Forces acting on a gyroscope base when trying to turn it

[refusenik]

I am studying the influence of a gyroscopic effect of wheels/tyres on a steering system in a car. I can't seem to get my head around few concepts. It boils down to a simple problem.

Consider a free spinning gyroscope wheel with horizontally aligned axis.
This axis is mounted in a vertical fork style support that itself can rotate in a vertical fixed support. That's it.

If you imagine a bicycle with front wheel suspended above ground that's about right. The bicycle frame is attached to e.g. a wall but handlebar is free to rotate.

Now is the problem. Suppose the wheel has moment of inertia I around its horizontal axis and spins with angular velocity w. What I want to know is what extra effort would be required to turn the fork support around vertical axis with e.g. angular speed y.

I know that in this case the spinning wheel creates a moment acting on a fork support

M = I * w * y

but it is acting around horizontal axis (at right angle to the spin axis) and therefore is fully counteracted by reaction of vertical support (ball bearings.)

Will there be any extra effort required at all (compared to turning the same system with non-spinning wheel?)

 

[Valerie Park]

Yes, there will be an extra effort required due to the gyroscopic effect of the wheel. This is because the angular momentum of the spinning wheel creates a torque on the fork support which must be overcome in order to turn the fork support. This torque is calculated as:Torque = I * w * yWhere I is the moment of inertia of the wheel, w is the angular velocity of the wheel, and y is the angular speed of the fork support. Therefore, the extra effort required to turn the fork support is proportional to the angular velocity of the wheel.