A uniform disc with radius r and a mass of m kg is mounted centrally on a horizontal axle of negligible mass and length of 1.5r. The disc spins counter-clockwise about the axle with angular speed ω, when viewed from the right-hand side bearing, Q. The axle precesses about a vertical axis at ω_p = ω/10 in the clockwise direction when viewed from above. Let R_P and R_Q (positive upwards) be the resultant reaction forces due to the mass and the gyroscopic effect, at bearings P and Q, respectively. Assuming ω²r = 300 m/s² and g = 10 m/s², the ratio of the larger to the smaller bearing reaction force (considering appropriate signs) is ______ .

Correct Answer: The ratio of the larger to the smaller bearing reaction force is -3.

Let's analyze the system step-by-step to find the bearing reactions at P and Q due to gravity and the gyroscopic couple:

1. Parameters given in the problem and diagram:
Mass of the disc, $m$ kg
Radius of the disc, $r$
Length of the horizontal axle, $l=1.5r$
Angular speed of spin, $\omega$ (counter-clockwise when viewed from the right-hand bearing, Q)
Angular speed of precession, ${\omega }_p=\frac{\omega }{10}$ (clockwise when viewed from above)
Given parameters: ${\omega }^{2}r=300{\text{m/s}}^2$ and acceleration due to gravity, $g=10{\text{m/s}}^2$

2. Reactions due to the weight of the disc:
Since the disc is mounted centrally on the horizontal axle of negligible mass, the gravitational force $W=mg$ acts downwards at the center of the axle. Due to symmetry, the reaction force at each bearing directed upwards is:

$R_{P,\text{gravity}}=R_{Q,\text{gravity}}=\frac{mg}{2}=\frac{10m}{2}=5m$

3. Gyroscopic Couple (C):
The mass moment of inertia of the disc about the polar axis is:

$I=\frac{1}{2}m{r}^2$

The magnitude of the gyroscopic couple is given by:

$C=I\omega {\omega }_p$

Substitute the given values into the formula:

$C=\left(\frac{1}{2}m{r}^2\right)\omega \left(\frac{\omega }{10}\right)=\frac{mr\left({\omega }^{2}r\right)}{20}$

Given that ${\omega }^{2}r=300$:

$C=\frac{mr·300}{20}=15mr$

4. Bearing reactions due to the reactive gyroscopic couple:
Applying the right-hand rule for gyroscopic motion:
- The spin angular velocity vector points from P to Q (to the right).
- The precession angular velocity vector points vertically downwards.
- This results in a reactive gyroscopic couple that acts in a direction tending to lift the bearing at Q and press down the bearing at P.
Let $R_g$ be the magnitude of the reaction forces at the bearings P and Q due to this couple:

$R_g·l=C\Rightarrow R_g·\left(1.5r\right)=15mr$

$R_g=\frac{15mr}{1.5r}=10m$

Therefore, the reaction forces due to the gyroscopic effect are:
At Q (upward): $R_{Q,\text{gyro}}=10m$
At P (downward): $R_{P,\text{gyro}}=-10m$

5. Total reactions at bearings P and Q:
Adding the gravitational and gyroscopic components:

$R_Q=R_{Q,\text{gravity}}+R_{Q,\text{gyro}}=5m+10m=15m$

$R_P=R_{P,\text{gravity}}+R_{P,\text{gyro}}=5m-10m=-5m$

6. Ratio of the reaction forces:
The ratio of the larger bearing reaction force to the smaller bearing reaction force is:

$\frac{R_Q}{R_P}=\frac{15m}{-5m}=-3$