# Right hand rule for angular momentum



## DemonD

The right hand rule I'm referring to is the one where if you wrap your right hand fingers along the direction of the rotation your thumb when pointing straight out will give you the direction of the angular momentum. 

My physics-English might be a bit shoddy, but hopefully you know what I'm talking about.

The thing I never understood was why that is always the case? Why does it always go in that direction?


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## Adrift

My physics textbook (Resnick & Halliday) says it's "by convention". I guess that means that someone decided that this was the case and it stuck.


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## EmpireConquered

Ah, I remember studying this, and forgetting this.


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## DemonD

EmpireConquered said:


> Ah, I remember studying this, and forgetting this.


I have vague memories of studying this as well, but clearly I didn't understand it back then either.


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## Sat Nam

I know some physics, but not enough to really help you with this question lol. I could make up something that sounds really convincing, but I'll abstain lol.

Maybe @FlightsOfFancy can help you. I remember reading something of his regarding him studying high level physics. If I'm mistaken, don't shoot me lol 

Sat Nam


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## FlightsOfFancy

Sat Nam said:


> I know some physics, but not enough to really help you with this question lol. I could make up something that sounds really convincing, but I'll abstain lol.
> 
> Maybe @_FlightsOfFancy_ can help you. I remember reading something of his regarding him studying high level physics. If I'm mistaken, don't shoot me lol
> 
> Sat Nam


Yeah I saw this; I really don't have a definitive answer but I'm not a full physicist. There's an 'Ask a Physicist ' thread here.

As far as I know it is just the nature of the cross product that so happens to be aligned with the right hand in such a way. I think this is actually more of a theoretical math question (I'm fairly certain it has to do with the alignment the cross product) of the so I'm throwing it at @ Mr. Meepers


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## Sat Nam

FlightsOfFancy said:


> Yeah I saw this; I really don't have a definitive answer but I'm not a full physicist. There's an 'Ask a Physicist ' thread here.
> 
> As far as I know it is just the nature of the cross product that so happens to be aligned with the right hand in such a way. I think this is actually more of a theoretical math question (I'm fairly certain it has to do with the alignment the cross product) of the so I'm throwing it at @Mr. Meepers


I appreciate the quick response : ) I was a science major and I took honors physics in high school, but I'm not really qualified to speak on the why's of physics lol. A lot of it was interesting, but I just did the work for my A's LOL

I read your long post yesterday (I think it was yesterday lol), but I didn't really know what to say so I just said nothing  I hope you're well : )

Btw, @Mr. Meepers is a hilarious name and it makes me say "meep!" when I see it lol. 

Sat Nam


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## Mr. Meepers

Is this the rule you are talking about?








Angular Momentum

This is merely a convention .... As you can see, the thumb is always perpendicular to the tangential velocity of the "points" of the thing (in this case a disk" that is rotating ... If the angular velocity is constant and the axis of rotation is the same (Note: in a three dimensional space, since rotation of anything with volume requires two dimensions, the axis of rotation exists and is unique), then we can create a constant vector along the axis of rotation to help us describe what is going on. ... Really, it is just a convention

Another way to look at it is to think about the angular motion of a particle (moving in a perfect circle with constant speed). You have the radius and tangential velocity ... If you take the vector product (cross-product ... just a binary operation defined in mathematics to be a certain way) of the radius and the tangential velocity, you will get the same vector no matter where the particle is on that point ... This lets us be able to create a nice definition where angular moment = Radius X Linear Momentum ... Remember, with vector multiplication, order matters and we get a third vector perpendicular to the other two (by definition of a cross product) ... I can't find the rule associated with this, but I can tell you that if you put your pointer finger in the direction of the radius/position vector (pointing away from the origin) and your middle finger in the direction of the tangential velocity, and make your thumb perpendicular to both, it will point in the same direction as the right hand rule above. ... You can say that the rule comes from vector multiplying the radius before the linear momentum.

So, the reason why is just because it was a convenient definition to talk about the world around us.




FlightsOfFancy said:


> As far as I know it is just the nature of the cross product that so happens to be aligned with the right hand in such a way. I think this is actually more of a theoretical math question (I'm fairly certain it has to do with the alignment the cross product) of the so I'm throwing it at @Mr. Meepers


Yup, although I have not seen any of the right/left hand rules since my freshman year in college (had to look them up lol)
Right-hand rule - Wikipedia, the free encyclopedia

Also, for some reason, your mention did not work 



Sat Nam said:


> Btw, @_Mr. Meepers_ is a hilarious name and it makes me say "meep!" when I see it lol.
> 
> Sat Nam



Thank you ... Meep ^__^


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## DemonD

@Mr.Meepers
Thank you for that explanation. I'd be lying if I said I understood most it, but I realize there's a limit for how simple things can be made.

If I understand the word 'convention' correctly, the force isn't necessarily exerted in the direction of the thumb, it might just as well go the precise opposite, only physicists have agreed to only speak of it in the one direction?


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## FlightsOfFancy

DemonD said:


> @_Mr._Meepers
> Thank you for that explanation. I'd be lying if I said I understood most it, but I realize there's a limit for how simple things can be made.
> 
> If I understand the word 'convention' correctly, the force isn't necessarily exerted in the direction of the thumb, it might just as well go the precise opposite, only physicists have agreed to only speak of it in the one direction?





*Now this is what I was going to say before but I was unsure of its propriety but I did it and it works (for this example): *
If you take your left hand, and make it sort of the 'left hand rule' That is have your fingers aligned so that your middle finger points to the right, your thumb points towards you, and your index towards the sky/ceiling. Notice, ALL the vectors point in the positive direction. This is NOT correct for the cross product as it should always be i -j k. If you mirror this with your right hand, however, you see that you do indeed have i and k positive, yet k in the negative direction. Therefore, the right hand rule is in accord with the mathematical convention of the cross product, taking into account sign.


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## DemonD

FlightsOfFancy said:


> *Now this is what I was going to say before but I was unsure of its propriety but I did it and it works (for this example): *
> If you take your left hand, and make it sort of the 'left hand rule' That is have your fingers aligned so that your middle finger points to the right, your thumb points towards you, and your index towards the sky/ceiling. Notice, ALL the vectors point in the positive direction. This is NOT correct for the cross product as it should always be i -j k. If you mirror this with your right hand, however, you see that you do indeed have i and k positive, yet k in the negative direction. Therefore, the right hand rule is in accord with the mathematical convention of the cross product, taking into account sign.


