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Why does the Sin of 0.036 degrees approximately equal 2 pi?

  1. Jan 20, 2017 #1
    1. The problem statement, all variables and given/known data
    The formula for centripetal acceleration (Ac) is $$Ac = \frac {4π^2r} {T^2},$$ where r = radius and T = period of rotation

    2. Relevant equations
    The above formula can be rearranged as follows: $$Ac = \frac {2π} {T} × \frac {2π} {T} × \frac {r} {1},$$ $$= \frac {2π} {T} × \frac {2πr} {T},$$ Setting the value of T at 1 (sec) gives the following:$$Ac = \frac {2π} {sec} × \frac {2πr} {sec}$$
    The second term is simply the magnitude of the tangential velocity of an object in uniform circular motion. The numeric value of the first term is ##\frac {6.283185} {sec}##. When looking at this problem from the standpoint of vectors, this first term is the factor applied to the magnitude of the tangential velocity to derive the radial (inward) acceleration. Trigonometrically, when breaking the circular motion into very small segments, the sin of the angle formed by the initial and final velocity vectors as the circle segments become smaller and smaller asymptotically approaches 2π.

    For example, dividing the circular motion into 100 segments (3.6° each) yields ##\frac {6.279052×10^{-2}} {sec×10^{-2}}##, 1000 segments (0.36° each) yields ##\frac {6.283144×10^{-3}} {sec×10^{-3}}## and 10000 segments (0.036° each) yields ##\frac {6.283185×10^{-4}} {sec×10^{-4}}##.


    3. The attempt at a solution

    What I can't figure out is whether this approximate equality of values between the Sin of the angle formed by the initial and final velocity vectors and 2π is just a happy mathematical coincidence or whether there is some subtle connection there that I'm just not seeing. I don't much believe in happy mathematical coincidences, so I figure I'm missing something.

    BTW, this isn't a homework problem. I'm just curious and I've got waaay too much time on my hands. :smile:
     
    Last edited by a moderator: Jan 20, 2017
  2. jcsd
  3. Jan 20, 2017 #2
    What is sin here ? Is it sine function ?
    Sine function can't be greater than 1.
     
  4. Jan 20, 2017 #3
    The vector method for determining centripetal acceleration breaks the circular path into small segments and then uses the sine of the angle formed by the initial and final velocity vectors to determine the factor (numerically, ##{\frac {2π} {T}}##) which, when applied to the tangential velocity yields the magnitude of the centripetal acceleration.

    In my example, for instance, dividing the circle into 1000 segments produces a value for the sine of 0.36° of 0.006283144. This value, since it is for a segment only ##\frac {1} {1000}## T, must be multiplied by 1,000 - yielding 6.283144.
     
  5. Jan 20, 2017 #4
    $$\bbox[#FFBF00, 10px, Border: 2px solid black] {\sin x \to x \ as \ x \to 0}$$
    So when you divide a circle by 1000, you are making the angle x very very small. Thus the sine of x approaches x because x is very very small.
    When you multiply ##\sin(x)## by 1000, you get roughly ##360^{\circ}## or ##2\pi## because well, you divided 360 by 1000 to get x and Sine of x is roughly x.

    Notice that it is not ##2\pi##, it is roughly ##2\pi##.
     
    Last edited: Jan 20, 2017
  6. Jan 20, 2017 #5
    Yes, I noticed that the sine asymptotically approaches the absolute value 6283185.... (2π) as the angle diminishes.

    Hmm........I'll have to think on your reply for a while. I'm trying to get my head around it but I'm not quite grasping the concept.:confused:
     
  7. Jan 20, 2017 #6
    Ok. Here is a formula $$\sin x° \approx x°$$ for very small ##x##.

    Now we need to find ##\sin 0.36^\circ##.
    Using the formula, we get $$\sin 0.36^\circ \approx 0.36^\circ$$
    Multiplying by 1000 on both sides, $$1000\sin 0.36^\circ \approx 1000 * 0.36^\circ = 360^\circ \implies 1000\sin 0.36^\circ \approx 360^\circ = 2\pi$$

    Isn't this what you are saying ?
     
    Last edited by a moderator: Jan 20, 2017
  8. Jan 20, 2017 #7

    Ray Vickson

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    You are giving the OP incorrect and misleading information. The approximation ##\sin \theta \approx \theta## holds only when angles are measured in radians, not degrees. So, if a small angle measures ##x## degrees, that is ##\pi x/180## radians, and so the sin is approximately ##\pi x/180##; in other words, when we measure angles in degrees we get ##\sin x \approx \pi x/180## for ##x \to 0##.

