Question

The radius of curvature of a railway track at a place, where the train is moving at a speed of 72 km h−1 is 625 m. The distance between the rails is 1.5 m. Find the angle and the elevation of the outer rail, so that there may be no side pressure on the rails. Take = 9.8 ms−2.

Answer 1

The radius of curvature of a railway track at a place, where the train is moving at a speed of 72 km/h is 625 m. the distance between the rails is 1.5 m

To find : find the angle and elevation of the outer rail, so that there may be no side pressure on he rails.

solution : to find the angle, θ

using formula, tanθ = v²/rg

where v = 72 km/h = 72 × 5/18 = 20 m/s

r = 625 m and g = 9.8 m/s²

so, tanθ = (20)²/(625 × 9.8) = 0.0653 ......(1)

⇒θ = tan¯¹ (0.0653) = 3.74°

Therefore angle is 3.74°

now using, tanθ = elevation of the outer rail/distance between rails.

⇒0.0653 = elevation of outer rail/1.5 m [ from eq (1). ]

⇒elevation of outer rail = 0.0653 × 1.5 m = 0.098 m = 9.8 cm

Therefore the elevation of outer rail is 9.8 cm