Displacement for rotation along some axis

Question

If I rotate something along some axis which has direction $vec Omega$, at an small angle $epsilon$ and if the position of the body is $vec R$ then according to my book the displacement will be $$delta R=epsilonvecOmegatimesvec R$$
I am unable to understand how this happened. I will be helpful if someone illustrate it to me.

Frederic · Accepted Answer

The rotation direction is perpendicular to $vec{Omega}$ and $vec{R}$ and is given by $vec{v} = frac{vec{Omega}timesvec{R}}{||vec{R}||}$. Note that $vec{v}$ is a unit vector. The vector $vec{R}$ rotates in this direction and thus $vec{v} parallel delta R$.

If you draw a triangle with bases $vec{R}$ and $delta R$, you find $tan(epsilon) = frac{||delta R||}{||R||}$. This means that the amplitude of the displacement is given by $||delta R||= tan(epsilon)||vec{R}||$.

So, now we know the amplitude and direction of the displacement which gives
$$delta R = ||delta R||vec{v} = tan(epsilon)||vec{R}||frac{vec{Omega}timesvec{R}}{||vec{R}||} = tan(epsilon)(vec{Omega}timesvec{R}) ,.$$

For small angles, the tangent can be approximated by $tan(epsilon) approx epsilon$ which gives the result
$$delta R =epsilon(vec{Omega}timesvec{R}) ,.$$

Eli · Answer

I have this answer for you ?
from this equation:
$$vec v=vec omega times vec RquadRightarrowquadfrac{d vec R}{dt}=frac {d vec varphi}{dt}times vec R$$
thus :
$$delta vec R=delta vec varphi times vec Rtag 1$$
for equation (1)  equal to
$$delta vec R=epsilonvecomegatimes vec R =epsilon frac {delta vec varphi}{delta t}times vec Rtag 2$$
$epsilon$ must be equal to $delta t$.  This doesn't make sense for me!

Displacement for rotation along some axis

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