    Thus, ##\sin 0.36^\circ \approx 0.36 \pi /180 = .0062832.## Check it out on your calculator, but be sure to choose the degrees/radians setting correctly.
     
  9. Jan 20, 2017 #8
    I mentioned ##\sin 0.36^\circ \approx 0.36\color{red}{^\circ}##.
    By definition, ##x^\circ = x\pi/180##.
    Substitute ##0.36^\circ##, you will get the answer.

    I did not mention the end answer as a real number with no unit. I mentioned the unit as degrees, you are converting that unit to radians. (units here does not mean as what it means in physics but rather as a constant multiplied to it).

    Would say the same if wrote ##\sin 0.36^\circ \approx 0.36\color{red}{\alpha}##, where ##\alpha = \pi/180## ??


    I don't think I have mentioned ill information other than I should have written degree symbol in my formula.
     
  10. Jan 20, 2017 #9
    Thanks Ray and Buffu. I was having trouble imagining why the the sin 0.36°, 0.036°, 0.0036°...asymptotically approaches 2π. You each explained it quite well from your different perspectives. Ray, your explanation was particularly helpful in getting me to see why the numbers "work out".:smile:

    :dademyday:
     
  11. Jan 20, 2017 #10

    Ray Vickson

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    Of course if you wrote ##\sin 0.36^\circ \approx 0.36\color{red}{\alpha}## I would have no objection, since that is exactly what I wrote in a slightly different notation.
     
  12. Jan 20, 2017 #11
    So then you do you agree that I have not mentioned ill-information ?
    I think I have just made it a bit ambiguous.
     
  13. Jan 20, 2017 #12

    Ray Vickson

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    No, I don't quite agree (almost, but not quite); however, enough has been said on the subject. The main point was to not confuse the OP, who seems to be struggling with some basic concepts.
     
  14. Jan 20, 2017 #13
    Should I prove it ?

    I took degrees instead of radians because I thought it might confuse. But I messed it up.
     
  15. Jan 20, 2017 #14

    haruspex

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    In the standard view, that does not really mean anything. The left hand side is just a number. To justify your use of degrees on the right you would need a convention that says degrees are a unit of dimensionless numbers, different from 1; namely, π/180. You could then have degrees of anything. Sometimes it would be useful. 90 degrees of oranges, for example.
    As it happens, there is a forum article showing how angles could be assigned a dimension, albeit a rather strange one. In this view, the sine function both takes an angle as argument and returns an angle as result. (Cosine doesn't.). That would also make your equation valid. See https://www.physicsforums.com/insights/can-angles-assigned-dimension/
     
  16. Jan 21, 2017 #15
    Degrees are dimensionless. Do I need to prove ##2##
    is dimensionless before saying ##2 \times 2 = 4## ?
    Same for degrees just that there is ##\pi/180## instead of ##2##

    I have read this Insight. I like your Insight, But it is of no use now because I am not assigning a dimension (as used in physics) to angles.
     
  17. Jan 21, 2017 #16

    Ray Vickson

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    Let me repeat: the OP seems to be struggling a bit with some of the related concepts, and may possibly be laboring under some mis-understandings. The best service we can provide him is to try to have him "unlearn" incorrect concepts before they become too fixed in his mind. It would serve him well to understand the distinction between degrees and radians, and to understand important basic properties of trig functions.

    Anyway, whether or not angle measurements have "dimensions" (they already have units!) is not something you or I can decide; it is a matter of convention and international agreement after a couple of centuries of use and possible past argumentation.
     
  18. Jan 22, 2017 #17
    Fair enough.

    But ##\sin x^\circ \approx x^\circ## is perfectly valid statement.
     
  19. Jan 22, 2017 #18

    haruspex

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    Not in the conventional view, in which there are two ways of regarding the sine function.

    Treated as purely an algebraic function on the reals, defined by a power series, it takes a number as argument and produces a number as result. For that function, sin(x)≈x for small x. But it would not make sense to attach a degree symbol to either x.

    Treated as a function that takes an angle α as argument and returns a pure number, sin(α)≈"the number of radians in α". It now makes sense to attach a degree symbol to the argument to sine, but it still makes no sense to attach one to the result.

    To make your claim valid, you need a new convention. The "angular dimension" proposal is one way; defining x° as the number πx/180 is another.
     
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