Question

A cylindrical disc of a gyroscope of mass, $ m = 15kg $ and radius $ r = 5cm $ spins with an angular velocity $ \omega = 330rad{s^{ - 1}} $ . 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 $ {\varphi _m} = {20^ \circ } $ . Find the maximum value of the gyroscopic forces exerted by the axle on the bearings in $ N $ .

Answer 1

Hint : In order to solve this question, we are going to first find the moment of the inertia then, from that the torque is found which gives the precession for the disc. After finding the rate of change of the precession, we get the maximum torque and hence the corresponding gyroscopic force.
Formula used: The moment of inertia,
 $ I = \dfrac{1}{2}m{r^2} $
The angular momentum is given by,
 $ L = m{r^2}\omega $
The gyroscopic force
 $ F = \dfrac{{\pi m{r^2}\omega {\varphi _m}}}{{Tl}} $

Complete Step By Step Answer:
It is given that
The mass of the cylindrical disc of gyroscope, $ m = 15kg $
Radius $ r = 5cm $
Angular velocity $ \omega = 330rad{s^{ - 1}} $
Distance between the bearings, $ l = 15cm $
Now as we know that the moment of inertia,
 $ I = \dfrac{1}{2}m{r^2} $
The angular momentum is given by,
 $ L = m{r^2}\omega $
The axis oscillates about a horizontal axis making an instantaneous angle
 $ \varphi = {\varphi _m}\sin \dfrac{{2\pi t}}{T} $
Now as the axle is forced to oscillate about an axis,
Time period, $ T = 1.0s $
Amplitude, $ {\varphi _m} = {20^ \circ } $
This means that there is a variable precession with a rate of precession,
 $ \dfrac{{d\varphi }}{{dt}} $
The maximum value of this is $ \dfrac{{2\pi {\varphi _m}}}{T} $
The torque on the axle when the precession maximum is $ \tau = I\omega \dfrac{{d\varphi }}{{dt}} $
Putting the values in this, we get
 $ \tau = \dfrac{1}{2}m{r^2}\omega \times \dfrac{{2\pi {\varphi _m}}}{T} = \dfrac{{\pi m{r^2}\omega {\varphi _m}}}{T} $
The corresponding gyroscopic force can be calculated as $ \dfrac{\tau }{l} $
This implies, $ F = \dfrac{{\pi m{r^2}\omega {\varphi _m}}}{{Tl}} $
Putting the values and further solving
 $ F = \dfrac{{3.14 \times 15 \times {5^2} \times {{10}^4} \times 330 \times {{20}^ \circ }}}{{1 \times 15 \times {{10}^2}}} = 90N $ .

Note :
Precession is the phenomenon in which the axis of a spinning object describes a cone in space when an external torque is applied to it. The gyroscopic forces make the spinning object to maintain its orientation of the rotation motion. This is commonly seen in spinning objects but all rotating objects show this.