The axis of gyroscope is tilted at an angle of ${30}^{\circ }$to the vertical (figure 11.110). The rotor has a radius of 15cm , mass 3kg , moment of inertia is $0.06Kg/m^2$and spins on its axis at $30rad/s$. It is supported in a cage (not shown) in such a way that without an added weigh it does not process. Then a mass of 0.2 kg is hung from the axis at a distance of 18 cm from the center of the rotor.

a. Viewed from above, does the gyroscope process in a (1) clockwise or a (2) counterclockwise direction? That is, does the top end of the axis move (1) out of the page or (2) into the page in the next instant? Explain your reasoning

b. How long does it take for the gyroscope to make one complete procession?

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 $\omega$is,

$\omega =\frac{{\pi }_{CM}}{L_{rot}}$

Here, ${\pi }_{CM}$is the torque on the center of mass and $L_{rot}=I\omega$

Here, l is the moment of inertia and $\omega$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 ${30}^{\circ }$with the lever arm of length r . The torque is

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

The angular speed of the precession is,

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

Hence, the angular momentum is 0.098 rad/s

There are $2\pi$radians in a single revolution.

$$\begin{gathered} \omega =\frac{2\pi }{t} \\ t=\frac{2\pi }{\omega } \\ t=\frac{2\pi }{0.098} \\ t=64s \end{gathered}$$

Hence, it takes 64 seconds to complete a revolution.