# Angular momentum / conservation of momentum questions

qsduahuw
Homework Statement:
See image below.
Relevant Equations:
N/A
I thought the answer is B because the angular momentum in conserved in all 3 pictures.

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Delta2

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Homework Statement:: Answer to this question:

Relevant Equations:: N/A

I thought the answer is B because the angular momentum in conserved in all 3 pictures.

Conservation of angular momentum can only be assumed if no external torque operates. As the student manipulates the wheel, what external forces act on the student-wheel system? What torques might they exert?

Delta2
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I am not an expert in angular momentum but something tells me that states B and C have same magnitude (and opposite direction ) of angular momentum so the correct answer must be one from (a) or (d)

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I am not an expert in angular momentum but something tells me that states B and C have same magnitude (and opposite direction ) of angular momentum so the correct answer must be one from (a) or (d)
I find the image in post #1 too blurry to be sure, but it looks to me that a, b and d all have B and C equal.

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I find the image in post #1 too blurry to be sure, but it looks to me that a, b and d all have B and C equal.
Yes, right, but the OP implies that he has been told that (b) is not the correct answer and so does your post implies an external torque so AM is not conserved.

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Yes, right, but the OP implies that he has been told that (b) is not the correct answer and so does your post implies an external torque so AM is not conserved.
I was careful only to say that it might not be conserved.

jbriggs444
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I was careful only to say that it might not be conserved.
I agree it cannot be conserved. In state A I think angular momentum is in the horizontal direction , while in state b and C in the vertical direction.

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I thought the answer is B because the angular momentum in conserved in all 3 pictures
Angular momentum is a vector. For it to be conserved, it must have the same magnitude and direction in all three pictures. Does it?

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I agree it cannot be conserved. In state A I think angular momentum is in the horizontal direction , while in state b and C in the vertical direction.
Be a bit careful with that claim about the direction of the angular momentum in (b) and (c). If it is defined at all, it is vertical. But is it defined at all?

Delta2 and kuruman
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Be a bit careful with that claim about the direction of the angular momentum in (b) and (c). If it is defined at all, it is vertical. But is it defined at all?
Sorry no clue why it is not defined?. We have some sort of double rotation: We have the wheel spinning and the person and wheel rotating around the platform center, kind of like Earth spinning around itself and rotating around the sun, why it should not be defined?

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Sorry no clue why it is not defined?. We have some sort of double rotation: We have the wheel spinning and the person and wheel rotating around the platform center, kind of like Earth spinning around itself and rotating around the sun, why it should not be defined?
What is the direction of a zero displacement?

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What is the direction of a zero displacement?
Sorry due to my chronical disease I haven't slept well and I got no clue what you try to imply.

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Sorry due to my chronical disease I haven't slept well and I got no clue what you try to imply.
Consider what kind of torques that could act on the system given that it is supported by a (presumably frictionless) support rotating freely around its axis and what this implies for the vertical component of the angular momentum.

Edit: Also note that the objection people are raising is that the direction of angular momentum may be undefined. Not that angular momentum itself is undefined.

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Edit: Also note that the objection people are raising is that the direction of angular momentum may be undefined. Not that angular momentum itself is undefined.
Sorry I still don't understand. If I take the following integral $$\int \vec{r}\times\vec{v} dm$$ over the whole region of our system (spinning wheel+person+rotating platform) isn't this integral well defined? Isn't it the angular momentum of the system?

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Sorry I still don't understand. If I take the following integral $$\int \vec{r}\times\vec{v} dm$$ over the whole region of our system (spinning wheel+person+rotating platform) isn't this integral well defined? Isn't it the angular momentum of the system?
Pay attention to the edit from @Orodruin in #13

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Sorry I still don't understand. If I take the following integral $$\int \vec{r}\times\vec{v} dm$$ over the whole region of our system (spinning wheel+person+rotating platform) isn't this integral well defined? Isn't it the angular momentum of the system?
As I wrote, angular momentum is well defined. It is its direction that is not.

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As I wrote, angular momentum is well defined. It is its direction that is not.
Ok let me ask you this, how can a vector be well defined, if its direction is not well defined?

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Ok let me ask you this, how can a vector be well defined, if its direction is not well defined?
Let me ask you the same question. Which unique vector is this true for?

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Ok let me ask you this, how can a vector be well defined, if its direction is not well defined?
Did you try to answer my question in post 11?

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Let me ask you the same question. Which unique vector is this true for?
It is true for all vectors I know of. A force is well defined if both direction and magnitude is well defined. Same for the velocity of a particle. Same from every other vector.

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Did you try to answer my question in post 11?
I don't even understand your question. What a zero displacement has to do with this problem?

I feel a piece of information is very obvious to you guys but it's not obvious to me at all. I need to try to sleep, i ll come back in 2-3hours, hopefully with a clearer mind.

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It is true for all vectors I know of. A force is well defined if both direction and magnitude is well defined. Same for the velocity of a particle. Same from every other vector.
A vector is well defined if its direction and magnitude are both well defined. Yes, that is correct. However, the reverse implication does not hold. There is a condition under which a vector can be well defined without its direction being well defined.

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Oh ok i see the zero vector.

jbriggs444