Do Opposite Rotating Discs Cancel Precessional Forces in Motorcycle Wheels?

  • Thread starter Thread starter top dawg
  • Start date Start date
  • Tags Tags
    Precession Vector
AI Thread Summary
The discussion focuses on the vector math of gyroscopic precession, specifically whether two opposite rotating discs of equal speed and mass can cancel precessional forces, particularly in the context of motorcycle wheel design. The inquiry highlights a lack of resources addressing the direction of rotation as a vector in existing analyses. Participants emphasize the need for mathematical clarity to resolve differing opinions on the subject. The conversation also touches on the external torque responsible for precession. Overall, a deeper understanding of the physics involved is sought to inform practical applications in motorcycle steering.
top dawg
Messages
2
Reaction score
0
Trying to find the vector math for gyroscopic precession - all web sites I've seen that deal with analysis don't include direction of rotation as a vector and I can't find a discussion in my texts on the subject.

The question I'm trying to answer is whether two opposite rotating discs of equal speed/mass will cancel precessional forces. This is being looked at in terms of a motorcycle wheel patent to make steering easier at high speed.

Everybody seems to have their own opinion on the subject and so it seems the only way to settle the question is with the math.

Thanks for any help
Robert
 
Physics news on Phys.org
Your angular velocity vector lies along the axis of rotation (perpendicular to the rotating wheel) in the same direction as the angular momentum vector (cross product of the radius vector and the linear velocity vector at some given instant of time).

What force is providing the external torque that's causing the precession?

Here's a link on precession that may help: HyperPhsyics - Precession Torques
 
Thanks Bob - the references mentioned will give me a little more info as well
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top