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Angular Acceleration Equations

  1. Oct 4, 2014 #1
    According to what I've read, I was told that
    1) s = r*theta
    2) angle("theta"): theta(t) in radians
    3) angular velocity ("omega"): omega(t) = d/dt theta(t) in rads/s
    4) angular acceleration ("alpha"): alpha(t) = d/dt omega(t) in rad/(s^2)
    However I do not understand the concept of this. I understand that s is the arc length, r is radius, but why is r multiplied by the angle? For example shouldn't it be rsintheta, rcostheta, etc?
    Also if someone could explain what 2, 3 and 4 represent. Is it the derivative of the functions at t?
    For example f(x) = omega, and then you would take f'(x) to get alpha?
    Also, is angular velocity/acceleration the velocity/acceleration as it changes relative to the angle? Or simply put the change in the angle? Would this mean that the velocity changes as the angle changes? I'd like some help understanding this please
    Edit: I'm really confused so my explanation might not be very good. I would also like to ask, if s = rtheta, then if I had an angle of let's say as an example... pi/3. Then would I multiply r by pi/3 to get a value of s? (if theta = pi/3. s = r(pi/3) right? )
     
    Last edited: Oct 4, 2014
  2. jcsd
  3. Oct 4, 2014 #2

    vela

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    Consider a unit circle (r=1), and say you start at the point (1,0) where it intersects the x-axis. Now imagine you go along the circle for 90 degrees (in the counter-clockwise direction) so that now you're at the point (0,1), where the circle intersects the y-axis. The formulas
    \begin{eqnarray*}
    x &= r \cos\theta = (1)\cos 90^\circ = 0 \\
    y &= r \sin\theta = (1)\sin 90^\circ = 1
    \end{eqnarray*} give you the coordinates of the point. On the other hand, you have ##s=r\theta = (1)\frac \pi 2 = \frac \pi 2##. This is the distance you travelled along the circle, that is, the arc length. It should make sense that in this case, it turns out to be one-fourth of the circumference, ##2\pi / 4 = \pi/2##.

    It helps to draw an analogy with linear quantities (the stuff you learned about first). In the linear case, you had the displacement ##x##, the velocity ##v=\frac{dx}{dt}##, and the acceleration ##a=\frac{dv}{dt}##. Similarly, you now have the angular displacement ##\theta##, the angular velocity ##\omega = \frac{d\theta}{dt}##, and the angular acceleration ##\alpha = \frac{d\omega}{dt}##. The angular displacement represents how much you've rotated from some arbitrary orientation just like the regular displacement represents where you are relative to some point you arbitrarily designate as the origin.

    In the linear case, you could have a constant velocity so that ##v## didn't change even as ##x## did. Similarly, in the angular case, you can have constant angular velocity, and ##\omega## will not change even though ##\theta## does.
     
  4. Oct 4, 2014 #3
    Okay I think that makes sense with what I partially assumed. However could you further elaborate with an example for your explanations on the angular equations? Since it only defines the derivative, I'm hoping to get some sort of explanation involving it's application in a question. For example similar to how a linear quantity question 'could' use an equation of V = D/t where when you derive it you could get that it is equal to the acceleration, how would it apply to an application of questions?
     
  5. Oct 4, 2014 #4

    vela

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    Your textbook surely has worked examples involving angular quantities.
     
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