework Equations

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

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

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

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

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.

 

[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 V_c and the angular speed ω at the instant the block collides with the table. I don't believe V_c = ωl is correct. You are correct that the center of mass only has a vertical component of velocity during the fall.

 

[Wavefunction]

Homework Statement

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?

Homework Equations

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

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

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

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

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)

$\frac{mg\sqrt{2}l}{2}=\frac{mV^2}{2}+\frac{1}{2}\frac{2ml^2\omega^2}{3} + \frac{mgl}{2}$

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

Then: $\frac{mg(\sqrt{2}-1)l}{2}=\frac{m\omega^2\frac{l^2}{2}}{2}+\frac{ml^2\omega^2}{3}$

$= \frac{g(\sqrt{2}-1)}{2}=[\frac{\frac{l}{2}}{2}+\frac{l}{3}]\omega^2$
$= \frac{g(\sqrt{2}-1)}{2}=[\frac{1l}{4}+\frac{l}{3}]\omega^2$
$= \frac{g(\sqrt{2}-1)}{2\frac{7l}{12}}=\omega^2$
$= \frac{6(\sqrt{2}-1)g}{7l}=\omega^2 = \omega_{a}^2$

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 $\omega_{b}^2=\omega_{a}^2$

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.

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

 

[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?

 

[Wavefunction]

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?

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?

 

[Wavefunction]

Homework Statement

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?

Homework Equations

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

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

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

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

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 $V=\frac{\omega l}{2}$?
$g(\sqrt{2}-1) =\frac{\omega^2 l}{4}+ \frac{l}{6}\omega^2$
$g(\sqrt{2}-1) =[ \frac{l}{4}+ \frac{l}{6}]\omega^2$
$\frac{g(\sqrt{2}-1)}{\frac{10l}{24}} = \omega^2$
$\frac{12g(\sqrt{2}-1)}{5l} = \omega^2=\omega_{b}^2$

Finally, since $\frac{5}{12}>\frac{3}{2} \Rightarrow \omega_{b}>\omega_{a}$

where I have used the fact that since the red dot has zero vertical velocity that implies that the blue dot has zero vertical velocity: this implies that $V=\omega\frac{l}{2}$

Okay so how does that look now?

 

[TSny]

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