Net reaction of ground on wheels due to gyroscopic couple, due to wheels and the dead weight and centrifugal force of a vehicle negotiating a curve is

This question was previously asked in

MPSC AMVI Official Paper 2: Set A/2013

Option 2 : decreased on inner wheels and increased on outer wheels

CT 3: Building Materials

2962

10 Questions
20 Marks
12 Mins

__Explanation:__

When a four-wheeler takes a turn, the axis of rotating parts (i.e. wheels and engine) undergoes precession and the gyroscopic couple will be acting on these rotating parts.

Also as the vehicle is moving along a curved path, the centrifugal force will also be acting on the vehicle. This force will also give a couple.

The total reaction at the wheels will be due to the weight of the vehicle, due to centrifugal force and due to the gyroscopic couple.

1) The weight of the vehicle is equally divided among four wheels, then weight on each wheel is W/4 in a downward direction and the reaction of the ground at each wheel due to the weight W = W/4 in an upward direction.

2) Couple due to centrifugal force.

The vehicle is moving along a curved path, hence a centrifugal force will also be acting on the vehicle in the outward direction. This force acts through the center of mass of the vehicle (i.e. through CG).

Here the magnitude of centrifugal force is P.

Therefore, couple due to centrifugal force, Cc = P × h

where h = distance from ground level to CG

The above is the overturning couple acting on the vehicle. The resisting couple will be acting on the wheels which is equal to the couple due to equal and opposite forces acting on outer or inner wheels.

Therefore, the force on each inner wheel is P/4 in an upward direction.

3) Gyroscopic effects on four-wheel vehicles taking a left turn

W = weight of the vehicle, \(\frac{W}{4}\) = reaction from the road on each wheel, Iw = Moment of inertia of each wheel, IE = Moment of inertia of engine flywheel, ωw = Angular velocity of each wheel, ωE = Angular velocity of each engine, ωP = Angular velocity of precision

The gyroscopic couple on a vehicle is,

C = Cw + CE

C = 4 × IW × ωW × ωp ± IE × ωE × ωP

**Due to gyroscopic couples, there will be vertical reactions on wheels, on outer wheels, it would be positive, and on inner wheels, it would be negative.**