Didn't understand that either, well I did a bit, I think. But that was my question, why is the right hand correct, and not the left hand?


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## FlightsOfFancy

DemonD said:


> Didn't understand that either, well I did a bit, I think. But that was my question, why is the right hand correct, and not the left hand?


Do this:

Align your fingers on your left hand so that: your thumb points towards you, your index finger points towards the ceiling. It looks like you're pointing a gun upwards. Then extend your middle finger to make the 'third vector'. 

If you do this, you will see that all your fingers point in the positive direction if you think of the x/y being your middle/index finger, respectively and the z being the vector towards you. This is incorrect because it is not aligned with the convention of the mathematical cross product, which states that the middle vector (the y) need point opposed to the x and z; this is a mathematical convention in itself that the left-hand rule breaks. This is why they tell you that it's x -y z, or i -j k. 

Now if you mirror this on your right hand; you should notice that you still have the x and z positive, but your middle finger points in the negative direction for that vector. Perfect. This is in line with x -y k, or put simply, having two vectors point in their positive direction while one points in the negative (a feat we couldn't do with the left)


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## DemonD

FlightsOfFancy said:


> Do this:
> 
> Align your fingers on your left hand so that: your thumb points towards you, your index finger points towards the ceiling. It looks like you're pointing a gun upwards. Then extend your middle finger to make the 'third vector'.
> 
> If you do this, you will see that all your fingers point in the positive direction if you think of the x/y being your middle/index finger, respectively and the z being the vector towards you. This is incorrect because it is not aligned with the convention of the mathematical cross product, which states that the middle vector (the y) need point opposed to the x and z; this is a mathematical convention in itself that the left-hand rule breaks. This is why they tell you that it's x -y z, or i -j k.
> 
> Now if you mirror this on your right hand; you should notice that you still have the x and z positive, but your middle finger points in the negative direction for that vector. Perfect. This is in line with x -y k, or put simply, having two vectors point in their positive direction while one points in the negative (a feat we couldn't do with the left)


Now I get it. Thank you for your patience!


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## Mr. Meepers

DemonD said:


> @_Mr._Meepers
> Thank you for that explanation. I'd be lying if I said I understood most it, but I realize there's a limit for how simple things can be made.
> 
> If I understand the word 'convention' correctly, *the force isn't necessarily exerted in the direction of the thumb*, it might just as well go the precise opposite, only physicists have agreed to only speak of it in the one direction?





DemonD said:


> Didn't understand that either, well I did a bit, I think. But that was my question, why is the right hand correct, and not the left hand?


It is not a force, it is angular momentum. ... You said you understood @FightsofFancy's explanation so I don't need to explain the right hand vs the left (it is just a sign difference) and you may know vector multiplication, but there was a larger point I wanted to make, but I guess I was not very clear lol. back, when I was an undergraduate in college, a fellow physics major asked me why a vector times a vector creates another vector. He wanted to understand how when we multiply two "physical" vector, we get a third "physical" vector. Now, luckily, I had asked myself that question before and it helped me to understand what are new laws of physics and what are things created by the language we use to define our physical terms in. ... And this is an instance where it is just normal newtonian physics, but we created a language that allows us to talk about angular motion and we are just using the same physics we used for linear motion .... In the case of why a vector times a vector equals another vector (in a vector product), it is important to realize that in these equations the vectors represent different quantities (such as radius and momentum) and this "third vector", in this case angular momentum, is just a term we defined to fit this equation ... and the reason a vector times a vector is another vector has nothing to do with physical reality at all, it (vector multiplication) is just something we created. Same goes with the direction of angular momentum. We could have defined it to be a vector that points in any direction (although, our definition may have to be dependant on how we create the coordinate system and then rotating the coordinate system would change it and that is just very messy), but we chose the right hand out of convenience.
That may sound obvious, but I know how hazy the line between physics and mathematics can be.



FlightsOfFancy said:


> *Now this is what I was going to say before but I was unsure of its propriety but I did it and it works (for this example): *
> If you take your left hand, and make it sort of the 'left hand rule' That is have your fingers aligned so that your middle finger points to the right, your thumb points towards you, and your index towards the sky/ceiling. Notice, ALL the vectors point in the positive direction. This is NOT correct for the cross product as it should always be i -j k. If you mirror this with your right hand, however, you see that you do indeed have i and k positive, yet k in the negative direction. Therefore, the right hand rule is in accord with the mathematical convention of the cross product, taking into account sign.


This is true, but you can make left hand rules .... For convenience let's keep the thumb to be the direction we are looking for ... right now you are looking for i*j (middle finger times the index if we keep your hand pose) = -k but not getting -k ... but if you do j*i (Index times middle) = k and we get k

I like this image for showing the right hand rule (the curling of the fingers one) and how it relates to the order of vector multiplication 








- Vector Product of Vectors

Now, I was going to put that in my last post, but I was worried that someone may think vector A is rotating into vector B and think that this is the reason for the right hand rule in the First Post of this thread, which it is not ... The reason we happen to be able to use the right hand rule in the OP is because of the vector equation of angular momentum for a point particle:
L= r*p where L=angular momentum, r= radius, p= linear momentum
And when we look at the direction of rotation, vector p is 90 degrees "ahead" of vector r and that is how we can relate curling our fingers in the direction of rotation and getting the direction of angular momentum and relating it to the image above.

