Question

A uniform disc with radius r and a mass of m kg in mounted centrally on a horizontal axie of negligible mass and length of 1.5r. The disc spins counter-clockwise about the aise with angular speed $\omega$, when viewed fromthe right-hand side bearing, Q. The axis precesses about a vertical axis at ${\omega }_p=\omega /10$ 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 bearing P and Q, respectively.
Assuming $\omega r=300m/s^{2}andg=10m/s^2$, the ratio of the larger to the smaller bearing reaction force(considering appropriate signs ) is

Answer 1

Active gyroscopic couple $=I\omega {\omega }_p$
$=\frac{m{r}^2}{2}\times \omega \times \frac{\omega }{10}=\left(\frac{m{\omega }^2{r}^2}{20}\right)$
Reaction force at bearing P due to gyroscopic couple is ($R_{p_m})$
$R_{p_m}=\frac{m{\omega }^2{r}^2}{20\times 1.5r}$
$=\frac{m{\omega }^2{r}^2}{20\times 1.5r}=\frac{m{\omega }^{2}r}{30}=10m$ N (upward)
Reaction force at bearing Q due to gyroscopic couple is ($R_{Q_m})$
$R_{Q_m}=\frac{m{\omega }^2{r}^2}{20\times 1.5r}=\frac{m\omega r}{30}$
Reaction force at bearing P due to gravity
$=\frac{mg}{2}$ (downward) = 5m N
Reaction force at bearing Q due to gravity
$=\frac{mg}{2}=5$m N (upwared)
Reaction at bearing P,
$R_P=10mN-5mN=5mN$ [Upward]
Reaction at bearing Q,
$R_Q=10mN-5mN=-15mN$ [Upward]
$\frac{R_Q}{R_P}=\frac{-15}{5}=-3$