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Fastest velocity possible for a 90 degree turn

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data
    What is the fastest possible velocity for a bicyclist travelling down a path with a width of 'w', can have while they go around a 90 degree corner, with a curvature with a radius of 'rc'? Friction = static friction [itex]\mu[/itex]s.


    2. Relevant equations
    Fac=(mV2)/2
    Ff=[itex]\mu[/itex]FN


    3. The attempt at a solution
    So I've started off by making a diagram of the path and the potential bike path, I've made it such that the radius rc at 45 degrees is the x-axis. I've moved the axis back to account for the new radius of the bikers path, which we can call radius effective. From there I'm able to go 45 degree in either direction but I'm confused as to where to go from here.

    This question was given as a 'challenge question' just now and it's got me stumped, any hints would be appreciated.
     
  2. jcsd
  3. Oct 11, 2011 #2

    PeterO

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    Does this path have a genuine right angle in it, so the cyclist can approach at the edge of the track, turn across the apex and finish at the edge of the track again?

    Given the cyclist will lean into the corner to achieve ANY turn, is there space to lean over the edge of the track?
     
  4. Oct 11, 2011 #3
    Yes, in order to acheive the max potential velocity the cyclist will start at the outside edge (r+w/2) and come to the inside of the track (r-w/2) in the apex with room to lean outside of the track. I would post a diagram but I'm on my blackberry at the moment. The geometry had me confused when I was working on this problem earlier.
     
  5. Oct 11, 2011 #4

    PeterO

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    Not sure why you are using R + w/2 and r - w/2??

    I would start by drawing a circle centred on the origin. [eg x2 + y2 = 9
    I am only now using the positive quadrant.
    Draw a horizontal tangent to the top, and a vertical tangent to the right.

    Establish the "45o position" [where y = x intersects the circle] and draw a horizontal and vertical line to that point.

    You then see the track, and the bend the rider will take.

    The track width is w and try to proceed from there.
     
  6. Oct 11, 2011 #5
    Here is a crude diagram I've drawn in paint since I have no scanner. This is what I've got so far basically:
    [PLAIN]http://img607.imageshack.us/img607/4168/diagramd.jpg [Broken]

    So this is a representation of the curve in the track. The radius goes from the axis to the middle of the track. That is the red line. Now the yellow line is the line which goes all the way too the far edge of the track (where the cyclest would come into the turn). This is 1/2 of 'w' width + the original radius. The cyclist would then come to the very mid point of the curve which has the length of the green line. This would be original radius - 1/2 of the 'w' width.

    Now this can be shown by extending the where back some to a point such that it is now the axis of the cyclist instead of the axis of the centre of the track. This is show with the purple line (just the extension I didn't want to make it all sloppy and draw lines over lines). So the curve would go from new centre point out to the far edge of the track, to the inside of the middle of track, and then back out to the outside of the track.

    From that I've obtained that if I choose a point which is at 45o to the 'origin' on the far edge of the track that it will be equal to rc-w/2+'a' ('a' is the value of how far back the new axis goes) and this would give the effective radius. (where the cyclist will travel) The curve of this will differ from the original curve (necessarily so or the cyclist could just stay in the middle and go thru the track just as fast right?)
    I also obtain that thru the Pythagorean theorem that effective radius = √(y2+x2) (where y and x are the co-ordinate values of the point on the edge of the graph I've chosen) I can rewrite this in terms of 'a' and then my head hurt and that's when I came here :D

    I wish I could scan the paper I worked on it'd be much easier to show the work considering the diagrams are somewhat necessary.

    I'll try to rework it out again and post back.
     
    Last edited by a moderator: May 5, 2017
  7. Oct 11, 2011 #6

    PeterO

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    You are showing the bend as a curve - probably correctly, I haven't seen the question. I was thinking it was going to be a perfect right angle - like a city intersection but with tracks narrower than a road.

    You would want to show more of the straights approaching this bend.
    As a cyclist, I would begin my curve well before the ends of the quadrant you have shown, and end my curve beyond what you have shown.

    Like you I can't draw pictures on this forum.

    The centre of the circular path I would follow is down and to the right of the centre of your circles - like yours is shown, and finding that point is the key.
     
    Last edited by a moderator: May 5, 2017
  8. Oct 12, 2011 #7
    Hmm does the curvature really matter? The ratio should always be the same as a function of the radius? If so how were you suggesting going about the question? Would your method not only solve for the middle of the track I'm not certain how to get it in terms of "w" width.
     
  9. Oct 12, 2011 #8
    Shouldn't you just find the force applied by the static friction and then set that equal to the centripetal acceleration to solve for v?
    you would just have to find the radius of the turn in terms of the width of the path.
     
  10. Oct 12, 2011 #9
    Yeah, that's what I'm trying to do. The forces part of the problem is not any big deal but the radius is messing with my brain.
     
  11. Oct 13, 2011 #10
    In the problem they give you rc don't they?
     
  12. Oct 13, 2011 #11
    No. Just the knowledge that it is the radius of the curvature of the turn.
     
  13. Oct 13, 2011 #12
    hmm okay. I guess you would have a circle inside of a square if the path is a 90 degree turn. The largest radius yields the highest speed, so the circle will touch the square halfway between vertices. You might be able to use that geometry to solve. You would just have to put the length of the side in terms of w.
     
  14. Oct 14, 2011 #13
    Yeah, if you look at the diagram above you can visualize what the curve will look like. I don't think the actual curvature matters much because it'll be a ratio right. So the answer obtained should work for ALL curvatures (so long as the turn is on a 90o angle.

    The problem is I'm struggling to get it in terms of 'w'. My T.A. was helping me but even she struggled with the question, eventually giving up and telling me it 'was possible' but she 'forgot how she did it before'. Which is cool, it's just I don't like starting a difficult problem and not finishing it.
     
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