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Fundamentals Of Physics
Found in: Page 91

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Short Answer

The fast French train known as the TGV (Train à Grande Vitesse) has a scheduled average speed of216km/h . (a) If the train goes around a curve at that speed and the magnitude of the acceleration experienced by the passengers is to be limited to 0.050g, what is the smallest radius of curvature for the track that can be tolerated? (b) At what speed must the train go around a curve with a 1.00km radius to be at the acceleration limit?

a) Smallest radius of curvature of the trackis 7.3km.

b) Maximum allowed speed of the train is80km/h.

See the step by step solution

Step by Step Solution

Step 1: Given data

1) Average speed of the train is 216km/h

2) Acceleration experience by the passengers is limited to 0.050g

Step 2: Understanding the concept of centripetal acceleration

When an object moves in a circular orbit, it will experience a centripetal acceleration. The magnitude of this acceleration depends on the speed of the object as well as the radius of curvature. Using the formula for centripetal acceleration, we can find the radius of the curvature of the track if the velocity is known, or we can find the velocity if the radius of curvature is known.



Where acis the centripetal acceleration, vis the average speed, and ris the radius of curvature.

Step 3: Conversion of units

First, we need to convert the units of given speed from km/h to m/s


v=216kmh =216kmh×1000m3600s =60m/s

Step 4: (a) Calculating the smallest radius of curvature

To determine the smallest radius of curvature, we need to use formula forcentripetal acceleration.




Substitute 60m/sthe for v, and9.8m/s2 for g in the above equation.


Therefore, the smallest radius of curvature is7.3km .

Step 5: (b) Calculating the required speed for 1.00 km radius of curvature

Now, the radius of curvature available is given as,

r=1km =1000m


Substitute the values,

=0.050×9.8m/s2×1000mv=22.1m/s 80km/h

Hence, the speed at the curvature should be approximately 80km/h.

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