I feel like I just made things more complicated  ... If you understood what was said before this post, and got confused by this post, just ignore this post


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## FlightsOfFancy

Mr. Meepers said:


> It is not a force, it is angular momentum. ... You said you understood @FightsofFancy's explanation so I don't need to explain the right hand vs the left (it is just a sign difference) and you may know vector multiplication, but there was a larger point I wanted to make, but I guess I was not very clear lol. back, when I was an undergraduate in college, a fellow physics major asked me why a vector times a vector creates another vector. He wanted to understand how when we multiply two "physical" vector, we get a third "physical" vector. Now, luckily, I had asked myself that question before and it helped me to understand what are new laws of physics and what are things created by the language we use to define our physical terms in. ... And this is an instance where it is just normal newtonian physics, but we created a language that allows us to talk about angular motion and we are just using the same physics we used for linear motion .... In the case of why a vector times a vector equals another vector (in a vector product), it is important to realize that in these equations the vectors represent different quantities (such as radius and momentum) and this "third vector", in this case angular momentum, is just a term we defined to fit this equation ... and the reason a vector times a vector is another vector has nothing to do with physical reality at all, it (vector multiplication) is just something we created. Same goes with the direction of angular momentum. We could have defined it to be a vector that points in any direction (although, our definition may have to be dependant on how we create the coordinate system and then rotating the coordinate system would change it and that is just very messy), but we chose the right hand out of convenience.
> That may sound obvious, but I know how hazy the line between physics and mathematics can be.
> 
> 
> 
> This is true, but you can make left hand rules .... For convenience let's keep the thumb to be the direction we are looking for ... right now you are looking for i*j (middle finger times the index if we keep your hand pose) = -k but not getting -k ... but if you do j*i (Index times middle) = k and we get k
> 
> I like this image for showing the right hand rule (the curling of the fingers one) and how it relates to the order of vector multiplication
> 
> 
> 
> 
> 
> 
> 
> 
> - Vector Product of Vectors
> 
> Now, I was going to put that in my last post, but I was worried that someone may think vector A is rotating into vector B and think that this is the reason for the right hand rule in the First Post of this thread, which it is not ... The reason we happen to be able to use the right hand rule in the OP is because of the vector equation of angular momentum for a point particle:
> L= r*p where L=angular momentum, r= radius, p= linear momentum
> And when we look at the direction of rotation, vector p is 90 degrees "ahead" of vector r and that is how we can relate curling our fingers in the direction of rotation and getting the direction of angular momentum and relating it to the image above.
> 
> I feel like I just made things more complicated  ... If you understood what was said before this post, and got confused by this post, just ignore this post


I always thought the cross product to only make sense when we dealt with physical quantities. For example, when I think or torque, I can visualize why the vectors create another by simply looking at a wrench untighten a bold (if we allow the loosening to be in the k dir than the bold is in the i j; it would make no sense for another orthogonal vector to not be 'made' in this case). And since conservation of angular momentum is the absence of a torque (e.g. the torque set it in motion and no external forces are present after...e..g. spinning the bolt), the bolt in our case would continue infinitely upwards (untightening forever). I don't get how it makes sense in the purely mathematical form (but again, I caution that I am not a pure mathematician). 

I think, as you said, the multiplication is arbritary and in fact there ARE uses for the left hand rule when dealing with magnetized objects that behave somewhat erradically; for example, there are materials with negative refractive indicies that would indeed result in an i j k (positive vector) occurrence. 

I have a question; do you find that some mathematical formulations are easier understood via physics? For example, as with the cross product, I didn't understand it until physics stuff. With the dot product (how much force is in the same dir), I didn't understand it until I understood work. Do you think that physics is an elucidation of mathematical principle or that the average scientist doesn't have enough theory of math to comprehend it when abstracted from reality (For example, I don't think we could understand 13-dimensional space from a physical standpoint)?


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## bigtex1989

The cross product (which is what you use to calculate angular momentum) has 3 parts to the definition.

1. It is perpendicular to both vectors being crossed.
2. The magnitude is given by the area of parallelogram formed by the vectors being crossed.
3. The direction is in accordance with the right hand rule.

So by definition, you MUST use the right hand rule for angular momentum. The short answer is then "by convention". 

If the question then becomes, "why did mathematicians define the cross product that way?" we get into a slippery slope of why anything is defined any which way. There is no GOOD reason to prefer right-handed coordinate systems to left handed. Most things do happen to work out fine in left handed coordinate systems as long as you stay left handed. Then you can just do a transform at the end. Long story short, the answer is there needed to be a standard (much like the metric system) so some important guys in the field at the time decided that since most people are right handed, they would use that coordinate system. It makes it annoying though when you are trying to write stuff down XD


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## Mr. Meepers

FlightsOfFancy said:


> I always thought the cross product to only make sense when we dealt with physical quantities. For example, when I think or torque, I can visualize why the vectors create another by simply looking at a wrench untighten a bold (if we allow the loosening to be in the k dir than the bold is in the i j; it would make no sense for another orthogonal vector to not be 'made' in this case). And since conservation of angular momentum is the absence of a torque (e.g. the torque set it in motion and no external forces are present after...e..g. spinning the bolt), the bolt in our case would continue infinitely upwards (untightening forever). I don't get how it makes sense in the purely mathematical form (but again, I caution that I am not a pure mathematician).
> 
> I think, as you said, the multiplication is arbritary and in fact there ARE uses for the left hand rule when dealing with magnetized objects that behave somewhat erradically; for example, there are materials with negative refractive indicies that would indeed result in an i j k (positive vector) occurrence.
> 
> I have a question; do you find that some mathematical formulations are easier understood via physics? For example, as with the cross product, I didn't understand it until physics stuff. With the dot product (how much force is in the same dir), I didn't understand it until I understood work. Do you think that physics is an elucidation of mathematical principle or that the average scientist doesn't have enough theory of math to comprehend it when abstracted from reality (For example, I don't think we could understand 13-dimensional space from a physical standpoint)?


Well I suppose whether the nut is going up or down depends on which side of the bolt is facing you lol. 
Now I am a little confused what you are cross multiplying as you talked about torque and then talked about getting rid or torque.
Now, torque = Radius X Force by it's very definition (i.e. a concept defined already using the cross product), but I don't see how that physical example helps us to understand that Radius X Force = - Force X Radius ... Nevermind, I suppose I can see that if you switch their components ... but that still does not say "*why*" a vector times a vector is another vector ... and that has to do with actually creating a binary operation * over a 3 dimensional vector space (i.e. it is a function from C^6 to C^3 .... C^6 being the six dimensional complex space, in other words the cartesian product of C^3 with itself).
^^ That is actually why a vector times a vector is a vector ... it is because vector multiplication is a binary operation in a 3-dimensional space ... it has nothing to do with the physical world, but we are able to use it to describe the physical world. That is what I mean ... I suppose you could say the same thing for any word, but some words correspond to physical quantities in nature or specific characteristics ... binary operations do not. A binary operation gives meaning to a sentence ... 5+5 has a different meaning than 5*5, but "+" and "*" clearly are not real. ... Torque=Radius*Force, so the concept of torque does not give us new properties, it just redefines the properties we have (although, if you forget both the force and the radius, but remember the torque, you will lose information, but it is still useful to look at the information differently). 

