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Car acceleration around a corner

  1. Oct 10, 2007 #1
    A car is going around a quarter-circle from north to east. When halfway around the car's acceleration is (2m/s^2 , 15 degrees south of east). At this point, what is the COMPONENT of acceleration (1) tangent to the circle and (2) perpendicular to the circle?

    Well I know that if I draw a line Tangent to the circle and then a line Perpendicular to the Tangent line I have created a X-Y coordinate system that is tilted. After this point I am stuck and not sure where to go.

    Any help would be great.
     
  2. jcsd
  3. Oct 11, 2007 #2

    learningphysics

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    Draw the circle... with the tangent and perpendicular drawn at the point of interest... draw a radius from the center to the point of interest... draw the vector 2m/s^2 at 15 degrees south of east... the two components of this vector along the perpendicular and the tangent are what you need...

    so you need the angle between the tangent line to the circle and the 2m/s^2 line... what is this angle?

    draw the horizontal line through the cente of the circle... what is the angle between this line and the radius to the point of interest?
     
  4. Oct 11, 2007 #3
    The horizontal line that goes via the circle and the perpendicular line has an angle of 45 degrees. Is that what you were asking about?
     
  5. Oct 11, 2007 #4

    learningphysics

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    yes. exactly. can you find the angle between the tangent to the circle and the 2m/s^2 at 15 degrees south of east vector?
     
  6. Oct 11, 2007 #5
    Well yes, because the line that is perpendicular to the tangent line is 90 degrees therefore you could draw a line in both the south and east directions which would be 45 degrees. So, if directly east is 45 degrees, another 15 degrees south of east would make that vector 60 degrees from our tangent line.
     
  7. Oct 11, 2007 #6

    learningphysics

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    wait... from the point of interest... we have a line going east and a line that is tangent... the angle between these two lines is 45 degrees. The 2m/s^2 line is in between these two...
     
  8. Oct 11, 2007 #7
    I knew this would get confusing because you dont have a drawing in front of you. I am pretty sure that from our point of interest which lies on the tangent to circle line which is also are x coordinate TO the acceleration vector which is 15 degrees south of east is 60 degrees from the tangent line.
     
  9. Oct 11, 2007 #8

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  10. Oct 11, 2007 #9
    well can u draw it in the second quadrant because that is how it was assigned to us. so draw the point of interest in the upper left hand corner so its like at 135 degrees
     
  11. Oct 11, 2007 #10

    learningphysics

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    Oh, I see. yes, you're right. Sorry about that. Then the angle is 60 degrees... so you should be able to get the 2 components using 2m/s^2 and 60 degrees...
     
  12. Oct 11, 2007 #11
    Does this file show up for you
     

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  13. Oct 11, 2007 #12

    learningphysics

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    it says attachment pending approval. it usually takes a while for attachments to get approved...

    but I think you should be able to do the problem using the 60 degree angle.
     
  14. Oct 11, 2007 #13
    The green line is 15 degrees south of the blue line, the red line is the line perpendicular to the black tangent line. So from the tangent line to the green line should be 60 degrees.
     
  15. Oct 11, 2007 #14
    im not sure how though? please help
     
  16. Oct 11, 2007 #15

    learningphysics

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    What is the component of the 2m/s^2 along the tangent?

    What is the component of the 2m/s^2 along the perpendicular?

    Use the 60 degree angle.
     
  17. Oct 11, 2007 #16
    By using ...

    A_x = A cos60
    A_y = A sin60

    and plugging 2 in for A

    Is this how you solve for the components???

    Help!!!
     
  18. Oct 11, 2007 #17

    learningphysics

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    Yes. The 2 m/s^2 is the hypoteneuse of a right triangle with angle 60.

    2cos60 is the tangential acc.
    2sin60 is the perpendicular acc.
     
  19. Oct 11, 2007 #18
    A_x = Acos60
    A_y = Asin60

    Where A is 2

    Is this how you get the components??

    Help!!
     
  20. Oct 11, 2007 #19
    Woops sorry

    Should I be in radians or degrees?
     
  21. Oct 11, 2007 #20
    2 cos 60 = 1
    2 sin 60 = 1.73
     
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