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Skateboarder in halfpipe (Uniform Circular Motion Problem)

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data

    Public skateboarding parks often include a well in the shape of a half cylinder (half-pipe). The skateboarder's path traces out a semicircular arc, with the midpoint of the arc at the lowest point. When starting from rest at one of the upper edges of the arc (at height h), the horizontal speed attained at the bottom can be shown to be the same as for vertical free-fall, v=[tex]\sqrt{2gh}[/tex]. At the bottom point, by what factor does the normal force from the skateboard on the skateboarder's feet exceed his weight?

    My comments:

    The force diagram of the skater at the bottom of the half pipe in relation to an x-y axis with horizontal and vertical directions: Normal force is pointing straight up, toward the center of the half-pipe and weight is pointing straight down, away from the center of the half pipe. In a separate diagram, centripetal acceleration is pointing straight up, toward the center of the half-pipe and centripetal force is pointing straight down, away from the center of the half pipe.



    2. Relevant equations

    Summation of Forces on the x-axis = 0

    weight (W) on the x-axis = 0

    Summation of Forces on the y-axis = m (v[tex]_{2}[/tex]/r)

    Weight (W) on the y-axis = -mg
    Normal (N) = mg

    3. The attempt at a solution

    N - W = m (v[tex]_{2}[/tex]/r);
    N -mg = m (v[tex]_{2}[/tex]/r);
    N = mg(1 + 2h/r)
     
    Last edited: Feb 27, 2010
  2. jcsd
  3. Feb 27, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi Jim01! Welcome to PF! :smile:

    (have a square-root: √ and try using the X2 tag just above the Reply box :wink:)

    Yes, your method is fine … you've used Ftotal = ma, and you've carefully kept the acceleration diagram away from the forces diagram, so you don't get them mixed up! :smile:

    (oh, except the bracket should be after the "r")

    Just one comment …
    There's no such thing as centripetal force (unless you're in a rotating frame, like in a car going round a corner, in which case it's centrifugal force, pointing away from the centre), only weight, normal force, etc. :wink:
     
  4. Feb 27, 2010 #3
    Thank you for the welcome and your reply tiny-tim. Also, thanks for the list of symbols.

    As far as the centripetal force, I thought every time you had centripetal acceleration you had centripetal force acting in the opposite direction? So by rotating frame, you mentioned a car going around a turn. I'm guessing it would also include some sort of structure in space rotating in order to produce gravity? I'm confused on when centripetal force will be there and when it will not. For instance, will there be centripetal force when a car goes over a hill or if a pail of water is swung in a vertical circle?

    I seem to have no problem drawing the force diagrams, nor do I seem to have problems figuring out the equations for each force. The main problem I have with these type of problems is trying to figure out how to put them together to come up with the correct summation of forces.
     
    Last edited: Feb 27, 2010
  5. Feb 27, 2010 #4

    tiny-tim

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    Hi Jim01! :smile:
    i] "centripetal" means toward the centre (from Latin peto, meaning I seek), so there's no centripetal anything away from the centre (of curvature)!

    ii] thinkF = ma … so the force must always be in the same direction as the acceleration, mustn't it? :wink:

    iii] all the forces combined will produce a resultant force which must of course be centripetal, so you could call that resultant the centripetal force … but that leads to "double accounting", and is just plain confusing. :redface: Forget it!
    Yes, all these examples will have a centrifugal force (not a centripetal one!!), but only if you choose the non-inertial rotating frame as your frame of reference.

    Centrifugal force is a "fictitious" force, invented to make a non-inertial frame of reference work.

    See the PF Library on centrifugal force :smile:
    "factor" means multiplicative.

    So A exceeds B by a factor of (A-B)/B. :wink:
     
  6. Feb 27, 2010 #5
    Of course that makes complete sense both mathematically and linguistically. However, this raises some confusion due to physics problem my professor put on the board involving one of those amusement park rides which spins around in a circle, pinning the occupants against the wall before dropping the floor out from beneath them.

    In that problem, static friction points straight up, perpendicular with the center of the circle, weight points straight down, perpendicular with the center of the circle. For a person to remain stationary, these two forces must equal zero. The normal force points inward towards the center of the circle and equals mv2/r. However, for there to be a normal force pointing towards the center of the circle, mustn't there also be a force pointing the opposite way which also must be equal and opposite to the normal force (I believe these two forces must also equal zero for a person to remain stationary)? I could of sworn that the professor called this force the centripetal force. I am obviously wrong, so what is that force?


    I will do that.
     
  7. Feb 27, 2010 #6

    tiny-tim

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    Nope … or rather, it depends which frame you're using.

    Using an inertial frame, the person isn't stationary, he's accelerating towards the centre, and so that acceleration must equal the normal force (which you could call the centripetal force … centripetal force = mass time centripetal acceleration)

    Using the non-inertial rotating frame, the person is stationary, so the normal force (inward) must be balanced by the equal and opposite centrifugal force (outward).

    Just remember …
    you never have both centripetal acceleration and centrifugal force in the same frame (ie in the same equation).​
    If the professor meant that the centripetal force keeps the person stationary, he's completely wrong, it keeps the person accelerating!

    If he meant centrifugal force, that's ok. :smile:
     
  8. Feb 27, 2010 #7
    Thank you very much for your time and patience in explaining this to me. I really appreciate it.

    Please forgive my thick headedness, I just want to make sure I understand this. Would it be correct to say that in circular motion problems not involving friction, tension, or a rotating frame that there are only three forces: weight, normal and the summation of weight and normal?
     
  9. Feb 27, 2010 #8

    tiny-tim

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    No.

    There are only two forces: weight and normal.

    Thinking of the total (net) force as an extra force is exactly what leads to "double accounting". :wink:
     
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