As far a mathematical concept of it ... Let
Vector A = (Ai, Aj, Ak)
Vector B = (Bi, Bj, Bk)
Let the matrix
|a1 a2| be denoted as {a1,a2,a3,a4}
|a3 a4|
Det (A) = Determinant of A
AXB= Det({Aj,Ak,Bj,Bk})*i* + Det({Ak,Ai,Bk,Bi})*j* + Det({Ai,Aj,Bi,Bj})*k* 
= (Det({Aj,Ak,Bj,Bk}), Det({Ak,Ai,Bk,Bi}), Det({Ai,Aj,Bi,Bj}))

The easier way to remember it is |A|*|B|*cos(theta) in the direction of the right hand rule from A to B (the curling right hand rule)


Hmmmm ... I often find physicists' explanations of no more than 3-dimensional calculus mathematics simpler ... but there is no physics used ... it is just the informal geometrical interpretation of the concept that makes it easier to understand ... but, for the most part, I have found that understanding the mathematics on a more fundamental level can improve understand of physics and can make learning certain concepts feel redundant (I mean when two different fields of classical physics show some similarities and their new concepts from the same mathematical technique feels similar).

I think many scientist have the ability to understand abstract mathematics, but if we try to understand mathematics from physics itself, you would lose a lot of abstractness ... I mean, you probably won't have to worry about existence theorems, undefined cases, cases that don't exist in reality. ... Here is one, I don't think you can find a function that is defined over all real numbers that is continuous at only one point using physics ... And there are areas of mathematics that are applicable to fields other than physics or not applicable at all.
Now, I think a physicist's informal geometric approach to mathematics can greatly aid in the understanding of parts of mathematics, even if it does lack rigour (mathematicians need more visual concept pictures lol)


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## DemonD

Mr. Meepers said:


> * *
> 
> 
> 
> 
> It is not a force, it is angular momentum. ... You said you understood @FightsofFancy's explanation so I don't need to explain the right hand vs the left (it is just a sign difference) and you may know vector multiplication, but there was a larger point I wanted to make, but I guess I was not very clear lol. back, when I was an undergraduate in college, a fellow physics major asked me why a vector times a vector creates another vector. He wanted to understand how when we multiply two "physical" vector, we get a third "physical" vector. Now, luckily, I had asked myself that question before and it helped me to understand what are new laws of physics and what are things created by the language we use to define our physical terms in. ... And this is an instance where it is just normal newtonian physics, but we created a language that allows us to talk about angular motion and we are just using the same physics we used for linear motion .... In the case of why a vector times a vector equals another vector (in a vector product), it is important to realize that in these equations the vectors represent different quantities (such as radius and momentum) and this "third vector", in this case angular momentum, is just a term we defined to fit this equation ... and the reason a vector times a vector is another vector has nothing to do with physical reality at all, it (vector multiplication) is just something we created. Same goes with the direction of angular momentum. We could have defined it to be a vector that points in any direction (although, our definition may have to be dependant on how we create the coordinate system and then rotating the coordinate system would change it and that is just very messy), but we chose the right hand out of convenience.
> That may sound obvious, but I know how hazy the line between physics and mathematics can be.
> 
> 
> 
> This is true, but you can make left hand rules .... For convenience let's keep the thumb to be the direction we are looking for ... right now you are looking for i*j (middle finger times the index if we keep your hand pose) = -k but not getting -k ... but if you do j*i (Index times middle) = k and we get k
> 
> I like this image for showing the right hand rule (the curling of the fingers one) and how it relates to the order of vector multiplication
> 
> 
> 
> 
> 
> 
> 
> 
> - Vector Product of Vectors
> 
> Now, I was going to put that in my last post, but I was worried that someone may think vector A is rotating into vector B and think that this is the reason for the right hand rule in the First Post of this thread, which it is not ... The reason we happen to be able to use the right hand rule in the OP is because of the vector equation of angular momentum for a point particle:
> L= r*p where L=angular momentum, r= radius, p= linear momentum
> And when we look at the direction of rotation, vector p is 90 degrees "ahead" of vector r and that is how we can relate curling our fingers in the direction of rotation and getting the direction of angular momentum and relating it to the image above.
> 
> I feel like I just made things more complicated  ... If you understood what was said before this post, and got confused by this post, just ignore this post


I may have been a bit liberal with the use of the word "Understand".

What I "understood" was ++=+ is bad while ++=- is alright. What I got out of that was that the reason you can't have a "left hand rule" is because it is blocked mathematically. 

I barely know what a vector is, much less how to count them.

If it is not a force then why does it have a physical effect on things?

I thought i'd show you where the question comes from. I got addicted to this channel, but he only explains half of what I need know in his videos. 

* *












That vid gave me more questions than clarity. 



Mr. Meepers said:


> That may sound obvious...


Now you're just making fun of me...:sad:


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## FlightsOfFancy

DemonD said:


> I may have been a bit liberal with the use of the word "Understand".
> 
> What I "understood" was ++=+ is bad while ++=- is alright. What I got out of that was that the reason you can't have a "left hand rule" is because it is blocked mathematically.
> 
> I barely know what a vector is, much less how to count them.
> 
> If it is not a force then why does it have a physical effect on things?
> 
> I thought i'd show you where the question comes from. I got addicted to this channel, but he only explains half of what I need know in his videos.
> 
> * *
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> That vid gave me more questions than clarity.
> 
> 
> 
> Now you're just making fun of me...:sad:


As @MrMeepers said, my explanation of torque is kind of dirty.

But imagine the bolt on an xy plane; when you tighten or untigthen it, it moves in a third direction ,perpendicular to the plane. It's a physical occurence that just so happens to be in line with the cross product, as the cross product guarantees ortonormal vectors. 

With angular momentum, torque is basically the 'force' 

T = Fxr 
or 
T= Fsin(thetha)r 

Basically, this is saying "what is ortogonal to these vectors; that must be the vector and dir in which this moves" which is correct. When you tighten a bolt, it moves in a direction orthogonal to F and r. Since it's ortogonal, why not give it another name...torque. The sin usage is saying "I need something that has a 90 degree overlap," again because when you tigthen the bold, the direction it moves is 90 degrees away from the xy surface. 

