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Newton's second law : circular motion

  1. Dec 6, 2015 #1
    1. The problem statement, all variables and given/known data
    A car travels along the perimeter of a vertical circle (radius = 0.25 km) at a constant speed of 30 m/s. What is the magnitude of the resultant force on the 60-kg driver of the car at the lowest point on this circular path?
    Question options:
    1) 0.37 kN
    2) 0.80 kN
    3) 0.22 kN
    4) 0.59 kN
    5) 0.45 kN


    2. Relevant equations
    F_gravity=mg
    F_centripetal=m(a_centripetal)
    a_centripetal = m (v^2/r)

    3. The attempt at a solution
    F=mg + ma_centripetal = m (g + a_centripetal) = 60 kg (9.8 m/s^2 + (30m/s)^2/(250m)) = 804N = 0.8kN
    but the answer I got was different from the answer i got with the question which is 0.22kN
    can someone please explain how is it 0.22kN?
     
  2. jcsd
  3. Dec 6, 2015 #2

    cnh1995

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    They have asked the resultant force and not the normal reaction. What should be the "resultant" force in this case?
     
  4. Dec 6, 2015 #3
    Perhaps you might want to reconsider why you are adding the centripetal and gravitational forces together? What is the relationship between centripetal force and resultant force?
     
  5. Dec 6, 2015 #4
    "What is the magnitude of the resultant force on the 60-kg driver"
    it asked about the force exerted on the driver, so by Newton's third law, the magnitude of the force exerted on the driver is equal to the magnitude of the force the driver exerts on the track (action reaction pair) which in this case will be the normal force plus centripetal force as the driver is at the lowest point.


    please correct me if am wrong.
     
  6. Dec 6, 2015 #5

    cnh1995

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    To move in a perfect circle, the resultant force should always be equal to the centripetal force.
     
  7. Dec 6, 2015 #6
    I am adding because at the lowest point the gravitational and centripetal forces are in the same direction, therefore the resultant will be the addition of the two.
     
  8. Dec 6, 2015 #7

    cnh1995

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    \
    How so? Centripetal force is always directed towards the center of the circular path.
     
  9. Dec 6, 2015 #8
    sorry about that my mistake, but should i subtract then? that did not give me a correct answer either.
    in my calculation the answer is equal to the centripetal force only, so it doesn't matter where on the trak the object is, it will be the same answer?
     
  10. Dec 6, 2015 #9

    cnh1995

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    Resultant force will always be the same i.e. centripetal force. Normal reaction will change depending on the position of the object.
     
  11. Dec 6, 2015 #10
    al
    alright, thanks for putting up with me :)
     
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