Calculating Train Speed Using Centripetal Force | 15° Angle, 150m Radius

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SUMMARY

The discussion centers on calculating the speed of a subway train using centripetal force principles, specifically in the context of a 150m radius unbanked curve and a 15-degree angle of a hanging strap. The correct gravitational acceleration is established as 9.8 m/s², which is essential for accurate calculations. The initial calculation incorrectly used 35.28 km/h as gravitational acceleration, leading to dimensional inconsistencies. The correct approach confirms that the train's speed exceeds 35 km/h, with the final calculation yielding a speed of 4.44 km/h based on proper unit conversions and consistent application of physics principles.

PREREQUISITES
  • Understanding of centripetal force and its formula F=ma=m(v²/r)
  • Knowledge of gravitational acceleration (9.8 m/s²) and its units
  • Familiarity with unit conversion between metric and imperial systems
  • Basic principles of dimensional analysis in physics
NEXT STEPS
  • Study the application of centripetal force in real-world scenarios
  • Learn about dimensional analysis and its importance in physics calculations
  • Explore the implications of unbanked curves on train safety and speed limits
  • Investigate the effects of different angles on centripetal force calculations
USEFUL FOR

Physics students, engineering professionals, and anyone involved in transportation safety and dynamics will benefit from this discussion, particularly those focused on the calculations of forces acting on moving vehicles.

Slam

Homework Statement


There is a subway derailed. Radius of an unbanked curve is 150 m. An unused strap hangs at a 15 degrees angle to the vertical just before the accident. Did the train exceed 35 km/h and what speed was it at just before the accident.

Homework Equations


F=ma=m(v^2/r)

The Attempt at a Solution


The angle of the Normal force is 75 degrees counterclockwise to the horizontal axis. The x-component of Force is m(v^2/r)=Ncos75
The y-component of Force is 0=Nsin75-mg
m=(Nsin75)/g Substitute this in for m in the x-component
((Nsin75)/g)(v^2/r)=Ncos75
v^2=g*r*tan75
g=35.28 km/h
r=.150 km
v=4.44 km/h
 
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You have the wrong numerical value for g. Probably resulting from using the wrong physical dimension. The gravitational acceleration is 9.8 m/s^2, not 9.8 m/s. The unit km/h is a unit of velocity, not of acceleration.

You should note that your equation would not be dimensionally consistent if g had dimension L/T.
 
Slam said:
g=35.28 km/h
This is dimensionally and numerically incorrect..
Work with consistent units: 1km = 1000m
1hr = 3600 sec.
That's why they invented the SI unit system! :smile:
Everything else you did looks right.
The answer is way higher than what you came up with.
 
Slam said:
g=35.28 km/h
As others have noted, your problem is a failure to keep track of units. If you multiply 9.8m/s2 by 3.6 (km/h)/(m/s) you get 35.28 km/h/s. Multiplying that by .15 km yields units of km2/h/s, not (km/h)2.
 

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