Question

A cylindrical disc of a gyroscope of mass $m=15kg$ and radius $r=5.0cm$ spins with an angular velocity $\omega =330rad/s$. The distance between the bearings in which the axle of the disc is mounted is equal to $l=15cm$. The axle is forced to oscillate about a horizontal axis with a period $T=1.0s$ and amplitude ${\phi }_m={20}^{\circ }$. Find the maximum value of the gyroscopic forces exerted by the axle on the bearings in $N$.

Answer 1

The moment of inertia is $\frac{1}{2}m{r}^2$ and angular momentum is $\frac{1}{2}m{r}^2\omega$.
The axle oscillates about a horizontal axis making an instantaneous angle.
$\phi ={\phi }_m\sin \frac{2\pi t}{T}$
This means that there is a variable precession with a rate of precession $\frac{d\phi }{dt}$.
The maximum value of this is $\frac{2\pi {\phi }_m}{T}$.
When the angle between the axle and the axis is at its maximum value, a torque
$I\omega \Omega$$=\frac{1}{2}m{r}^2\omega \frac{2\pi {\phi }_m}{T}=\frac{\pi m{r}^2\omega {\phi }_m}{T}$ acts on it.
The corresponding gyroscopic force will be $\frac{\pi m{r}^2\omega {\phi }_m}{lT}=90N$