Question 13.6: A homogeneous circular disk (mass M, radius R) rotates with ...
A homogeneous circular disk (mass M, radius R) rotates with constant angular velocity ω about a body-fixed axis passing through the center. The axis is inclined by the angle α from the surface normal and is pivoted at both sides of the disk center with spacing d. Determine the forces acting on the pivots.
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
I_{1} = σ \int\limits_{0}^{2π}{\int\limits_{0}^{R}{y^{2}r drd\varphi}} = σ \int\limits_{0}^{2π}{\int\limits_{0}^{R}{ r^{3} sin^{2} \varphi drd\varphi}}= \frac{1}{4} σR^{4} \int\limits_{0}^{2π}{sin^{2} \varphi d\varphi}
= \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).