With angular momentum, it is essentially the same idea because Torque is what gives rise to angular momentum. Instead of F we now use p 

L = p(mv) *r 

this is still saying the same thing. When I spin like a die on its side (I apply a torque, analogous to a force) then let go (no torque now on the spinning die, analogously, no forces), the die stays upright because of conservation of momentum. But what you did was spin it in the x/y plane, yet it is still spinning 'upright'; hence, you have "created" another vector quantity 90 degrees away from the surface. Physicist said well shit, that's just the cross product, and the pure mathematicians said "you're welcome bitches". Life was merry after that.


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## Mr. Meepers

DemonD said:


> I may have been a bit liberal with the use of the word "Understand".
> 
> What I "understood" was ++=+ is bad while ++=- is alright. What I got out of that was that the reason you can't have a "left hand rule" is because it is blocked mathematically.
> 
> I barely know what a vector is, much less how to count them.
> 
> If it is not a force then why does it have a physical effect on things?
> 
> I thought i'd show you where the question comes from. I got addicted to this channel, but he only explains half of what I need know in his videos.
> 
> * *
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> That vid gave me more questions than clarity.
> 
> 
> 
> Now you're just making fun of me...:sad:


No, I was not making fun of you ... I was being more self conscious lol ... And I assumed you had a higher level understanding in math and physics.

Now, there are forces involved, but angular momentum is not a force ... but a force applied at a certain radius away can affect angular momentum. ... Since you don't know vectors very well, I will just say that the right hand rule works because we defined angular momentum to use the right hand rule.

Now, what is force? Net Force is the mass of an object times the acceleration (F=m*a) (Note: the acceleration has a direction attached to it)... A net force is left over force after all the other forces cancelled each other out. 
Newton's 3rd law is basically that an object will keep its momentum (mass times velocity and velocity is speed with a direction) unless acted upon by an outside net force

Now, think about the Earth revolving around the sun (I believe it happens in an oval shape, but you can image in moving in a perfect circle with a still Sun for simplicity). ... It is always changing direction to revolve around the Sun, so the velocity is changing, which means the momentum is changing ... which means there is a Force acting on it ... kind of like a tether making sure the constant *speed* object can't leave ... so it is an inward force ... This is true for all rotating bodies ... every molecule in that bike tire is feeling an inward force ... but the force is not so great as to bring the molecules closer to the center of the wheel, nor is it so weak that the molecules fly off from their speed (okay, in real life some would, but we are going to assume we can't deform the wheel for simplicity's sake).
Now the angular momentum for each bike molecule is the distance away from the center times the speed*(1)* of the molecule and we use the right hand rule for information about the tilt of the rotational axis and whether or not the spin is clockwise or counter clockwise (... it always looks counter clockwise when looking down at the thumb).
And the angular momentum for the whole wheel is just the sum of all the angular momentums of all the molecules in the bike wheel.

Now, as we can see, angular momentum is not a force, but it is a very useful definition. ... Now, there are two ways to do that ... slowing down the rotation (requires applying a force(or forces) away from the center that opposes the rotation) or tilting the axis of rotation (which adds to the change of direction of all the molecules and requires the force (or forces) to be felt away from the center of gravity) ... basically to change the angular momentum (the direction of the axis of rotation or how fast it is rotating), you need to introduce a torque.

Now, gravity adds some torque to the wheel (the origin is where the rope meets the wheel and gravity is acting downward) ... but the spinning wheel already has angular momentum perpendicular to the torque applied by gravity and a perpendicular torque can only change the axis of rotation so the wheel's axis of rotation is constantly changing.

Edit: The word speed should be momentum


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## DemonD

Mr. Meepers said:


> No, I was not making fun of you ... I was being more self conscious lol ... And I assumed you had a higher level understanding in math and physics.


No. 

In school I understood a lot of the basics in physics. Like F=m*a that you mentioned, friction, gravitational pull and even waves to a degree. But as soon as things started spinning I became that stupid kid in class. And that still holds true, my brain just refuses to understand it.

My suspicion is that torque is what fries my brain

I sort of understand centrifugal(or petal, I have no idea) force. One force goes in one direction, but the tether, or wall, forces a direction change for that force. And that's why it seems like it pushes outward.

If I understood the video right, the wheel has two sets of angular momentum. First the one for the wheel that keeps it upright, and the second that comes from gravity pulling the wheel down causing an angular momentum perpendicular to the first one. 

But, if angular momentum is not a force, then why does the wheel care? Why does it stay upright and why does it start circling?

If the tilt had been everything the thumb denoted then I would've had no problems, but now it is the "angular momentum", which I don't even know what it is anymore.

I understand the concept of the angular momentum being the distance from the center and the speed(didn't know that before you told me though). But why does it fuck off in that orthogonal direction?

@_FlightsOfFancy_ The reason I can't think of it in terms of bolts is because they have grooves that physically force them to go in one direction or the other depending on the torques direction.


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## FlightsOfFancy

DemonD said:


> No.
> 
> In school I understood a lot of the basics in physics. Like F=m*a that you mentioned, friction, gravitational pull and even waves to a degree. But as soon as things started spinning I became that stupid kid in class. And that still holds true, my brain just refuses to understand it.
> 
> My suspicion is that torque is what fries my brain
> 
> I sort of understand centrifugal(or petal, I have no idea) force. One force goes in one direction, but the tether, or wall, forces a direction change for that force. And that's why it seems like it pushes outward.
> 
> If I understood the video right, the wheel has two sets of angular momentum. First the one for the wheel that keeps it upright, and the second that comes from gravity pulling the wheel down causing an angular momentum perpendicular to the first one.
> 
> But, if angular momentum is not a force, then why does the wheel care? Why does it stay upright and why does it start circling?
> 
> If the tilt had been everything the thumb denoted then I would've had no problems, but now it is the "angular momentum", which I don't even know what it is anymore.
> 
> I understand the concept of the angular momentum being the distance from the center and the speed(didn't know that before you told me though). But why does it fuck off in that orthogonal direction?
> 
> @_FlightsOfFancy_ *The reason I can't think of it in terms of bolts is because they have grooves that physically force them to go in one direction or the other depending on the torques direction.*


then you understand it :ninja:. If it were the grooves that dictated solely the movement, then it would be unidirectional.

