Understanding Vector Rotation and Derivation of Angular Velocity Formula

[sherlockjones]

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

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

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 \vec{A}\frac{d\theta}{dt} or the angular velocity of \vec{A}


Thanks

 

[HallsofIvy]

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

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

Any plane vector of constant length r can be written \vec{A}= rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}. Then
\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})
You can use cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) and sin(a+b)= sin(a)cos(b)+ cos(a)sin(b) 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 2- 2cos(\Delta\theta). I was surprised when I saw that the terms involving only \theta rather than /Delta/theta cancel out! Of course, the square root of that does give the half angle formula.

 

[sherlockjones]

it was actually the change in \vec{A} in the time interval t to t + \Delta t. So wouldn't it be: \Delta \vec{A} = \vec{A}(t+\Delta t)-\vec{A}(t)?