How can angular displacement be larger than 2pi?

  • #1
adjurovich
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Angular displacement is defined as the angle between initial and final radius vector of some arbitrary point on an object undergoing rotation. I’ve seen that some problems include angular displacement bigger than 2 ##\pi## radians. Also, I would note this example:

Let us start with definition of angular velocity:

##\vec{\omega} = \dfrac{d\vec{\theta}}{dt}##

##\int_\vec{\theta_1}^\vec{\theta_2}d \vec{\theta} = \int_{0}^{t} \vec{\omega}dt##

Let’s say that angular velocity is given by linear function: ##\vec{\omega} = \vec{\omega_0} + \vec{\alpha} t##

By integrating, we obtain the final equation:

##\Delta{\vec{\theta}} = \vec{\omega_0}t + \dfrac{1}{2} \vec{\alpha} t^2##

This equation poses no limitations to values of angular position nor angular displacement. Hence it could result in angular displacement bigger than 2 ##\pi## radians. What am I missing?
 
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  • #2
Why do you have a problem w/ angular displacement > 2PI ? Consider a watch spring that is would up 10 x 2PI radians (that is, 10 full turns of the spring stem). Is that a problem? If the spring tension is proportional to the number of turns it's wound, would you like like to reduce 10 x 2PI to zero?
 
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  • #3
phinds said:
Why do you have a problem w/ angular displacement > 2PI ? Consider a watch spring that is would up 10 x 2PI radians (that is, 10 full turns of the spring stem). Is that a problem? If the spring tension is proportional to the number of turns it's wound, would you like like to reduce 10 x 2PI to zero?
If we think of simple case of spinning wheel, would it make sense? Indeed you would reduce 20 radians to zero because angle between initial and final position vector is, indeed, zero? An arbitrary point on wheel would return to its initial position after every full rotation. Angular distance however should still add up tho
 
  • #4
adjurovich said:
If we think of simple case of spinning wheel, would it make sense? Indeed you would reduce 20 radians to zero because angle between initial and final position vector is, indeed, zero? An arbitrary point on wheel would return to its initial position after every full rotation. Angular distance however should still add up tho
Obviously, BUT ... you have totally not answered my question.
 
  • #5
phinds said:
Obviously, BUT ... you have totally not answered my question.
In the specific case you mentioned, I wouldn’t reduce it. What I am trying to say is that by being driven by the logic, I could (probably) solve any textbook problem. However, I am more interested about the general case
 
  • #6
adjurovich said:
... being driven by the logic, I could (probably) solve any textbook problem. However, I am more interested about the general case
What is more general than being able to solve any problem?
 
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  • #7
adjurovich said:
I am more interested about the general case
Have we not shown in this thread that there IS no "general case"?
 
  • #8
phinds said:
Have we not shown in this thread that there IS no "general case"?
What I got from this thread is an explanation that angular displacement can have values “whichever you need for the problem”. At least it seems like that, maybe I missed the point? If there wasn’t the “general case” for such basic stuff, learning physics would be half memorizing or let’s say more precisely learning each case and interpreting it. This would lead to losing any rigor if I am correct? The short version of my question would be: how does it makes sense for angular displacement to be larger than 2 pi on “spinning wheel” if the arbitrary observed point came back to its initial position? I will also mention that in textbook I used it was stated that angular displacement cannot be larger than 2 pi and later on the author simply ignored the statement and in solution of one of the problems it was larger than 2 pi
 
  • #9
A.T. said:
What is more general than being able to solve any problem?
If someone hypothetically asked you what is angular displacement and which values can it have, what would you answer? Doing problems and understanding something deeply aren’t really the same things, I guess? Some textbooks are more math based and learning the important equations is probably enough for a good number of problems.
 
  • #10
adjurovich said:
I will also mention that in textbook I used it was stated that angular displacement cannot be larger than 2 pi

Well, it's wrong. What textbook are you using? It's hard for me to imagine that anyone who has done the very basic problems on rotational kinematics could even think that it cannot be larger than 2pi. Let alone, someone who writes a textbook...
 
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  • #11
You would not want to constrain the values from 0 to 2π when calculating work due to torque.
 
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  • #12
adjurovich said:
What am I missing?
You are missing the context. You have a clock on the wall and it tells you 'the time', modulo 12 hours. We can easily translate that to a 24hour day (after late primary school) or, by adding the number of revs, to the number of days / hours / years / months. to suit. The angle between the hands and 12.00 will be somewhere between zero and 2π but we can cope.
What sort of paradox are you looking for?

I had a friend who wrote to the Times, saying "In three days I shall be 1Gigasecond old, How should I celebrate?". Data presented in a strange way but the letter got published.
 
