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Circular motion under gravity; a car travelling on hills

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data

    A car moves along a straight, but hilly road, at a constant speed. There is a crest and dip in this road, both with a radius of 250m.

    a) As the car passes over the peak of the crest, the normal force is half the 16kN weight of the car. What is the normal force at the bottom of the dip?

    // Here I'm already noticing that circular acceleration must be equal to gravity. so that 16kN = 2N = m(g+ ac). The normal force is a contact force. It opposed what is imposed, given the surface it's sitting on. In this case, the normal force equals mg or the force towards the centre of the circle. But not both. Why is that?​


    b) What is the greatest speed that the car can move without leaving the road at the top of the crest.

    // Once you pass the crest, you're relying on gravity to maintain your circular motion. So if ac>g at the peak, then your velocity will be too great for gravity to keep you on that curve. Your maximum velocity is v = √(gr)​


    c) Moving at the speed found in (b), what would your normal force be at the bottom of the dip?

    // Given my confusion in (a), I'm having issues conceptualizing this one. But, I believe it should also be equal to mg, or mac, but not both.​
     
    Last edited: Oct 8, 2012
  2. jcsd
  3. Oct 8, 2012 #2

    collinsmark

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    Let's step back to the basics. I'm hoping that using this will alleviate any confusion.

    Newton's second law of motion, expressed a little more generally than it sometimes is, states:

    [tex] m \vec a = \sum_i \vec F_i [/tex]

    In other words, an object's mass times its acceleration is equal to the sum [that is, vector sum] of all forces acting on that object.

    So now ask yourself, "what are the forces acting on the car?"

    The force of gravity is the car's weight, and that's already given to you in the problem statement. The normal force is the only other force acting on the car. And the magnitude of that force is given to you as well.

    So just sum them up as vectors (don't forget to pay attention to their directions) and you know what your [itex] m \vec a [/itex] is.

    You've already demonstrated that you know how to find [itex] v [/itex] after you've calculated [itex] a [/itex], so I'll let you take it from there. :smile:
    That looks right to me :approve:

    But just for practice (and clarity), can you find the same answer by using the method I discussed in above? (It's the same procedure, except the normal force is equal to zero. :wink:)
     
  4. Oct 8, 2012 #3
    I think you are off in your analysis of part a? Off by a sign I think? See below.
     

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