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Homework Help: Angular displacement in radians problem

  1. Jan 10, 2009 #1
    1. The problem statement, all variables and given/known data

    what is the angular displacement after 45 second if a wheel spins at 46.07669 radians/s

    2. Relevant equations

    3. The attempt at a solution

    i got 2073.45105 and i dont know why that is not right
  2. jcsd
  3. Jan 10, 2009 #2

    Doc Al

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    Looks OK to me.

    Is that the exact statement of the problem?
  4. Jan 10, 2009 #3
    the wheel makes 22 revolutions in 3 seconds

    thats the only other information
  5. Jan 10, 2009 #4


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    Biker, I wonder if you are supposed to express your answer as a number less than 2*pi (360 degrees). Who cares how many full turns there were?
  6. Jan 10, 2009 #5

    Doc Al

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    Is the speed constant? Or does it start from rest? (Unless you post the exact question--word for word--we can only guess if there's something you might have missed.)

    Your solution assumes a constant speed.
  7. Jan 10, 2009 #6


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    Like Doc Al said, if they mention that the disk started from rest and has constant angular acceleration, then the solution is different than what you did.

    On the other hand, if they say that it is rotating at constant angular velocity, your calculation is correct. But the problem may be one of sig figs. Do they say 22 revolutions in 3 seconds or 22 revolutions in 3.0 seconds? Or 22.0 revolutions in 3.00 seconds?
    Depending on th eprecision they want, the answer they are looking for could be 2070 rad or 2100 rad.
  8. Jan 11, 2009 #7
    Wait a minute! Displacement isn't the same as distance. Angular distance is what you have posted(2070 etc). Displacement is a vector which describes a change in position of a point or particle in reference to a previous position. In this case it's a rotational movement (I hope that's the term in English) and angular displacement, which means it can't be more than [tex]\pi[/tex] radians. For example if it has rotated [tex]\pi[/tex] / 2 radians that that's the angular displacement, if it has rotated 100[tex]\pi[/tex] radians, than the displacement is 0 because the object, point or whatever is rotating is again in the starting position after making 50 full turns. If it has rotated 3[tex]\pi[/tex] / 2 radians, the displacement is [tex]\pi[/tex]/2 radians, because it is only [tex]\pi[/tex]/2 radians away from the initial position. Don't confuse distance with displacement.
  9. Jan 11, 2009 #8

    Doc Al

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    Interesting point! (Which I suppose is what Delphi51 was getting at earlier.)

    See if they just want the change in angle (mod 2π). (It's worth a shot.)

    But I disagree that this is common usage, since angular motion is one dimensional. The parallel kinematic equations would be (for uniformly accelerated motion):

    [tex]x = x_0 + v_0t + 1/2 a t^2[/tex] (For linear motion.)

    [tex]\theta = \theta_0 + \omega_0t + 1/2 \alpha t^2[/tex] (For rotation.)

    Just like x is the linear displacement, so is θ the total angular "displacement" (not mod 2π).
  10. Jan 11, 2009 #9
    I googled angular displacement and got to this http://en.wikipedia.org/wiki/Angular_displacement" [Broken]. And it says the same thing. So I guess I was wrong. In a sense, angular motion is analogical to one-dimensional straight-line movement. So you are right about the displacement. I apologize for my intrusion.
    Last edited by a moderator: May 3, 2017
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