Question 13.6: A homogeneous circular disk (mass M, radius R) rotates with ...

The Euler equations read

$I_{1}  \dot{ω}_{1} − ω_{2}ω_{3}(I_{2} − I_{3}) = D_{1},$                                        (13.13)

$I_{2} \dot{ω}_{2} − ω_{1}ω_{3}(I_{3} − I_{1}) = D_{2},$                                      (13.14)

$I_{3}  \dot{ω}_{3} − ω_{1}ω_{2}(I_{1} − I_{2}) = D_{3},$                                      (13.15)

where $D = \left\{D_{1},D_{2},D_{3}\right\}$ represents the torque in the body-fixed system. We choose the body-fixed coordinate system in such a way that $n = e_{3}   and   e_{1}$ lies in the plane spanned by n,ω. For the principal momentum of inertia $I_{1}$, we have

$= \frac{1}{4} σR^{4}π = \frac{1}{4} \left(\frac{M}{πR^{2}}\right) R^{4}π = \frac{1}{4} MR^{2},$                            (13.16)

since the surface density σ is given by $σ = M/F = M/πR^{2}.$ And likewise for $I_{2}   and   I_{3}$

$I_{1} = I_{2} = \frac{1}{2} I_{3} = \frac{1}{4} MR^{2}.$                                            (13.17)

The components of the angular velocity vector are given by

$ω = \{ω_{1} = ω sin  α,ω_{2} = 0,ω_{3} = ω  cos  α\}.$                                         (13.18)

Because $\dot{ω} = 0$, inserting (13.17) and (13.18) into (13.13) to (13.15) yields

$D_{1} = D_{3} = 0         and         D_{2} =−ω^{2}  sin  α   cos  α \frac{1}{4} MR^{2}.$                                        (13.19)

Because $D = r×F$, in the pivots act equal but oppositely directed forces of magnitude

$|F| =\frac{|D_{2}|}{2d}=MR^{2}ω^{2} \frac{1}{4d} \left(\frac{1}{4} sin   2α\right) =MR^{2}ω^{2}   \frac{sin  2α}{16d}$                                      (13.20)

(see Fig. 13.13).