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Circular Motion. Roller-coaster mass of car and max speed

  1. Mar 5, 2015 #1
    1. The problem statement, all variables and given/known data
    A roller-coaster car speeds down a hill pas point A and then rolls up a hill past point B.
    A. Car speed = 20 m/s at point A. If the track exerts a normal force on the car of 2.06x10^4N at this point, what is the mass of the car? (account for gravitational force)
    B. What is the maximum speed the car can have at point B for the gravitational force to hold it on the track?
    Diagram(I made it myself): http://gyazo.com/0415a6ba75f87f0229e08aab0f68fbe1
    2. Relevant equations
    Circular motion
    Fc= (mv^2)/r
    Fn-mg= (mv^2)/r

    3. The attempt at a solution
    Completely lost.
     
  2. jcsd
  3. Mar 5, 2015 #2

    PhanthomJay

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    I don't know why you are lost because your relevant equations are quite relevant for part A, assuming point A is at the bottom of dip. Solve for m.
     
  4. Mar 5, 2015 #3
    I should have clarified.We did most of A in class, B. is where I'm struggling at most.
     
  5. Mar 5, 2015 #4

    PhanthomJay

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    Ok, got it. For part B, the question asks for the max speed at point B if the only force in the vertical direction acting on the coaster at that point on the top of the curve is the gravity (weight) force that is keeping it on the track in centripetal motion, without flying off it. You should be able to manipulate your relevant equations to solve for that speed.
     
  6. Mar 5, 2015 #5
    Can you draw a free-body diagram for the cart at point B to help you out? What forces are acting on the cart?
     
  7. Mar 5, 2015 #6
    I don't think I can solve this problem using fn-mg=(mv^2)/r. The Ms would just cancel out?
     
  8. Mar 5, 2015 #7
    "Can you draw a free-body diagram for the cart at point B to help you out? What forces are acting on the cart?"
    It would be G only. Correct?
    It would be G only, correct?
     
  9. Mar 5, 2015 #8

    PhanthomJay

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    if only the gravity force is acting , what is fn? Let the m's cancel if they do. Now watch your signage. Draw a free body diagram. Net centripetal force is always in direction of the centripetal acceleration,
     
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