Angular momentum of a rotating door

[bananabandana]

Homework Statement

A door ( a rod of length $L$, mass $M$) rotates with angular velocity $\omega$ about a point $H$, and approaches a stop at $S$. $H$ and $S$ are along the same line, and separated by a distance $s$. Show that the angular momentum of the door about the point $S$ just before it hits the stop is:

$$ \mathbf{L} = M \omega L \bigg( \frac{s}{2} - \frac{l}{3}\bigg) \mathbf{\hat{z}} $$

Homework Equations

The Attempt at a Solution

- Use definition of angular momentum and apply parallel axis theorem.[/B]
Since rotation is about the z axis - (i.e the axis through $H$) we have $\vec{\omega} = \omega \mathbf{\hat{z}}$ so:
$$ \mathbf{L} = I \omega \mathbf{\hat{z}} $$
Notation $I_{s}$ means "moment of inertia evaluated at S". The moment of inertia of a rod as described about the centre of mass is $\frac{M}{12}L^{2}$ so, by parallel axis theorem:
$$ I_{s} = \frac{M}{12}L^{2} +M\big( \frac{L}{2}-s\big)^{2} =\frac{ML^{2}}{3}+Ms^{2} -MLs$$

Therefore:
$$\mathbf{L} = M\omega\big(\frac{L^{2}}{3}+s^{2} -Ls\big)\mathbf{\hat{z}} $$
What has gone wrong??

 

[CWatters]

I think you are correct.

Is it really a lower case "l" in their answer?

 

[bananabandana]

I think you are correct.

Is it really a lower case "l" in their answer?

Ah no, sorry - it should be $L$ . This was part of a final year exam question, so I'm pretty sure I must be wrong somewhere!

 

[BvU]

Hi guys,

Therefore:

$$\mathbf{L} = M\omega\big(\frac{L^{2}}{3}+s^{2} -Ls\big)\mathbf{\hat{z}}$$​

I agree, provided you use the $\omega$ around S, not the $\omega$ around H ...

 

[bananabandana]

Hi guys,

I agree, provided you use the $\omega$ around S, not the $\omega$ around H ...

Why aren't the two the same? Surely the whole system is connected and therefore has the same $\omega$

 

[BvU]

No.

 

[BvU]

That's a bit short, I concede. Must admit I don't have the answer worked out in full, but this is what came to mind:

$$\theta {L\over 2} = - \theta' \left (s - {L\over 2}\right ) \Rightarrow \dot \theta {L\over 2} = -\dot \theta' \left (s - {L\over 2}\right )$$

(all for $\theta \downarrow 0$, of course)

The least it does is give us a minus sign for s > L/2

My memory of the definition is $\omega = {d\theta\over dt}$

H and S see different $\theta$ (or the same $\vec v_{\rm c.o.m.}$ from different distances)

PS I like the

H and S are along the same line

in the problem statement. Two points have this habit, so: does it mean something else ? ANd does that have a bearing on the problem ? I think not...

 

[haruspex]

The book answer is correct.
The easiest way is to consider the door's motion as the sum of the linear motion of its mass centre and a rotation about that centre.
The linear motion leads to an angular momentum about S. The angular momentum about the mass centre can then be added to this.
Be careful with signs.

Why aren't the two the same? Surely the whole system is connected and therefore has the same $\omega$

Yes.
Edit:

by parallel axis theorem:

The parallel axis theorem is for finding the moment of inertia about a point as a centre of rotation. S is not that.

 

[bananabandana]

The book answer is correct.
The easiest way is to consider the door's motion as the sum of the linear motion of its mass centre and a rotation about that centre.
The linear motion leads to an angular momentum about S. The angular momentum about the mass centre can then be added to this.
Be careful with signs.

Yes.
Edit:

The parallel axis theorem is for finding the moment of inertia about a point as a centre of rotation. S is not that.

Ah, of course. No, it's definitely not rotating about S. Thank you. Also realize now that the definition of the intertia tensor only involves motion about the centre of mass, so obviously it doesn't take the angular momentum relative to the origin at $S$ into account.

So then I get this- that the angular momentum about the centre of mass, $\mathbf{L^{*}}$ is given by:

$$ \mathbf{L^{*}} = I\omega \mathbf{\hat{z}} = -\frac{ML^{2}}{12}\mathbf{\hat{z}} $$

(It points down due to right hand rule)
The angular momentum of the centre of mass itself is then $\mathbf{L_{CM}}$:

$$ \mathbf{L_{CM}} = \mathbf{r_{CM}} \times \mathbf{p_{CM}} = M \ -\big(s-\frac{L}{2} \big) \mathbf{\hat{i}} \times -v_{CM} \mathbf{\hat{j}} = M\omega\big( s-\frac{L}{2}\big)\frac{L}{2} \mathbf{\hat{k}} $$
Where I've used the fact that the velocity of the centre of mass is the distance from the axis of rotation times the angular velocity $\mathbf{v_{CM}} = \omega\frac{L}{2}\mathbf{\hat{j}}$

So,yep:

$$ \mathbf{L} = \mathbf{L^{*}} + \mathbf{L_{CM}} = M\omega L\bigg( \frac{s}{2}-\frac{L}{3} \bigg)\mathbf{\hat{k}} $$Thanks!