topsquark
Senior Member
- Joined
- Aug 27, 2012
- Messages
- 2,370
In terms of the problem you were given
r˙=R˙RTr
To simplify things let's take motion in the xy plane so we have
R=⎝⎛cos(θ)sin(θ)0−sin(θ)cos(θ)0001⎠⎞
And we get
R˙RT=⎝⎛0θ˙0−θ˙00000⎠⎞
So
r˙=⎝⎛0θ˙0−θ˙00000⎠⎞r=⎝⎛0θ˙0−θ˙00000⎠⎞⎝⎛xy0⎠⎞=⎝⎛0x0−y00000⎠⎞θ˙
Notice that ω here will be in the z direction, so notice that
r×ω=⎝⎛xy0⎠⎞×⎝⎛00θ˙⎠⎞=⎝⎛0x0−y00000⎠⎞θ˙=R˙RTr=r˙
or v=r×ω.
I don't know why the author didn't finish the job properly. All he did was draw a parallel to the cross product and say "Here's an example and voila!" It was all there but for some reason he simply didn't pick the right example.
-Dan
r˙=R˙RTr
To simplify things let's take motion in the xy plane so we have
R=⎝⎛cos(θ)sin(θ)0−sin(θ)cos(θ)0001⎠⎞
And we get
R˙RT=⎝⎛0θ˙0−θ˙00000⎠⎞
So
r˙=⎝⎛0θ˙0−θ˙00000⎠⎞r=⎝⎛0θ˙0−θ˙00000⎠⎞⎝⎛xy0⎠⎞=⎝⎛0x0−y00000⎠⎞θ˙
Notice that ω here will be in the z direction, so notice that
r×ω=⎝⎛xy0⎠⎞×⎝⎛00θ˙⎠⎞=⎝⎛0x0−y00000⎠⎞θ˙=R˙RTr=r˙
or v=r×ω.
I don't know why the author didn't finish the job properly. All he did was draw a parallel to the cross product and say "Here's an example and voila!" It was all there but for some reason he simply didn't pick the right example.
-Dan