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Homework Help: 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]

  2. jcsd
  3. Oct 31, 2006 #2


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    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]?
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