All Problems

Dynamics of a solid body

Problem 1.285

A top of mass m=1.0 kgm=1.0 \mathrm{~kg} and moment of inertia relative to its own axis I=4.0 gm2I=4.0 \mathrm{~g} \cdot \mathrm{m}^{2}  spins  with  an  angular  velocity ω\begin{array}{lllll}\text { spins } & \text { with } & \text { an } & \text { angular } & \text { velocity } & \omega\end{array} =310rad/s.=310 \mathrm{rad} / \mathrm{s} . Its point of rest is located on a block which is shifted in a horizontal direction with a constant acceleration w=1.0 m/s2w=1.0 \mathrm{~m} / \mathrm{s}^{2}. The distance between the point of rest and the centre of inertia of the top equals l=10 cml=10 \mathrm{~cm}. Find the magnitude and direction of the angular velocity of precession ω\omega^{\prime}.

Reveal Answer
 1.285. ω=mlg2+w2/Iω= \text { 1.285. } \quad \omega^{\prime}=m l \sqrt{g^{2}+w^{2}} / I \omega= =0.8rad/s.=0.8 \mathrm{rad} / \mathrm{s} . The vector ω\omega^{\prime} forms the angle θ=arctan(w/g)=6\theta=\arctan (w / g)=6^{\circ} with the vertical.