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Why do all points on a uniform disc have the same angular velocity?

  1. Mar 31, 2014 #1
    ω = dΘ/dt

    if any points on a uniform disc have the same angular velocity then the corollary implies that any point on the disc must transverse 2π rad in the same amount of time.
    Is there a mathematical way to demostrate this or is this only demostratable purely by experiement?
  2. jcsd
  3. Mar 31, 2014 #2

    Simon Bridge

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    That is only the case for a rigid disk - each point has to have the same angular velocity because they are all connected to each other. If they had a different angular velocity the disk would break apart.
    It's the same reason all points in an object that moves in a line have the same velocity.

    To prove it mathematically, use basic geometry and assume the converse.
  4. Mar 31, 2014 #3


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    Could you choose a point an the disc that had a different angular velocity from the centre, for instance? Your problem with this could be that you are not 'feeling intuitively' what angular velocity actually is. It is not directly related to the forces that you might feel if you were at different points on the disc.
  5. Mar 31, 2014 #4
    Suppose you have a phonograph record, and you put a straight scratch on the record that runs from the center to the edge of the record. If you put it on a rotating turntable, does the scratch stay straight?

  6. Mar 31, 2014 #5


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    I was never really good at maths so I wouldn't know how to actually do this but...

    Perhaps you write an equation for the distance between any two points on a disc and then show that for the distance to be constant the angular velocity must be the same.
  7. Mar 31, 2014 #6
    The closest connection I have, or, "intuitively", is regardless the distance of a point from the rotational axis, the point transverses only a radian of 2π in one revolution.
    However, the point further away from the rotational axis has a much higher magnitude of tangential velocity from the idea that |vt| = ωr.
    Since ω remains constant for any point on a disc in a uniform circular motion, it is fairly obvious that |vt| is purely a function of r.
    Therefore, if r increases, |vt| changes correspondingly. (is 'proportional' a valid word to use in a mathematical sense?)
  8. Mar 31, 2014 #7
    By definition, one revolution is 2π radians. The only issue is do all the points of the disc complete 2π rad in the same amount of time? If they do not, then it is (by definition) not rigid. If the disc is rigid, they MUST.
    If the Earth were the center of the Universe, then some galaxies would be spinning around us at HUGE multiples of the speed of light (their tangental velocity). Such as system, under the known laws of Physics, would have exploded apart long ago...that is, it couldn't have ever formed.
    The equation |v| = ωr implies |v| ∝ r , and |v| ∝ ω. Although "directly proportional to" would be less ambiguous, since you may sometimes hear F ∝ r² based on Newton's Universal Law of Gravitation R, F = GmM/r² (where the actual proportionality is F ∝ 1/r² (as well as F∝G, F ∝ m, F ∝ M ). Proportionality between A and B implies the existence of a constant, k, such that A = kB in many areas of mathematics. I can't think of any off the top of my head, but I am pretty sure that the word is used with different meanings in different mathematical areas. Meaning that there are many many profoundly different types of mathematical objects than numbers, variables, and functions. If I were to say the dimension of A is proportional to the dimension of B, I hope you'd realize that you DON'T have much idea what that means. (In fact unless "dimension" is defined, the meaning here would be quite ambiguous to an expert in the field). But for simple equations, such as what you are talking about, yes proportional is correct and "correspond" has (afaik) NO clear meaning ..don't use it.
    BTW, v IS a function of ω, you are incorrect to claim it is ONLY a function of r. (I'm being picky, picky). When you hear that sort of loose talk, what is (probably) meant is v is only an interesting function of ω. FWIW.
    Last edited: Mar 31, 2014
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