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Q5.31:
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 $10^6$ kg. What provides the centripetal force required for this purpose — The engine or the rails ? What is the angle of banking required to prevent wearing out of the rail ?

Solution :

Step 1: Identifying Given Variables and Unit Conversion

To perform calculations in the SI system, we convert the speed from $\text{km/h}$ to $\text{m/s}$.

  • Radius of the track, $r = 30 \, \text{m}$
  • Speed of the train, $v = 54 \, \text{km/h} = 54 \times \left(\frac{5}{18}\right) \, \text{m/s} = 15 \, \text{m/s}$
  • Mass of the train, $m = 10^6 \, \text{kg}$
  • Acceleration due to gravity, $g \approx 9.8 \, \text{m/s}^2$

Step 2: Identifying the Source of Centripetal Force

In the case of a train on an unbanked circular track, the centripetal force required to keep the train in circular motion is provided by the lateral thrust of the outer rail against the wheel flanges [Per Newton’s Third Law of Motion]. The engine provides the tractive force (tangential force) to overcome rolling friction and air resistance, but it does not provide the radial centripetal force required for turning. The rails exert a normal reaction force on the flanges, which acts as the centripetal force.

Step 3: Calculating the Required Angle of Banking

To prevent the lateral wear of the rails, the track should be banked at an angle $\theta$ such that the necessary centripetal force is provided by the horizontal component of the normal reaction force. The formula for the banking angle is given by:

$\tan(\theta) = \frac{v^2}{rg}$

Step 4: Algebraic Substitution and Execution

Substitute the values into the banking equation:

$\tan(\theta) = \frac{(15)^2}{30 \times 9.8}$

$\tan(\theta) = \frac{225}{294}$

$\tan(\theta) \approx 0.7653$

Calculating the inverse tangent: $\theta = \arctan(0.7653) \approx 37.43^\circ$

Visual Representation of Banking Mechanics

Normal Force (N) mg θ

Data Summary Table

Parameter Value
Centripetal Source Lateral thrust of outer rail
Required $\theta$ $37.43^\circ$

Final Solution: The centripetal force is provided by the lateral thrust of the outer rail against the wheel flanges. The required angle of banking to prevent rail wear is approximately $37.43^\circ$.


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