Alright, try understanding it from a spinning top:






We apply a torque in the x/y direction to get it to spin, it stays upright.. Why? There are forces that would dictate that it should fall over (it is a heavy object). It doesn't because despite exterting spin force (torque) in the xy dir (you didn't add a pull or push force on the z), there is some quantity in the z direction. Since it is in the z direction, it must be a new vector, orthonormal to the xy..hence angular momentum conservation (in that third vector). The reason it succumbs is because of the distribution of weight/friction. The more the weight is distributed, the less likely it wants to spin (this is why skaters do that thing where they pull in and remain perfectly upright despite their high speeds).


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## DemonD

FlightsOfFancy said:


> then you understand it :ninja:. If it were the grooves that dictated solely the movement, then it would be unidirectional.
> 
> Alright, try understanding it from a spinning top:
> 
> 
> * *
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> We apply a torque in the x/y direction to get it to spin, it stays upright.. Why? There are forces that would dictate that it should fall over (it is a heavy object). It doesn't because despite exterting spin force (torque) in the xy dir (you didn't add a pull or push force on the z), there is some quantity in the z direction. Since it is in the z direction, it must be a new vector, orthonormal to the xy..hence angular momentum conservation (in that third vector). The reason it succumbs is because of the distribution of weight/friction. The more the weight is distributed, the less likely it wants to spin (this is why skaters do that thing where they pull in and remain perfectly upright despite their high speeds).


Something like the wheel doesn't have grooves, and it still behaves like that?

I have no problems understanding why the spinny-top stays up while spinning, as it has a "torque momentum" so to say. The mass goes round and round rather than tilting over. 

Assuming the one in the vid spins clockwise, that should mean the angular momentum is down, against the table. Here it makes sense that it is not a force, but it makes no sense that the momentum is down, rather than up. 

Also, imagine that you tilt it 90 degrees, hang the bottom end from a string or something, and spin it. Now the angular momentum goes sideways and the spinny top will start circling like the bicycle wheel. Why? Furthermore, if you were to only tilt it 45 degrees instead of 90 and set a spin, would it still stay in that angle(as long as it spins ofc)?


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## FlightsOfFancy

DemonD said:


> Something like the wheel doesn't have grooves, and it still behaves like that?
> 
> I have problems understanding why the spinny-top stays up while spinning, as it has a "torque momentum" so to say. The mass goes round and round rather than tilting over.
> 
> Assuming the one in the vid spins clockwise, that should mean the angular momentum is down, against the table. Here it makes sense that it is not a force, but it makes no sense that the momentum is down, rather than up.
> 
> Also, imagine that you tilt it 90 degrees, hang the bottom end from a string or something, and spin it. Now the angular momentum goes sideways and the spinny top will start circling like the bicycle wheel. Why? Furthermore, if you were to only tilt it 45 degrees instead of 90 and set a spin, would it still stay in that angle(as long as it spins ofc)?





> Something like the wheel doesn't have grooves, and it still behaves like that?


Yes.



> I have problems understanding why the spinny-top stays up while spinning, as it has a "torque momentum" so to say. The mass goes round and round rather than tilting over.


It does EVENTUALLY tilt over. Remember, there are other forces acting on the mass, namely gravity. The fact that it remains upright means that there is something keeping it such, staving off the gravitational effect that would have it fall over. You must agree that there is something in which is holding this against gravity? Since you spun it in the x y dir, the only way it would stay upright in the z dir is if there was some component force doing such. Indeed there is, the 'third' vector from the cross product explains this. If there were no such force in this direction, the cap would not stay upright as there would be no force to 'fight' for force of gravity.



> Assuming the one in the vid spins clockwise, that should mean the angular momentum is down, against the table. Here it makes sense that it is not a force, but it makes no sense that the momentum is down, rather than up.


It doesn't matter if its down against the table; the table is providing a normal force to keep it from 'going through the table'. This normal force begets friction, which is another force that slows down the spinning. The fact that it is opposed to the table in your example is a reason why the spinning would get slower and slower (frictional dissipation). 



> Also, imagine that you tilt it 90 degrees, hang the bottom end from a string or something, and spin it. Now the angular momentum goes sideways and the spinny top will start circling like the bicycle wheel. Why?


Because it does! 




Note, as long as he keeps it centered (with the wheel straight), he remains still. This is because he is countering the angular momentum force with his hands on both sides. 

However, when he turns it, he turns. Therefore, there must have been some force to set this into gear correct? The reason why is he is 'fighting' against the angular momentum by tilting it (he is providing a force to change the angular momentum). This is analogous to regular momentum...if we disturb it we get a resultant force as Force = change in momentum/time. By Newton's Third law, we get an opposite reaction, which is seen in him spinning. 

NOTE: This is NOT solely what keeps bikes in line; an experiment showed it had more to do with fictitious forces that are easiest seen in Lagrangian mechanics (not going into that here).



> Furthermore, if you were to only tilt it 45 degrees instead of 90 and set a spin, would it still stay in that angle(as long as it spins ofc)?


Yeap, I think you are having trouble because you are considering angular momentum to be distinct from linear momentum. It isn't. It's another type of momentum that is dependent on angle. Assuming no other forces were acting on this at 45 deg, yes it would spin at this angle. The reason why you wouldn't see this happen IRL is because this angle makes for greater gravitational opposition (mgsin(45) + NormalForceFriction) than the slowing down being due to Normal force friction (mgsin(90) + NormalForceFriction).


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## DemonD

FlightsOfFancy said:


> It does EVENTUALLY tilt over. Remember, there are other forces acting on the mass, namely gravity. The fact that it remains upright means that there is something keeping it such, staving off the gravitational effect that would have it fall over. You must agree that there is something in which is holding this against gravity? Since you spun it in the x y dir, the only way it would stay upright in the z dir is if there was some component force doing such. Indeed there is, the 'third' vector from the cross product explains this. If there were no such force in this direction, the cap would not stay upright as there would be no force to 'fight' for force of gravity.


Oops, I missed a crucial 'no' in my post.


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## FlightsOfFancy

DemonD said:


> Oops, I missed a crucial 'no' in my post.


then you understand the RHR 

the third vector comes in because it's normal to the spinning surface..

put your right hand in the air with the index pointing left, middle finger towards you.

your index finger is the radius of the object.

your middle finger is the velocity of spin 

your thump is the angular momemtum created, which is what keeps the item upwards. 

