91Ó°ÊÓ

The mass of the gyroscope is $M=120g$

role="math" localid="1650463286647" $R=8cm=0.08m$

The spin angular frequency of the gyroscope is role="math" localid="1650463469116" $\mathrm{Ó\neg }=1000rpm=104.7rad/s$

Consider a gyroscope as a hoop with a massless spoke.

The moment of inertia of a hoop is $I=M{R}^2…\left(1\right)$

The precession frequency is given by role="math" localid="1650462099722" $\mathrm{Î\copyright }=\frac{Mdg}{I\mathrm{Ó\neg }}…\left(2\right)$

From (1) and (2),

role="math" localid="1650462308740" $$\begin{gathered} \mathrm{Î\copyright }=\frac{Mdg}{M{R}^2\mathrm{Ó\neg }} \\ =\frac{Rg}{R^2\mathrm{Ó\neg }} \\ =\frac{g}{R\mathrm{Ó\neg }} \end{gathered}$$

Therefore, the expression for the precession frequency is$\mathrm{Î\copyright }=\frac{g}{R\mathrm{Ó\neg }}$.

Substitute the given values into $\mathrm{Î\copyright }=\frac{g}{R\mathrm{Ó\neg }}$.

role="math" localid="1650463338918" $$\begin{gathered} \mathrm{Î\copyright }=\frac{9.80}{0.08×104.7} \\ =1.17rad/s \end{gathered}$$

The period is given by $T=\frac{2\mathrm{Ï€}}{\mathrm{Î\copyright }}$.

role="math" localid="1650463361689" $$\begin{gathered} T=\frac{2\mathrm{Ï€}}{1.17} \\ =5.4s \end{gathered}$$

Therefore, the precession period is$5.4s$.