91Ó°ÊÓ

Linear momentum is the vector quantity and is defined as the product of the mass of an object m, and its velocity v.

The gyroscope’s precession moves in a clockwise direction. To understand this, look at the direction of the linear momentum of the gyroscope and the direction of the torque. The gyroscope is spinning clockwise around the shaft of the device, so from the right-hand rule, you can determine that the angular momentum points out of the page. Applying torque to angular momentum causes the angular momentum to move upwards in the direction of the torque. The entire gyroscope is forced to spin clockwise towards the z-axis so the angular momentum can point toward the direction of the torque.

The angular speed $\mathrm{Ó\neg }$is,

$\mathrm{Ó\neg }=\frac{{\mathrm{Ï€}}_{CM}}{L_{rot}}$

Here, ${\mathrm{Ï€}}_{CM}$is the torque on the center of mass and $L_{rot}=I\mathrm{Ó\neg }$

Here, l is the moment of inertia and $\mathrm{Ó\neg }$is the angular frequency.

$$\begin{gathered} L_{rot}=\left(0.06Kg.m^2\right)(30rad/s) \\ =1.8Kg.m^2/s \end{gathered}$$

Since the system is stationary without the weight, you only need to consider the torque caused by the weight. The force of the weight is and makes a ${30}^{∘}$with the lever arm of length r . The torque is

$$\begin{gathered} {\mathrm{π}}_{CM}=mgr\sin \left({30}^{∘}\right) \\ =(0.2kg)(9.8m/s^2)\left(18cm\left(\frac{1m}{100cm}\right)\right)\sin {30}^{∘} \\ =0.1764kg.m^2 \end{gathered}$$

The angular speed of the precession is,

$$\begin{gathered} \mathrm{Ó\neg }=\frac{0.1764kg.m^2}{1.8kg{m}^2{s}^{-1}} \\ =0.098rad/s \end{gathered}$$

Hence, the angular momentum is 0.098 rad/s

There are $2\mathrm{Ï€}$radians in a single revolution.

$$\begin{gathered} \mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{t} \\ t=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }} \\ t=\frac{2\mathrm{Ï€}}{0.098} \\ t=64s \end{gathered}$$

Hence, it takes 64 seconds to complete a revolution.