As for friction, you can imagine if you put something under your hand and spun it around like this, it would slow down of create heat. Same issue with the spinning thing; it slows down due to the friction, even though it doesn't go 'through' the table. If you could make the spinning cap infinitely thin and spin it infinitely fast, it would generate enough angular momentum to 'drill' through the Earth, provided the Earth cannot counterbalance the forces.


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## DemonD

FlightsOfFancy said:


> then you understand the RHR
> 
> the third vector comes in because it's normal to the spinning surface..
> 
> put your right hand in the air with the index pointing left, middle finger towards you.
> 
> your index finger is the radius of the object.
> 
> your middle finger is the velocity of spin
> 
> your thump is the angular momemtum created, which is what keeps the item upwards.
> 
> As for friction, you can imagine if you put something under your hand and spun it around like this, it would slow down of create heat. Same issue with the spinning thing; it slows down due to the friction, even though it doesn't go 'through' the table. If you could make the spinning cap infinitely thin and spin it infinitely fast, it would generate enough angular momentum to 'drill' through the Earth, provided the Earth cannot counterbalance the forces.


I still don't under stand why not the opposite direction, unless it is the mathematical reason stated earlier in the thread. I understand why it is on that axis, but not why the specific direction over the other.

I did understand your explanation on why the dude started spinning though. You should be a teacher.


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## FlightsOfFancy

DemonD said:


> I still don't under stand why not the opposite direction, unless it is the mathematical reason stated earlier in the thread.
> 
> I did understand your explanation on why the dude started spinning though. You should be a teacher.





> I still don't under stand why not the opposite direction, unless it is the mathematical reason stated earlier in the thread.


Wait why the right hand? Because it's the one that's aligned with physical assessment AND mathematical convention.

The reason why it's hard to see with angular momentum is because it's related more to the curl of the vector than the explicit cross product. The curl means you treat the x/y as a contiguous surface that spins, so the relation to the directions of x y z is irrelevant. To get a better picture of why the RHR works, you should investigate electromagnetism

F= qv xB 

where B is the magnetic field. 

Do the exercise I mentioned before and you will see this only works with the Right Hand. Since all of the vectors here can be treated discretely, we don't need worry ourselves with the ambiguity of the curl. 



> I did understand your explanation on why the dude started spinning though. You should be a teacher.


Umm thankx; I like helping people understand things. It also reinforces it for me. I wanted to be a Physical Chemist and profess but I sucked at life and decided to turn it down cause I didn't want to wait 5 yrs


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## DemonD

So what's gyroscopic precession?


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## FlightsOfFancy

DemonD said:


> So what's gyroscopic precession?


wrong explanation


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## Mr. Meepers

DemonD said:


> No.
> 
> In school I understood a lot of the basics in physics. Like F=m*a that you mentioned, friction, gravitational pull and even waves to a degree. But as soon as things started spinning I became that stupid kid in class. And that still holds true, my brain just refuses to understand it.
> 
> My suspicion is that torque is what fries my brain
> 
> I sort of understand centrifugal(or petal, I have no idea) force. One force goes in one direction, but the tether, or wall, forces a direction change for that force. And that's why it seems like it pushes outward.
> 
> If I understood the video right, the wheel has two sets of angular momentum. First the one for the wheel that keeps it upright, and the second that comes from gravity pulling the wheel down causing an angular momentum perpendicular to the first one.
> 
> But, if angular momentum is not a force, then why does the wheel care? Why does it stay upright and why does it start circling?
> 
> If the tilt had been everything the thumb denoted then I would've had no problems, but now it is the "angular momentum", which I don't even know what it is anymore.
> 
> I understand the concept of the angular momentum being the distance from the center and the speed(didn't know that before you told me though). But why does it fuck off in that orthogonal direction?
> 
> @_FlightsOfFancy_ The reason I can't think of it in terms of bolts is because they have grooves that physically force them to go in one direction or the other depending on the torques direction.


Oops I meant angular momentum is the radius time the linear momentum (for just circular motion), not just speed ... sorry about that. I was a bit sleepy when I wrote it. ... The right hand rule is just a convention to tell us what plane the particle is rotating on (if we are just looking at a single particle) and the direction of the spin ... In other words, it is a summary of the linear momentums and positions

So angular momentum is related to momentum and Torque is related to force (a force applied away from the center and just the component of that force that is perpendicular to the radius) ... Just like angular momentum is a summary of the linear momentum that is perpendicular to the radius and how far away from the radius it is, torque is a summary of the forces perpendicular to the radius at a certain radius away ... and we defined both to use the right hand rule (for consistency)
As I said in my last post, a change in angular momentum requires a torque



Meepster said:


> Now, as we can see, angular momentum is not a force, but it is a very useful definition. ... Now, there are two ways to do that ... slowing down the rotation (requires applying a force(or forces) away from the center that opposes the rotation) or tilting the axis of rotation (which adds to the change of direction of all the molecules and requires the force (or forces) to be felt away from the center of gravity) ... basically to change the angular momentum (the direction of the axis of rotation or how fast it is rotating), you need to introduce a torque.


We could even redefine torque to mean a change in angular momentum and have it stay consistent with the definition (a force is the change in linear momentum and the torque is the radius times force = the radius times the change in linear momentum that is unrelated to the inward, centripetal, force).
If that last statement is too confusing then forget that and think about this statement instead. If we apply a torque, we change the angular momentum, either by changing the amount of revolutions per second or changing the the axis it is rotating on

Let's re-watch this video:





So there is and angular momentum and a constant torque that is making the whole system spin around the rope ... remember, torque is like force

If the vocabulary is too hard let's try another approach


















Unfortunately I can't find a slow moving wheel gif  ... Picture would help so much with the explanation

So, imagine a counter-clockwise spinning wheel. ... Now imagine we push down on the axle at one end (like the effect of gravity and push up at the other end of the axle (the rope) .... These are two forces trying to create a rotation perpendicular to the rotation of the wheel ... so you are adding a perpendicular force to every particle on that wheel, except for the ones at the second axis of rotation at that moment in time.

