Question

A four-wheeled trolley car of mass 2500 kg runs on rails, which are 1.5 m apart and travels around a curve of 30 m radius at 24 km.hr. the rails are at the same level. Each wheel of the trolley is 0.75 m in diameter and each of the two axels is driven by a motor running in a direction opposite to that of the wheels at a speed of five times the speed of rotation of the wheels. The moment of inertia of each axle with gear and wheels is 18 kg-m² . Each motor with shaft and gear pinion has a moment of inertia of 12 kg-m² . The center of gravity of the car is 0.9 m above the rail level. Determine the vertical force excreted by each wheel on the rails taking into consideration the centrifugal and gyroscopic effects. State the centrifugal and gyroscopic effects on the trolley.

Answer 1

Given m = 2500 kg; x = 1.5 m; R = 30 m ; v = 24 km/h  = 6.67 m/s ;= 15 m/s ;  d_W = 0.75 m or r_w = 0.375 m; G = ω_E/ω_w = 5 ; I_w = 18 kg-m²; I_E = 12 kg-m²; h = 0.9 m

The weight of the trolley (W = m.g) will be equally distributed over the four wheels, which will act downwards. The reactions. The reaction between the wheels and the road surface of the same magnitude will act upwards.

Road reaction over each  wheel

= W/4 = m.g/4 = 2500 x 9.81/4 = 6131.25 N

We know that angular velocity of the wheels,

ω_w = v/r_W = 6.67/0.375 = 17.8 rad/s

and  angular velocity of precession,

 ω_p = v/R = 6.67/30 = 0.22 rad/s

Gyroscopic couple due to one pair of wheels and axle,

Gyroscopic couple due to one pair of wheels and gears,

Net Gyroscopic couple 

C = C_W  - C_E = 141 - 470 = -329 N-m

Due to this net Gyroscopic couple, the vertical reaction on the rails will be produced. Since C_E  is greater than C_W, therefore the reaction will be vertically downwards on the outer wheels and vertically upwards on the inner wheels. Let the magnitude of this reaction at each of the outer or inner wheel be P/2 newton.

P/2 = C/2x 329/2 x 1.5 = 109.7 N

We know that centrifugal force,

F_C = m.v²/R = 2500(6.67)²/30 = 3707 N

C_o = F_C x h = 3707 x 0.9 = 3336.3 N-m

This overturning couple is balanced by the vertical reactions which are vertically upwards on the outer wheel and vertically downwards on the inner wheels.

Let the magnitude of this reaction at each of the outer or inner wheels be Q/2 newton.

Q/2 = C_o/2x 3336.3/2 x 1.5 = 1112.1 N

We known that vertical force exerted on each outer wheel,

P_o = W/4 - P/2 + Q/2 = 6131.25 - 109.7 + 1112.1 = 7142.65 N

and vertical force exerted on each inner wheel.

P₁ = W/4 + P/2 - Q/2 = 6131.25 + 109.7 - 1112.1 = 5128.85 N