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Angle of average acceleration while turning a curve

  1. Mar 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A cyclist is initially heading exactly due North. He then initiates a turn to the West, the turn being a quarter circle with radius 12-m. He travels at a constant speed of 5.5-m/s during the turn. What is the direction of this average acceleration, measured anti-clockwise from East?

    2. Relevant equations
    Theta=S/r

    Where s is arc length.

    3. The attempt at a solution

    0.25*2*12*Pi/12=1.57

    This value seems way too small. Am I suppose to apply tan inverse to it which gives me 57 degrees?
     
  2. jcsd
  3. Mar 18, 2016 #2
    Hi sadpwner:

    I may be misinterpreting the problem statement, but it seems to me that "the direction of this average acceleration" equals the average of the direction of acceleration. If that is correct, then the actual speed is irrelevant, as long as it is a constant. What do you think?

    Regards,
    Buzz
     
  4. Mar 19, 2016 #3
    I think the acceleration direction should be constant as the angle should be the same throughout the circle. I am not required to find the speed. I just need the direction of the acceleration in degrees.

    I also just remembered that theta is measured in radians.

    In that case, the working should be 360-1.57*180=77.4 degrees
     
    Last edited: Mar 19, 2016
  5. Mar 19, 2016 #4

    haruspex

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    I do not follow your calculation. What logic leads to that?
    That is not the right way to convert from radians to degrees.

    What is the initial velocity as a vector? What about the final velocity? What vector do you need to add to the first to get the second?
     
  6. Mar 19, 2016 #5
    lol try 180*1.57?
     
  7. Mar 19, 2016 #6
    I just used theta=S/r formula.

    Should be 180*1.57?

    5.5m/s north initial. Then 5.5m/s west final. Using tan inverse it gives 45 degrees? However, that isn't the answer.
     
  8. Mar 19, 2016 #7

    haruspex

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    That will give you the angle the cyclist turned through. That is not the same as the direction of acceleration.
    When a planet goes in a circular orbit around its star, in which direction does it accelerate?
    No, that is not how you convert from radians to degrees. How many radians in a full circle? How many degrees in a full circle? What is the ratio between those two numbers?
    Now you have calculated the bearing from the start of the arc to the end of the arc, NW. Again, that is not the acceleration.

    Do you know how to find the resultant of two vectors?
    Can you write the average acceleration vector in terms of the initial and final velocity vectors (and time)?
     
  9. Mar 19, 2016 #8
  10. Mar 19, 2016 #9

    rcgldr

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    The north component of speed decreases from 5.5 m /s to 0 m/s, so a southward (270 degree) component of acceleration. The west component of speed increases from 0 m/s to 5.5 m/s, so a westward (180 degree) component of acceleration. The direction of average acceleration is south west == 225 degrees == 3.927 radians. Or using calculus:

    ##\int_{180}^{270} \ \theta \ d\theta / 90 = 225##
     
    Last edited: Mar 19, 2016
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