Dynamics cheat sheet

14.2.3 Derivation for τ = Iω in 3D using principle axes
The above derivation simplifies now since we will be using principle axes. In this case, all cross products of
moments of inertia vanish.
⎛
⎞
I xx 0 0
I =
⎜ 0 I yy 0
⎟
⎝
⎠
0 0 I zz
Hence
⎡
⎤
A
A
{ ⎛ }} ⎞ ⎛ ⎞{
⎛ ⎞ { ⎛ }} ⎞ ⎛ ⎞{
τ = d I xx 0 0
ω x
ω x
I xx 0 0
ω x
dt
⎜ 0 I yy 0
⎟ ⎜ω y ⎟
+
⎝
⎠ ⎝ ⎠
⎜ω y ⎟
⎝ ⎠ ×
⎜ 0 I yy 0
⎟ ⎜ω y ⎟
⎝
⎠ ⎝ ⎠
⎢
⎣ 0 0 I zz ω z
⎥
⎦ ω z 0 0 I zz ω z
⎛
⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞
I xx 0 0
α x
ω x
I xx ω x
=
⎜ 0 I yy 0
⎟ ⎜α y ⎟
⎝
⎠ ⎝ ⎠ + ⎜ω y ⎟
⎝ ⎠ × ⎜I yy ω y ⎟
⎝ ⎠
0 0 I zz α z ω z I zz ω z
⎛ ⎞ ∣ ∣ ∣∣∣∣∣∣∣∣∣ ∣∣∣∣∣∣∣∣∣
I xx α x
i j k
=
⎜I yy α y ⎟
⎝ ⎠ + det ω x ω y ω z
I zz α z I xx ω x I yy ω y I zz ω z
⎛ ⎞ ⎛
⎞
I xx α x
ω y (I zz ω z ) − ω z (I yy ω y )
=
⎜I yy α y ⎟
⎝ ⎠ + ⎜−ω x (I zz ω z ) + ω z (I xx ω x )
⎟
⎝
⎠
I zz α z ω x (I yy ω y ) − ω y (I xx ω x )
⎛ ⎞ ⎛
⎞
I xx α x
ω y ω z (I zz − I yy )
=
⎜I yy α y ⎟
⎝ ⎠ + ⎜ω x ω z (I xx − I zz )
⎟
⎝
⎠
I zz α z ω x ω y (I yy − I xx )
So, we can see how much simpler it became when using principle axes. Compare the above to
⎛
⎞ ⎛ ⎞ ⎛
⎞
I xx I xy I xz
α x
ω y (I zx ω x + I yz ω y + I zz ω z ) − ω z (I yx ω x + I yy ω y + I yz ω z )
⎜I yx I yy I yz ⎟ ⎜α y ⎟
⎝
⎠ ⎝ ⎠ + ⎜ω x (I zx ω x + I yz ω y + I zz ω z ) − ω z (I xx ω x + I xy ω y + I xz ω z )
⎟
⎝
⎠
I zx I yz I zz α z ω x (I yx ω x + I yy ω y + I yz ω z ) − ω y (I xx ω x + I xy ω y + I xz ω z )
So, always use principle axes for the body fixed coordinates system!
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