TGV Train Circular Motion Calculations

Click For Summary
SUMMARY

The discussion focuses on calculating the radius of curvature for the TGV train's track and the required speed for a specific radius while maintaining a passenger acceleration limit of 0.050g. The formula used is a = v²/r, where 'a' is acceleration, 'v' is velocity, and 'r' is radius. The correct calculations reveal that the smallest radius of curvature tolerable at 216 km/h is approximately 72 km, while the speed required to maintain the acceleration limit at a 1.03 km radius is approximately 25.92 km/h. The misunderstanding lies in the conversion of acceleration units and the application of the formula.

PREREQUISITES
  • Understanding of circular motion physics
  • Familiarity with the acceleration formula a = v²/r
  • Ability to convert units between km/h and m/s
  • Knowledge of gravitational acceleration (g = 9.8 m/s²)
NEXT STEPS
  • Review the derivation and application of the formula a = v²/r in circular motion
  • Practice unit conversions between km/h and m/s for various physics problems
  • Explore the implications of acceleration limits in transportation engineering
  • Investigate real-world applications of circular motion calculations in train design
USEFUL FOR

Students studying physics, particularly in mechanics, transportation engineers, and anyone interested in the dynamics of high-speed trains like the TGV.

shell4987
Messages
43
Reaction score
0

Homework Statement


The fast French train known as the TGV (Train Grande Vitesse) has a scheduled average speed of 216 km/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? (in km) (b) At what speed must the train go around a curve with a 1.03 km radius to be at the acceleration limit? (in km/h)

Homework Equations


a=v squared/r


The Attempt at a Solution


I tried this problem and for part (a) i got 72 km which didn't turn out to be right, i used the acceleration formula and converted the units to m/s then reconverted them back to km/hr for acceleration, i had it equal to 0.050 m/s squared and for velocirty i had 60 m/s...

for part (b) i used 0.050 m/s squared as the acceleration again and then 1030m as the radius and solved, then converted it back to km/hr and got the answer to be 25.92km/hr

I don't know what I'm doing wrong, can anyone help me out? thanks.
 
Physics news on Phys.org
0.05g means 0.05*(9.8 m/s^2).
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
Replies
10
Views
4K
Replies
7
Views
3K
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K