A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10⁶ kg. What provides the centripetal force required for this purpose - The engine are the rails? What is the angle of banking required to prevent wearing out of the rail?
Answers
Answered by
7
Hey mate,
# Answer- 36.87°
## Explaination-
# Given-
r = 30
v = 54 km/h = 15 m/s
m = 106 kg
# Solution-
a) Here, centripetal force is provided by the lateral thrust of the rail on the wheel. The reaction force here causes wear and rear of the rail.
b) Angle of banking -
Angle of banking is given by,
tanθ = v^2/rg
tanθ = 152/(30×10)
tanθ = 225/300
tanθ = 0.75
Taking inverse,
θ = tan-1 (0.75)
θ = 36.87°
Therefore, the angle of banking should be 36.87° to prevent wear and tear of tyres.
Hope this helps you...
# Answer- 36.87°
## Explaination-
# Given-
r = 30
v = 54 km/h = 15 m/s
m = 106 kg
# Solution-
a) Here, centripetal force is provided by the lateral thrust of the rail on the wheel. The reaction force here causes wear and rear of the rail.
b) Angle of banking -
Angle of banking is given by,
tanθ = v^2/rg
tanθ = 152/(30×10)
tanθ = 225/300
tanθ = 0.75
Taking inverse,
θ = tan-1 (0.75)
θ = 36.87°
Therefore, the angle of banking should be 36.87° to prevent wear and tear of tyres.
Hope this helps you...
Answered by
8
Explanation:
Radius of the circular track, r = 30 m
Speed of the train, v = 54 km/h = 15 m/s
Mass of the train, m = 106 kg
The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail
The angle of banking θ, is related to the radius (r) and speed (v) by the relation:
tan θ = v2 / rg
= 152 / (30 × 10)
= 225 / 300
θ = tan-1 (0.75) = 36.870
Therefore, the angle of banking is about 36.87°.
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