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Homework Help: Extreme Confusion of centripetal acceleration force and normal force

  1. Jun 28, 2013 #1
    Hi guys, first off, thanks for answering my previous question...

    Now here's another one. I got confused by my school's homework a bit...so there's a car that travels over a humpback bridge with a radius of 45m, so basically it's a semicircle bridge, and so I'm supposed to calculate the maximum speed of the car if the wheels are to just stay in contact with the bridge. I don't get what the question mean when it says "just in contact with the bridge" does it mean that normal contact force will be zero? I imagine it to be right on top of this bridge, and checked the answer. I know the net force is the one that gives the centripetal force, and so

    mg - N (where N is just the magnitude of normal force alone) must be greater or equal to mv^2/r. Is this correct? By the way, when I say greater than, I'm already assuming the car doesn't have constant speed along the bridge as it travels right?

    Next, the answer told me that for the wheels to stay "just in contact with the bridge", the N at the top of the bridge will be zero. This, I don't get it at all, why would there even be N=0? The car is always in contact with the bridge. Does "just in contact" mean N=0? If that is the case, is N= 0 at other parts of the bridge?

    Next, there's this ferry wheel model type of question, so imagine a object moving in loops around the Ferris wheel circumference (inner rim of the Ferris wheel of course, not on it). I understand that at the bottom of the loop, the N force must be greater than mg such that there will be a net centripetal force upwards turning the object. Then, at this bottom point, N must be 2mg such that 2mg-mg of the object will give a centripetal mg force on the object. THEN, at the top of the loop, it will make sense for N to be zero since the weight can entirely provide the centripetal force, but this is different from the car question above, so why is the car enjoying N=0 too? And there is something really funny about this Ferris wheel model as well, why must the net centripetal force be mg (at the top and bottom of the loop)?? There is just no explanation, I searched everywhere but there's nothing. What about at the sides of the loop? Must the centripetal force in this Ferris wheel model be mg too? These are extremely confusing qns to me..

    Any help would be greatly appreciated, thanks a lot!!!
  2. jcsd
  3. Jun 28, 2013 #2

    Andrew Mason

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    Just stick to the free-body diagram. There are two force vectors (2 forces. mg and N, and the vector representing the sum of those two forces, which is equal to the mass x acceleration. The only acceleration is centripetal.

    If the car were to go over at too great a speed and leave the surface it would be because the centripetal force required to keep it following the road cannot be supplied by mg - N. But at that point N = 0 since it is not touching. So Fc=mg = mv^2/r where r>45m. If it is just touching the road but not putting any force on the road (N = 0) gravity is able to supply the needed centripetal force ie. Fc = mg = mv^2/r where r=45m.

    Last edited: Jun 28, 2013
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