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For the Doc JK some more questions using Centripetal Force

  1. Oct 5, 2006 #1
    For the Doc...JK....some more questions using Centripetal Force

    A motorcycle has a constant speed of 23.1 m/s as it passes over the top of a hill whose radius of curvature is 172 m. The mass of the motorcycle and driver is 364 kg. Find the magnitude of (a) the centripetal force and (b) the normal force that acts on the cycle.

    I am confused on this one, I don't quite understand Vertical centripetal force as well as horizontal, any direction I should take to get started?


    one last one:

    A block is hung by a string from the inside roof of a van. When the van goes straight ahead at a speed of 18.5 m/s, the block hangs vertically down. But when the van maintains this same speed around an unbanked curve (radius = 157 m), the block swings toward the outside of the curve. Then the string makes an angle θ with the vertical. Find θ.

    V=18.5 m/s
    r = 157

    I am confused on this one as well...it seems like there should be some tension involved or atleast a weight, but I am not succeeding.

  2. jcsd
  3. Oct 5, 2006 #2


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    Staff: Mentor

    Think about the definition of centripetal force. a.) What is keeping the rider and cycle on the ground? b.) What mass is being pulled down onto to the cycle.

    On the mass on the sting, there is tension in the string, which is at some angle. Acting on the mass is gravity (weight) acting down, and a lateral force caused by the circular motion of the vehicle.

    The vertical component of tension in the string acts opposite gravity, the lateral component of tension acts opposite the mv2/r of the circular motion.
  4. Oct 5, 2006 #3
    In both of these problems you must realize the sum of the forces = 0 When the motorcycle reaches the very top of the hill all force components are in the Y-direction and they sum to 0 (the motorcycle isn't moving up and down at that point so they must equal 0). that being said you know the 3 forces gravity + normal force + ceptriptal force = 0. Just plug the values & proper signs and you should be good.

    The second one is similar. The sum of the forces in the x/y direction equal 0. There are 3 forces acting on the mass: tension, gravity, centripital. Break each into x/y components and then set Fx = 0 and Fy = 0. At that point you will have 2 equations and 2 unknowns. You should not be concerned about mass as it cancels out.
  5. Oct 6, 2006 #4
    thanks, this was extremely helpful. But I am still confused as to where to start.

    with my given information:

    m = 364 kg
    r = 172 m
    v = 23.1 m/s

    So in using this information, I found the Centripetal force using:

    Fc = mv^2/r =

    Fc = 364(23.1^2)/172 = 1129.26 <----- this is correct.

    I was under the impression that Normal Force (Fn) =

    Fn = mg = 364(-9.80) = -3570 <----------- This number is incorrect. (So is the positive value for it)

    Is there something that I am over looking?
  6. Oct 7, 2006 #5
    Fn = mg only when there are 2 forces acting in the Y-direction (and when the object isn't moving in the y-direction). There is a third force here. The reason FN = MG is because...

    Fy = mg - Fn
    0 = mg - fn
    Fn = mg

    with 3 forces

    Fy = Fn + centripitalforce - MG
    0 = Fn + centripitalforce - MG
    Fn = mg - centripitalforce

    You can try to think of the normal force as the force a scale would read. Yes, in general when you stop on a scale it will read what is equiv to mass*gravity. However, if you imagined a scale at the top of the hill that could read the motorcyles weight, you would realize that as the motorcycle goes faster and faster the scale would read its weight as less and less. Same principle if you put a scale in an elevator. If you stood on the scale while the elevator was at rest it would read m*g, but if the elevator accelerates downward it would read a lesser value. That value is the normal force.
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