Question

Find the angle of inclination with respect to the vertical of a two wheeler negotiating a turn. Given: combined mass of the vehicle with its rider 250 kg; moment of inertia of the engine flywheel 0.3 kg-m² ; moment of inertia of each road wheel 1 kg-m² ; speed of engine 5 times that of road wheels and in the same direction; height of center of gravity of rider with vehicle 0.6 m; two wheeler speed 90 km/hr; wheel radius 300 mm; radius of turn 50 m. 

Answer 1

Given m = 250 kg; I_E = 0.3-m² ; 1 kg -m² ; I_W = 1 kg-m²; ω_E =5 ω_E 5 ω_W or G = ω_E/ω_W = 5; h = 0.6 m; v = 90 km/h = 25/s; r_W = 300 mm = 0.3 m; R = 50 m

Let θ = Angle of inclination with respect to the vertical of a two wheeler.

We know that gyroscopic couple,

Since the overturning couple must be equal to the balancing couple for equilibrium condition therefore

2021 cos θ = 1471.5 sinθ

tan θ = sinθ/cos θ = 2021/1471.5 = 1.3734 or θ = 53.94°