A massive uniform rigid circular disc is mounted on a frictionless bearing at the end E of a massive uniform rigid shaft AE which is suspended horizontally in a uniform gravitational field by two identical light inextensible strings AB and CD as shown, where G is the center of mass of the shaft-disc assembly and 𝑔 is the acceleration due to gravity. The disc is then given a rapid spin 𝜔 about its axis in the positive x-axis direction as shown, while the shaft remains at rest. The direction of rotation is defined by using the right-hand thumb rule. If the string AB is suddenly cut, assuming negligible energy dissipation, the shaft AE will

The correct option is: rotate slowly (compared to 𝜔) about the negative z-axis direction.

1. Initial State and Coordinate Setup:
Based on the coordinate system provided in the diagram:
• The shaft $AE$ lies along the positive $x$-axis.
• The vertical direction is represented by the $z$-axis (positive upwards, along which gravity acts downwards with acceleration $g$).
• The $y$-axis is perpendicular to the $x$- and $z$-axes, defining a right-handed system.
• The disc at end $E$ spins rapidly at angular velocity $\omega$ in the positive $x$-axis direction. Thus, the spin angular momentum vector ${\overrightarrow{L}}_s$ is:

${\overrightarrow{L}}_s=I\omega \hat{i}$

where $I$ is the moment of inertia of the disc about its spin axis.

2. Effect of Cutting String AB:
• Originally, the shaft is supported by two strings: one at $A$ (string $AB$) and one at $C$ (string $CD$).
• The center of mass of the shaft-disc assembly, $G$, is located to the left of the support point $C$ (towards point $A$).
• When the string $AB$ is cut, the point $C$ acts as the instantaneous pivot point.
• The force of gravity ${\overrightarrow{F}}_g$ acts downwards at $G$:

${\overrightarrow{F}}_g=-mg\hat{k}$

• The position vector of the center of mass $G$ relative to the pivot $C$ is in the negative $x$-direction:

${\overrightarrow{r}}_{G/C}=-d\hat{i}$

where $d$ is the horizontal distance between $G$ and $C$.

3. Torque Due to Gravity:
The gravitational torque $\overrightarrow{\tau }$ about the pivot point $C$ is:

$\overrightarrow{\tau }={\overrightarrow{r}}_{G/C}\times {\overrightarrow{F}}_g=\left(-d\hat{i}\right)\times \left(-mg\hat{k}\right)=mgd\left(\hat{i}\times \hat{k}\right)$

Since $\hat{i}\times \hat{k}=-\hat{j}$, the torque vector is directed along the negative $y$-axis:

$\overrightarrow{\tau }=-mgd\hat{j}$

4. Gyroscopic Precession:
The torque causes a change in the angular momentum vector over time according to:

$\overrightarrow{\tau }=\frac{d\overrightarrow{L}}{dt}$

For a system precessing at an angular velocity of precession ${\overrightarrow{\Omega }}_p$, the relationship between torque and precession is:

$\overrightarrow{\tau }={\overrightarrow{\Omega }}_p\times {\overrightarrow{L}}_s$

Let ${\overrightarrow{\Omega }}_p={\Omega }_z\hat{k}$ represent the precession about the vertical $z$-axis. Substituting the vectors:

$-mgd\hat{j}=\left({\Omega }_z\hat{k}\right)\times \left(I\omega \hat{i}\right)=I\omega {\Omega }_z\left(\hat{k}\times \hat{i}\right)=I\omega {\Omega }_z\hat{j}$

Solving for ${\Omega }_z$:

${\Omega }_z=-\frac{mgd}{I\omega }$

The negative sign indicates that the precession angular velocity vector points in the negative z-axis direction. Since the disc's spin $\omega$ is very rapid, the magnitude of the precession angular velocity $\left|{\Omega }_z\right|$ is small compared to $\omega$. Therefore, the shaft $AE$ will rotate slowly about the negative $z$-axis direction.