1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rotating Vectors

  1. Oct 31, 2006 #1
    Lets say we have a vector [tex] \vec{A}(t) [/tex] with a constant magnitude [tex] A [/tex]. Thus [tex] \vec{A}(t) [/tex] can only change in direction (rotation). We know that [tex] \frac{d\vec{A}}{dt} [/tex] is always perpendicular to [tex] \vec{A} [/tex]. This is where I become stuck:

    [tex] \Delta \vec{A} = \vec{A}(t+\Delta t)-\vec{A}(t) [/tex]
    [tex] |\Delta \vec{A}| = 2A\sin\frac{\Delta \theta}{2} [/tex].

    How do we get the trigonometric expression on the right in the second equation? It looks like some type of half/double angle formula. Eventually we are supposed to get [tex] \vec{A}\frac{d\theta}{dt} [/tex] or the angular velocity of [tex] \vec{A} [/tex]


    Thanks
     
  2. jcsd
  3. Oct 31, 2006 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Interesting! I was sure that couldn't possibly be correct when I started calculating it!

    First, what you have is for
    [tex] \Delta \vec{A} = \vec{A}(\theta+\Delta \theta)-\vec{A}(\theta) [/tex]
    that is, with [itex]\theta[/itex] not t. And, of course, there is a typo in the second equation: you mean [itex]|\vec{A}|[/itex], not just [itex]A[/itex].

    Any plane vector of constant length r can be written [itex]\vec{A}= rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}[/itex]. Then
    [itex]\vec{A(\theta+ \Delta\theta)}- \vec{A(\theta)}= r((cos(\theta+ \Delta\theta)- cos(\theta))\vec{i}+ (sin(\theta+ \Delta\theta)- sin(\theta))\vec{j})[/itex]
    You can use [itex]cos(a+ b)= cos(a)cos(b)- sin(a)sin(b)[/itex] and [itex]sin(a+b)= sin(a)cos(b)+ cos(a)sin(b)[/itex] to expand those. I won't do all of the calculations here (writing all of that in LaTex would be more tedious than the calculations!) but squaring each those and summing reduces to [itex]2- 2cos(\Delta\theta)[/itex]. I was surprised when I saw that the terms involving only [itex]\theta[/itex] rather than [itex]/Delta/theta[/itex] cancel out! Of course, the square root of that does give the half angle formula.
     
  4. Oct 31, 2006 #3
    it was actually the change in [tex] \vec{A} [/tex] in the time interval [tex] t [/tex] to [tex] t + \Delta t [/tex]. So wouldnt it be: [tex] \Delta \vec{A} = \vec{A}(t+\Delta t)-\vec{A}(t) [/tex]?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?