All Problems

Dynamics of a solid body

Problem 1.283

A top of mass m=0.50 kgm=0.50 \mathrm{~kg}, whose axis is tilted by an angle θ=30\theta=30^{\circ} to the vertical, precesses due to gravity. The moment of inertia of the top relative to its symmetry axis is equal to II =2.0 gm2=2.0 \mathrm{~g} \cdot \mathrm{m}^{2}, the angular velocity of rotation about that axis is equal to ω=350rad/s,\omega=350 \mathrm{rad} / \mathrm{s}, the distance from 'the point of rest to the centre of inertia of the top is l=10 cm.l=10 \mathrm{~cm} . Find: (a) the angular velocity of the top's precession; (b) the magnitude and direction of the horizontal component of the reaction force acting on the top at the point of rest.

Reveal Answer
1.283. (a) ω=mgl/Iω=0.7rad/s\omega^{\prime}=m g l / I \omega=0.7 \mathrm{rad} / \mathrm{s} (b) F=mω2lsinθ=10mNF=m \omega^{\prime 2} l \sin \theta=10 \mathrm{mN}. See Fig. 11.