91Ó°ÊÓ

Mass of the gyroscope $\mathit{M}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{140}\mathit{k}\mathit{g}$moment of inertia about its axis $\mathit{I}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathit{k}\mathit{g}\mathbf{.}{\mathit{m}}^{\mathbf{2}}$

Mass of the frame $\mathit{m}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{250}\mathit{k}\mathit{g}\mathbf{}$

The horizontal distance between the center of mass and the pivot is $\mathit{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{04}\mathit{m}$

The upward force exerted by the pivot is

$$\begin{gathered} F_N=(M+m)g \\ =(0.140\mathrm{kg}+0.0250\mathrm{kg})(9.80\mathrm{m}/{\mathrm{s}}^2) \\ =\boxed{1.62\mathrm{N}} \end{gathered}$$

Hence, the force is$1.62N.$

Let us take the angular speed of the rotor as $\mathrm{Ó\neg }$.

Then the precession angular speed of the axis,

localid="1667989962728" $$\begin{gathered} \mathrm{Î\copyright }=1rev/2.20s \\ =\frac{2\mathrm{Ï€}rad}{2.20s} \\ =2.856rad/s \end{gathered}$$

The precession angular speed of the gyroscope is

localid="1667990061937" $\mathrm{Î\copyright }=\frac{(M+m)gr}{1\mathrm{Ó\neg }}$

Therefore the required angular speed of the gyroscope

localid="1667990356475" $$\begin{gathered} \mathrm{Ó\neg }=\frac{(\mathrm{M}+\mathrm{m})\mathrm{gr}}{1\mathrm{Î\copyright }} \\ =\frac{(0.140\mathrm{kg}+0.025\mathrm{kg})(9.80\mathrm{m}/{\mathrm{s}}^2)(0.04\mathrm{m})}{(1.20×{10}^{(}-4)\mathrm{kg}.\mathrm{m}2\left)\right(2.856\mathrm{rad}/\mathrm{s})} \\ =188.92\mathrm{rad}/\mathrm{s} \\ =\frac{(188.92\mathrm{rad}/\mathrm{s})(60.0\mathrm{s}/\min )}{2\mathrm{Ï€}} \\ =\boxed{1.80×{10}^3\mathrm{rev}/\min } \end{gathered}$$

Hence, the angular speed is $1.80×{10}^{3}rev/min.$

When the rotor is rotating, the initial angular momentum $\left(L_1\right)$is pointed toward positive $x$. The torque due to weight is pointed toward positive $y$.

Thus the torque causes the angular momentum to presses around the pivot.