Problem-Smallest radius of curvature

  • Thread starter petal5
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  • #1
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I've been stuck on this for ages and would appreciate help on how to do it:

On a train, the magnitude of the acceleration experienced by the passengers is limited to 0.050g.If the train is going round a curve at a speed of 220km/hr what's the smallest radius of curvature that the curve can have without exceeding the maximum allowed acceleration on the passengers.
 

Answers and Replies

  • #2
Doc Al
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What have you done so far? Hint: What kind of acceleration is experienced when going around a curve?
 
  • #3
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so far i've attempted to solve using v=r w
 
  • #4
Doc Al
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Here's another hint: What have you learned about centripetal acceleration?
 
  • #5
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em,in circular motion direction constantly changes.Centripetal acceleration is the resulting center directed acceleration.

Do I need the formula: a=v^2/r ?
 
  • #6
Doc Al
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petal5 said:
Do I need the formula: a=v^2/r ?
Yes you do! :wink:

When going around a curve, the acceleration is centripetal--that's what they are talking about in this problem.

Be sure to convert everything to standard units before calculating the radius.
 
  • #7
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I'm getting an answer of 74.42km for the radius.Does this sound about right to you?
 
  • #8
Doc Al
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How did you arrive at that number?
 
  • #9
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I said 0.050g=v^2/r (taking v to be 61m/s)
 
  • #10
Doc Al
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You are forgetting to divide by g, which equals 9.8 m/s^2.
 
  • #11
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Thanks for all your help!So should my equation be: v^2/r=(0.050)(9.80)
 
  • #12
Doc Al
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That's right.
 
  • #13
OlderDan
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petal5 said:
Thanks for all your help!So should my equation be: v^2/r=(0.050)(9.80)
That looks good. Solve the equation for r. You know v. Be careful with your units.
 

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