Paper 2, Section II, A

Classical Dynamics | Part II, 2012

Consider a rigid body with principal moments of inertia I1,I2,I3I_{1}, I_{2}, I_{3}.

(a) Derive Euler's equations of torque-free motion

I1ω˙1=(I2I3)ω2ω3,I2ω˙2=(I3I1)ω3ω1I3ω˙3=(I1I2)ω1ω2\begin{aligned} &I_{1} \dot{\omega}_{1}=\left(I_{2}-I_{3}\right) \omega_{2} \omega_{3}, \\ &I_{2} \dot{\omega}_{2}=\left(I_{3}-I_{1}\right) \omega_{3} \omega_{1} \\ &I_{3} \dot{\omega}_{3}=\left(I_{1}-I_{2}\right) \omega_{1} \omega_{2} \end{aligned}

with components of the angular velocity ω=(ω1,ω2,ω3)\boldsymbol{\omega}=\left(\omega_{1}, \omega_{2}, \omega_{3}\right) given in the body frame.

(b) Show that rotation about the second principal axis is unstable if (I2I3)(I1I2)>0\left(I_{2}-I_{3}\right)\left(I_{1}-I_{2}\right)>0.

(c) The principal moments of inertia of a uniform cylinder of radius RR, height hh and mass MM about its centre of mass are

I1=I2=MR24+Mh212;I3=MR22.I_{1}=I_{2}=\frac{M R^{2}}{4}+\frac{M h^{2}}{12} \quad ; \quad I_{3}=\frac{M R^{2}}{2} .

The cylinder has two identical cylindrical holes of radius rr drilled along its length. The axes of symmetry of the holes are at a distance aa from the axis of symmetry of the cylinder such that r<R/2r<R / 2 and r<a<Rrr<a<R-r. All three axes lie in a single plane. Compute the principal moments of inertia of the body.

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