I haven't eaten yet, so I'll come back and try to finish the explanation, but it is tough to do without the definitions of torque and angular momentum (there is a lot going on and torque and angular momentum summarizes what is going on nicely ... but there is a lot of linear momentums of the particles, the center seeking forces, ... the effect of gravity and the rope and how the component vectors create a torque and how the torque effects which is a changing force to the rotating particles affects the direction of movement of the particles, ... I need to get something to eat lol)

That being said, as the video below says, it is not intuitive and I notice you are getting hung up on the language and asking why the language is the way it is and looking for a physical explanation, when it is just the language (ie. man made), but having a grasp of the language can help you grasp the basic idea of what is going on and how it related to force and momentum






Edit: It occurs to me if we look at the two particles (for an infinitely thin bike) from the second axis of rotation (the one on the top and the one on the bottom of the wheel), the perpendicular forces created by gravity, the rope, and not allowing this object to deform (transferring that force rotationally) give perpendicular forces to the two particles and perpendicular forces try to make rotational motion ... looking down on the wheel the top particle wants to move along a circle away from the rope (where it's velocity is tangent to the circle) in one direction (clockwise, I think, I can't remember which way the wheel was rotating) and the bottom on wants to move along a circle closer to the rope going the same direction (also clockwise) and this causes the original axis of rotation to rotate in the movement of the procession ... (why it moves in a circle instead of spinning along two axis, theoretically has to do with fixing the ropes position and the second spin combined ... I suppose I could have mentioned that in an earlier post lol ... and not theoretically, balancing may play a role as well)


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## DemonD

@Mr.Meepers Yes, terminology appears to be one of the biggest obstacles when it comes to learning physics for me.

I thought from the second vid that torque might've caused a force forwards, but when I googled torque direction they just said that the direction of the torque is the same as the direction for angular momentum(right hand rule).


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## Mr. Meepers

DemonD said:


> @_Mr._Meepers Yes, terminology appears to be one of the biggest obstacles when it comes to learning physics for me.
> 
> I thought from the second vid that torque might've caused a force forwards, but when I googled torque direction they just said that the direction of the torque is the same as the direction for angular momentum(right hand rule).


The site is slightly wrong, The direction of the torque is the same as the direction of the change in angular momentum

Anyway, the second half of my post, I tried to explain it without using torque or angular momentum, but it might be hard to understand without pictures (and I don't have any pictures to show you what I mean)


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## DemonD

Mr. Meepers said:


> The site is slightly wrong, The direction of the torque is the same as the direction of the change in angular momentum
> 
> Anyway, the second half of my post, I tried to explain it without using torque or angular momentum, but it might be hard to understand without pictures (and I don't have any pictures to show you what I mean)


I'll disregard the site then.

Say there's a spinny-top. It is pinning in one place CCW making the angular momentum being directed upwards. As long as it spins, it remains upwards. Does this mean there is no direction of torque?

Also, this vid explains angular momentum, but there's no direction mentioned for it. I had no problems following this. 


* *












For the robot arm, the angular momentum would be towards us, correct?


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## Mr. Meepers

DemonD said:


> I'll disregard the site then.
> 
> Say there's a spinny-top. It is pinning in one place CCW making the angular momentum being directed upwards. As long as it spins, it remains upwards. Does this mean there is no direction of torque?
> 
> Also, this vid explains angular momentum, but there's no direction mentioned for it. I had no problems following this.
> 
> 
> * *
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> For the robot arm, the angular momentum would be towards us, correct?


A top like a dreidel? ... If the angular momentum is pointing straight up and constant, then, I imagine the center of mass would be on the axis of rotation and no torque would be applied gravity and there would be no torque on the system. ... If the top is tilted, then both gravity and the normal force from the surface it is on would create a torque and we would see something similar to the bike wheel demonstration.

That video was a good video, but he did not really talk about direction because he was dealing with a two dimensional space and even though he showed the vector equations, he was using scalars lol (but the meaning, in this case was equivalent, so no big deal)

Did he say whether the mass was spinning clockwise or counter-clockwise? ... If it was spinning counter-clockwise then yes, the angular momentum is towards us


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## DemonD

Mr. Meepers said:


> A top like a dreidel? ... If the angular momentum is pointing straight up and constant, then, I imagine the center of mass would be on the axis of rotation and no torque would be applied gravity and there would be no torque on the system. ... If the top is tilted, then both gravity and the normal force from the surface it is on would create a torque and we would see something similar to the bike wheel demonstration.
> 
> That video was a good video, but he did not really talk about direction because he was dealing with a two dimensional space and even though he showed the vector equations, he was using scalars lol (but the meaning, in this case was equivalent, so no big deal)
> 
> Did he say whether the mass was spinning clockwise or counter-clockwise? ... If it was spinning counter-clockwise then yes, the angular momentum is towards us


Ok, but he was correct? 

Not sure what a dreidel is. Here is a pic of a spinning top.

* *















So if the spinning top starts tilting up to 90 degrees, the torque direction is straight down at the end of the tilt? Or, the toque direction is the tangent of the motion of the angular momentum?

In your video in the previous post the professor said the angular momentum is chasing the torque. The torque was forward, but how is that, if the wheel goes round and round? And, does that differ from the direction of the torque I was asking about earlier? 

And why did it speed up when weights were added? Is that due to Torque=M*a*r ? where r is the length of the "arm".


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## Mr. Meepers

DemonD said:


> Ok, but he was correct?
> 
> Not sure what a dreidel is. Here is a pic of a spinning top.
> 
> * *
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> So if the spinning top starts tilting up to 90 degrees, the torque direction is straight down at the end of the tilt? Or, the toque direction is the tangent of the motion of the angular momentum?
> 
> In your video in the previous post the professor said the angular momentum is chasing the torque. The torque was forward, but how is that, if the wheel goes round and round? And, does that differ from the direction of the torque I was asking about earlier?
> 
> And why did it speed up when weights were added? Is that due to Torque=M*a*r ? where r is the length of the "arm".


Yes it was correct

Okay, dreidels and tops are similar then

* *

















So, if the tops axis of rotation is parallel to the ground, the force of gravity is towards the ground, then the torque from gravity is parallel to the ground, but perpendicular to the axis of rotation (it is just the spinning bicycle wheel example).
"Or, the toque direction is the tangent of the motion of the angular momentum?" ... I'm not sure what you meant here but the angular momentum will be spinning around, if you create a circle where the angular momentum is the radius vector, then, for this particular case, the torque vector will always be tangent to that circle.

IF by earlier, you mean the paragraph before it, no, the concepts are the same, you justed messed up the direction of the torque in that example (in these examples, the torque is parallel to the ground and perpendicular to the angular momentum)

Yes, It was due to increasing the torque (adding a mass, a radius r away in a gravitational field)


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