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Before giving the formula, note that the response of the gyroscope wheel will not be in opposite direction. For names of direction of rotation let me use the standard names of aircraft principle axes: roll, pitch and yaw. Take the spin axis of the gyro wheel as rolling, and apply a large torque that will pitch the gyro wheel fast. Then in response the gyro ...


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For a spinning flywheel, the external torque vector determines the rate of change of the angular momentum vector , L, which equals Iw. For a uniform disk, I, (The rotational inertia) = (½) m R^2, and the angular velocity vector, w, (in radians/sec) is defined as being along the axis of rotation. (You may want to look up the associated “right hand rules”.) ...


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This answer elaborates on the earlier answer by use Krishna. As user Krishna describes: if the top is constructed in such a way that it can remain upright statically (which of course means that it needs a foot that is wide enough), then if if spins sufficiently close to vertical it will slow down to the statically upright position. But of course the ...


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A top can spin in an upright position. If you manage to balance a top and let it spin, it will spin about its axis in an upright position. But, it will never fall down, because it is balanced. The friction would cause it to stop and it will be left standing after it loses its energy. But, if you spin it with a very slight (almost unnoticeable) initial angle,...


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if you neglect the centrifugal and Coriolis torques this means that your system is rotate slowly , you get the equations of motion. $$\ddot{\varphi}= -{\frac {\sin \left( \psi \right) T_{{\vartheta }}}{J_{{\vartheta }} \cos \left( \vartheta \right) }}+{\frac {\cos \left( \psi \right) { \it uu}_{{1}}}{\cos \left( \vartheta \right) J_{{\varphi }}}} $$ $$...


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Sudden removal of the precession inducing torque has the same effect as sudden onset of the precession inducing torque, but in opposite direction. In both cases the sudden change kicks in a nutation. Onset of precession: if the gyroscope wheel is spinning very fast then the nutation frequency is high, and the amplitude is small. In cases where the spin rate ...


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You didn't miss anything. In your question you analyze the dynamics correctly at the instant $t=0$. If you use the same argument, after the application of the force $F$ over the time interval $dt$ you will see that $$-F.R = I_y\dot \omega_y \implies \omega_y = -F.R \frac {dt}{I_y},$$ and then from the Euler equations $$M_x = I_x\dot \omega_x - (I_y - I_z)\...


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