Angular momentum of a rotating door

In summary, the conversation discusses the problem of finding the angular momentum of a door rotating about a point and approaching a stop. The solution involves using the definition of angular momentum and the parallel axis theorem, and taking into account the linear motion of the door's mass centre. The final answer is given as MωL(s/2−L/3)k^.
  • #1
bananabandana
113
5

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??
 
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  • #2
I think you are correct.

Is it really a lower case "l" in their answer?
 
  • #3
CWatters said:
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!
 
  • #4
Hi guys,
bananabandana said:
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 ... :rolleyes:
 
  • #5
BvU said:
Hi guys,

I agree, provided you use the ##\omega## around S, not the ##\omega## around H ... :rolleyes:
Why aren't the two the same? Surely the whole system is connected and therefore has the same ##\omega##
 
  • #6
No.
 
  • #7
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:
upload_2017-2-22_17-43-46.png
$$\theta {L\over 2} = - \theta' \left (s - {L\over 2}\right ) \Rightarrow \dot \theta {L\over 2} = -\dot \theta' \left (s - {L\over 2}\right )$$

(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
bananabandana said:
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...
 
  • #8
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.
bananabandana said:
Why aren't the two the same? Surely the whole system is connected and therefore has the same ##\omega##
Yes.
Edit:
bananabandana said:
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.
 
Last edited:
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  • #9
haruspex said:
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!
 
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1. What is angular momentum?

Angular momentum is a measure of an object's rotational motion, calculated by multiplying its moment of inertia by its angular velocity.

2. How is angular momentum related to a rotating door?

In the case of a rotating door, the angular momentum is determined by the door's moment of inertia, which depends on its mass, shape, and distribution of mass relative to its axis of rotation, as well as its angular velocity, which is the speed at which it rotates.

3. Why does a rotating door continue to spin even after it has been pushed?

This is due to the principle of conservation of angular momentum, which states that angular momentum remains constant unless acted upon by an external torque. When a force is applied to a rotating door, the torque causes a change in its angular velocity, but the total angular momentum remains the same, causing the door to continue spinning.

4. Can the angular momentum of a rotating door be changed?

Yes, the angular momentum of a rotating door can be changed by changing its moment of inertia or its angular velocity. For example, if the mass distribution of the door is changed, its moment of inertia will change, and its angular velocity will change in response to maintain a constant angular momentum.

5. How does friction affect the angular momentum of a rotating door?

Friction can act as an external torque on a rotating door, causing a decrease in its angular velocity and therefore its angular momentum. This is why doors typically slow down and eventually stop spinning after being pushed.

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