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A top of mass m=1.0kg and moment of inertia relative to its own axis I=4.0g*m^2 spins with an angular velocity omega=310rad//s. Its point of rest is located on a block which is shifted in a horizontal direction with a constant acceleration w=1.0m//s^2. The distance between the point of rest and the centre of inertia of the top equals l=10cm. Find the magnitude and direction of the angular velocity of precession omega^'. |
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Answer» Solution :The effective `g` is `SQRT(g^2+W^2)` inclined at angle `"tan"^-1w/g` with the vertical. Then with reference to the new "vertical" we proceed as in problem. Thus `omega^'=(mlsqrt(g^2+w^2))/(Iomega)=0*8rad//s`. The vector `overset(rarr')omega` forms an angle `theta="tan"^-1(w)/(g)=6^@` with the NORMAL vertical. |
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