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  • #13
weirdoguy said:
Well, it's wrong. What textbook are you using? It's hard for me to imagine that anyone who has done the very basic problems on rotational kinematics could even think that it cannot be larger than 2pi. Let alone, someone who writes a textbook...
That’s why I made this thread. I’ve seen in some problems and equations that it can (it has to) to be larger than 2 pi. The only thing I still don’t understand is how would you explain it geometrically. Also, how do you differ angular displacement from angular distance? It seems logical to me to say that angular displacement decreases if object starts spinning counterclockwise (after spinning clockwise), but angular distance keeps increasing in that case.
 
  • #14
I think this hinges on whether you are thinking of "angular displacement" as "radians turned" or "radians away from the origin". If it's "radians turned" then the answer is clearly that any value, positive or negative, is valid. If it's "radians away from the origin" then I would use ##\pm\pi## as the range, personally, but I guess other conventions aren't unreasonable.

As always, if there are multiple conventions in use, you just have to be careful about which one a given text is using.
 
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  • #15
adjurovich said:
Angular displacement is defined as the angle between initial and final radius vector of some arbitrary point on an object undergoing rotation. I’ve seen that some problems include angular displacement bigger than 2 ##\pi## radians.
...
This equation poses no limitations to values of angular position nor angular displacement. Hence it could result in angular displacement bigger than 2 ##\pi## radians. What am I missing?
Since rotation of a disc is a cyclic event, perhaps it is useful to consider the frequency, which will be a constant out of the integral.
For problems involving angular velocities and accelerations, always use ω=2πf.

 
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  • #16
adjurovich said:
That’s why I made this thread. I’ve seen in some problems and equations that it can (it has to) to be larger than 2 pi. The only thing I still don’t understand is how would you explain it geometrically.
Just comparing linear and angular measurements:

What we normally measure as surface-of-Earth-distances are actually lengths of arcs, but we still call them straight lines.
"The accepted measurement of the Earth's circumference today is about 24,855 miles."
https://oceanservice.noaa.gov/educa...of the,his calculations were quite remarkable.

In order to calculate the "linear" velocity of a vehicle rolling over that Earth's circumference, we need to know how many times a full go-around is achieved in certain period of time (the frequency with which the cycllic event happens).

If it takes exactly one hour, our vehicle has had a linear velocity of 24 855 miles/hour.
If it takes exactly half an hour, our vehicle has had a linear velocity of 49 710 miles/hour.

adjurovich said:
Also, how do you differ angular displacement from angular distance? It seems logical to me to say that angular displacement decreases if object starts spinning counterclockwise (after spinning clockwise), but angular distance keeps increasing in that case.
Again, comparing linear and angular measurements:
Please, revisit the concepts of linear distances and displacements. https://www.physicsclassroom.com/class/1DKin/Lesson-1/Distance-and-Displacement

😎
 
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  • #17
Ibix said:
I think this hinges on whether you are thinking of "angular displacement" as "radians turned" or "radians away from the origin".
It's just a matter of modulo arithmetic and being aware of the nature of the dimensions you're dealing with. The only problem can arise due to sloppy use of terms or poor communication - like I wrote earlier; for a full definition of the quantities involved, you rely on context or a formal declaration of variable types. I have to admit that I don't think I've come across a universal shorthand for angular displacement in radians which solves the OP's prroblem..

"Revs" plus a remainder makes life more straightforward. Thank god for Engineers!
 
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  • #18
Often you will be responsible for providing the context and communicating what is important or interpreting what assumptions others are making. There isn't a universally correct answer to some things like this. That is why it is important not just to have an equation, you must also understand where it came from; what it means; how you should use it for your situation.
 
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  • #19
Ibix said:
As always, if there are multiple conventions in use, you just have to be careful about which one a given text is using.
When doing path analyses in the complex plane of course the technique is to foliate the planes and make cuts to topologically isolate regions. From these are constructed the Riemann surfaces . I think this removes the ambiguity but I am not a heavy or facile practitioner of such techniques.
 
  • #20
sophiecentaur said:
It's just a matter of modulo arithmetic and being aware of the nature of the dimensions you're dealing with. The only problem can arise due to sloppy use of terms or poor communication - like I wrote earlier; for a full definition of the quantities involved, you rely on context or a formal declaration of variable types. I have to admit that I don't think I've come across a universal shorthand for angular displacement in radians which solves the OP's prroblem..

"Revs" plus a remainder makes life more straightforward. Thank god for Engineers!
I understand what you’re trying to say. I have one more subquestion: if a rotating body has positive z-projection of angular position of let’s say ##20 \pi ## radians and it turns in the counter-clockwise direction for ##20 \pi## radians, If I am correct its angular displacement is zero. Should also the work done be zero in that case, considering the torque acted on the body?
 
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  • #21
adjurovich said:
Should also the work done be zero in that case,
If the system does not change (and there are no so-called "frictional" forces i .e. no other degrees of freedom that can suck up energy) then the net work will be zero. You do positive work on, say, a torsional spring to wind it up: negative work is done by you as it (slowly) unwinds.
 
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