Gyroscopic Precession
GYROSCOPIC PRECESSION
Submitted by, AISON JOSEPH B110271ME GOKUL V NAIR B110073ME JOMY GEORGE B110036ME MATHEW JACOB B110199ME PRADEEP GOPI B110263ME
GYROSCOPIC COUPLE Consider the motion of a heavy disk of mass m. The disk is spinning with an angular velocity of . At the same time, the rod on which it is supported precesses about the vertical axis with an angular velocity . This system is in balance.
The reaction R of the vertical post is R = mg.
However there is a torque . It acts in the vertical plane XZ and its magnitude is mgr. It can be represented by a vector in +ve Y direction.
Now, by Euler's law, the rate of change of angular momentum is torque. Assume that the precession velocity is very small in comparison to spin velocity of the disk. The spin velocity may be written as , where I0 is the polar moment of inertia of the disk about X axis. is a rotating unit vector in the direction of X.
Now,
The direction of Ty is along , so it is consistent with the applied torque.
We have already discussed that applied torque is
.
Thus,
Giving us,
We observe that precesional angular velocity is inversely proportional to spin angular velocity. In case, the disk was not spinning, it would not have been possible to balance it on A. It would have started falling down in the vertical XZplane.
However, it does not happen here due to spin motion of the disk. If it has to fall in the vertical plane during spinning of the disk, there has to be a torque in the horizontal plane( represented by a vector along negative z-direction).
This is clear from the following figure:
When the disk has fallen by a small angle, the
Submitted by, AISON JOSEPH B110271ME GOKUL V NAIR B110073ME JOMY GEORGE B110036ME MATHEW JACOB B110199ME PRADEEP GOPI B110263ME
GYROSCOPIC COUPLE Consider the motion of a heavy disk of mass m. The disk is spinning with an angular velocity of . At the same time, the rod on which it is supported precesses about the vertical axis with an angular velocity . This system is in balance.
The reaction R of the vertical post is R = mg.
However there is a torque . It acts in the vertical plane XZ and its magnitude is mgr. It can be represented by a vector in +ve Y direction.
Now, by Euler's law, the rate of change of angular momentum is torque. Assume that the precession velocity is very small in comparison to spin velocity of the disk. The spin velocity may be written as , where I0 is the polar moment of inertia of the disk about X axis. is a rotating unit vector in the direction of X.
Now,
The direction of Ty is along , so it is consistent with the applied torque.
We have already discussed that applied torque is
.
Thus,
Giving us,
We observe that precesional angular velocity is inversely proportional to spin angular velocity. In case, the disk was not spinning, it would not have been possible to balance it on A. It would have started falling down in the vertical XZplane.
However, it does not happen here due to spin motion of the disk. If it has to fall in the vertical plane during spinning of the disk, there has to be a torque in the horizontal plane( represented by a vector along negative z-direction).
This is clear from the following figure:
When the disk has fallen by a small angle, the