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I Rotational Motion - Understanding the Idea

  1. Feb 6, 2017 #1
    1- I am trying to understand the rotational motion and In most books theres ##\vec θ## unit vector which its, ##\vec θ=(-sinθ,cosθ)##
    I can see that If ##\vec r## unit vector is,##\vec r=(cosθ,sinθ)## then ,

    ##\frac {d\vec r} {dt}=\vec θ##

    Like calculating ##\vec v=\frac {d\vec r} {dt}=R.\frac {dθ} {dt}.\vec θ##

    I dont understand why we use this ? Just to make things easier ?

    2-And we know that ##\vec a=R∝\vec θ+Rw^2(-\vec r)##
    here
    ##\vec {a_t}=R∝\vec θ## and ##\vec {a_r}=Rw^2(-\vec r)##

    I know that If ##a_t## is zero then it becomes uniform circular motion.But In main equations there is an non-zero angular acceleration.So it means do we have to exert ##\vec F=m(-R∝\vec θ)## to create a uniform circular motion ?

    Note:This is not a homework question so,I didnt ask there
     
  2. jcsd
  3. Feb 6, 2017 #2
    In general yes. It simplifies rotational motion a lot. For instance, consider uniform circular motion. If you were to stick to ##x## and ##y## coordinates, they would be constantly changing and they become difficult to work with. They are best suited to translational motion because of this. If you use ##r## and ##\theta## coordinates you can immediately see that one derivative is zero, and that the other is simply the angular speed. The reason we use ##\vec \theta## is to give that speed a direction.

    No we do need to apply such a force for uniform circular motion. The ##\vec {a_t}## has to be caused by a force. If there is no tangential force, then it is zero. The term appears in the formula because the formula is general, so it works for any value of ##\vec {a_t}##, including zero. You obtain the value of ##\vec {a_t}## from the value of ##\vec F_t##, not the other way round.
     
  4. Feb 6, 2017 #3
    I understand now.Thanks
     
  5. Feb 6, 2017 #4
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