# Angular displacement and the displacement vector

I've been recently introduced to angular displacement, and I'm a little bit confused about it. I think that displacement which is a vector and which is defined as the shortest distance between any two points is different from angular displacement. Angular displacement is measured in radian or in other words it measures the 'angle' that a rotating body goes through. However, displacement vector is measured in meter. Angular displacement has the dimension of $1$; however, displacement vector has the dimension $L$. Is there any concept of the-shortest-distance-covered in angular displacement? I think angular displacement may just mean the change of position of a rotating body w.r.t. the angle subtended by the arc, traveled by it, at the center of the circle around which it is rotating.

If a body is rotating about the axis of a circle, and it moves from point A to B making an arc of length S, then the distance it covered is given by $S = θr$. However, the shortest distance it could have covered to reach point B can be given by the length of the vector D.

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First things first- Angular displacement is quite different from displacement. It has dimension 1 as its unit is radians(plane angle), which is not a fundamental unit. Not only that but I think that it is not the change in $\theta$ but actually the total $\theta$ travelled.

For example, if some particle moves in a complete circle and stops at its initial position, its angular displacement is actually $2\pi$, and not 0.

And thus if you multiply $\theta$ with R, you get distance travelled by the particle and not its displacement. You can always find the displacement too, I guess.

Example: A particle has angular displacent $5\cdot \pi$ in time $t$. Find its displacement in time $t$.

We keep subtracting $2\cdot \pi$ until $$0 \leq θ < 2\cdot\pi$$

(we do this to cancel out the full circles travelled by the particle as when it completes a circle its displacement becomes 0)

thus we know it travelled 2 full circles and then pi radians. Thus the displacement = $2r$

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You have to rotate it, you cannot just move every point across your vector $\vec D$, that would be a simple displacement. I would think of it as winding up a spring (like in a clock) and noting how many revolutions I have wound it up. And I would measure that in radian (or without a unit, if you like).

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Why do we call it 'displacement' anyways? –  Samama Fahim Jun 3 '13 at 21:37
Since it can be positive and negative, and rolls back, if you turn into the opposite direction. Your one dimensional rotation is like a one dimensional displacement. Say you move only in the $x$-direction. –  queueoverflow Jun 4 '13 at 16:46