Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations waross on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!

Gyroscopic Forces 1

Status
Not open for further replies.

cmb042

Geotechnical
Apr 28, 2008
39
0
0
US
I am trying to calculate the force imparted on a rotor bearing when the rotor is at speed and the vehicle takes a hard turn. I can find the rotor moment of inertia and its angular momentum, but not sure how to find the force due to gyroscopic action. Most of the information I have found focuses on precession and the shape and rate of precession. Seems like there should be a fairly straightforward way to get force from momentum when you rotate the spinning rotor in some direction other than along the shaft axis.
 
Replies continue below

Recommended for you

It is indeed "fairly straightforward". For linear motion the equation to use would be
m.(v - u) = F.t
where underlining indicates a vector quantity.[ ] You doubtless remember this equation from your high school physics, as "change in momentum equals impulse".

This has an exactly analogous equation for rotational motion, as "change in angular momentum equals angular impulse":
I.(V - U) = T.t
where I is the moment of rotational inertia, U & V are the before and after rotational velocities, and T is the average torque required to effect the change over the time t.

Just make sure you get the directions of the rotational vectors correct (and remember to use radians as your measure of rotational displacement).
 
Thank you all, especially Hydtools. I was able to find the answer I was looking for in another reference which is backed up by Hydtools link. Basically the gyroscopic force is a moment that results from the cross product of the angular momentum and the velocity of procession in radians/s. The moment of procession and angular momentum are at right angles, so the strait forward answer is the product of the magnitudes equal the gyroscopic moment, then it is just distance to bearing from center of mass times this moment for force.
 
Status
Not open for further replies.
Back
Top