Question 21.7: The top shown in Fig. 21-20a has a mass of 0.5 kg and is pre......

Equation 21-30 will be used for the solution since the motion is steady precession. As shown on the free-body diagram, Fig. 21-20b, the coordinate axes are established in the usual manner, that is, with the positive z axis in the direction of spin, the positive Z axis in the direction
of precession, and the positive x axis in the direction of the moment
ΣM_{x} (refer to Fig. 21-16). Thus,

ΣM_{x} = –  I\dot{\phi²} \sin  θ \cos  θ    +   I_{z}\dot{\phi} \sin   θ(  \dot{\phi} \cos  θ   +  \dot{ψ})

4.905  N(0.05   m)  sin  60° = –  [1.20( 10^{-3} ) kg · m² \dot{\phi²}] sin 60° cos 60°

+ [0.45(10-3) kg · m²] \dot{\phi} sin 60° ( \dot{\phi} cos 60° + 100 rad/s)

or

  \dot{\phi²}   –    120.0\dot{\phi}   +    654.0    =   0                        (1)

Solving this quadratic equation for the precession gives

  \dot{\phi} = 114  rad/s (high precession)

and

  \dot{\phi} = 5.72 rad/s (low precession)

NOTE: In reality, low precession of the top would generally be observed, since high precession would require a larger kinetic energy.