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Vertical Circular Motion problem

  1. Oct 19, 2013 #1
    1. The problem statement, all variables and given/known data
    A motorcycle has a constant speed of 25.0m/s as it passes over the top of a hill whose radius of curvature is 126m. The mass of the motorcycle and driver is 342kg. Find the magnitudes of (a) the centripetal force and (b) the normal force the acts on the cycle


    2. Relevant equations
    Fc = (mv^2)/r
    W = mg


    3. The attempt at a solution
    Well for part a I got the right answer,
    Fc = (mv^2)/r
    Fc = 342(25^2)/r
    Fc = 1696N => 1.70 x 10^3N

    however for part b i got the wrong answer. this is my thinking
    Since centripetal force always points towards the centre, and since the motorcycle is at the top of the hill, then the net force in the y direction is equal to = N-Fc-W
    Therefore N = Fc + W
    N = 1696 + mg
    N = 1696 + (342)(9.8)
    N = 5047.6
    however the right answer is 1.66 x 10^3N

    help please
     
  2. jcsd
  3. Oct 19, 2013 #2

    PhanthomJay

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    I am not sure whether you are incorrectly trying to use newton's first law when it should be his second, or else just misunderstanding the centripetal force concept. You correctly calculated the centripetal force, which is the net inward force comprised of the weight and normal forces. The centripetal force is not a force of its own. It is a net force which is the vector sum of the normal and weight forces in the inward direction.
     
  4. Oct 19, 2013 #3
    Why is centripetal force not a force on its own? If centripetal force is the net inward force comprised of the weight and normal forces then,
    Fc = N-W
    N = Fc+W
    and i still get the same thing? since the motorcycle is on top of the hill then normal force is upward and weight is downward hence the "N-W"
     
  5. Oct 19, 2013 #4

    PhanthomJay

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    Net force always acts in the direction of the acceleration. The acceleration, centripetal in this example, is inward, or down. Thus the net force must be inward, or down. If the net force is down, the weight must be greater than the normal force, right?
     
  6. Oct 19, 2013 #5
    ok so if net force acts in the direction of acceleration, and centripetal acceleration is downward.
    then fc is negative. so, -Fc = N-W
    N = Fc-W

    ?? is this correct
     
  7. Oct 19, 2013 #6

    PhanthomJay

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    You are continuing to treat Fc as a force of its own. It is not. Try Newton 2:

    [itex] F_{net} = ma[/itex]
    [itex] F_{net} = mv^2/r[/itex]
    [itex] mg - N = mv^2/r[/itex]
    [itex] mg - N = F_c[/itex]

    Solve for N.

    Note that the centripetal force is the net inward force that is the vector sum total, or net if you will, or resultant , of all forces acting in the inward direction. In uniform circular motion, the acceleration is inward, and the net force is thus inward, toward the center of the circle. Because the net force acts toward the center of the circle, it is called a centripetal, or center seeking, force.
     
  8. Oct 19, 2013 #7
    In the equation Fc = mg - N
    why is n treated as a negative
     
  9. Oct 20, 2013 #8

    PhanthomJay

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    It is an arbitrary use of signage. N is treated as negative, up, and mg is treated as positive, down. Acceleration, which acts down, is then also treated as a plus.

    You can reverse signs, using up plus and minus down, and get the same result. Looks like you tried that
    but your algebra was not so good, should be N = W -Fc.

    I find it easier to choose the direction of the acceleration as plus, but it is a personal choice that helps with minus sign errors. The choice however is yours.
     
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