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Help finding Normal average acceleration of path.

  1. Dec 29, 2013 #1
    A particle moves along the curved path sown. if the particle has a speed of 40ft/sec at A at time ta and a speed of 44ft/sec at B at time tb, determine the average values of he acceleration of hte particle between A and B, bath normal and tangent to the path.

    I found the average tangent acceleration to the path fairly easy. But I am not sure how to go about getting the normal average acceleration. I Tried to used the average tangential acceleration to get an average acceleration by 20tan(26)-20tan(36), i just am not sure.
    [URL=http://s1341.photobucket.com/user/nebula-314/media/20131229_171505_zps59dd5611.jpg.html][PLAIN]http://i1341.photobucket.com/albums/o745/nebula-314/20131229_171505_zps59dd5611.jpg[/URL][/PLAIN]

    My integral says vdv. ignore that it should just be dv.
     
    Last edited: Dec 29, 2013
  2. jcsd
  3. Dec 29, 2013 #2

    haruspex

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    I don't understand how to answer that question either. There doesn't seem to be enough information.
    But can you answer this: if the speed change from v to v+dv in a short time dt, and the angle changes from θ to θ+dθ, what is the normal acceleration during dt?
     
  4. Dec 29, 2013 #3
    yea nothing was given about a radius of curvature or anything. But to answer your question it would be v*dθ.
    (there is nothing about dθ in this problem either)

    I first thought that I could try to find a Normal velocity to the path, but quickly realized that there is no such thing lol. I tried to apply some pretty simply trig but nothing came close to what the answer was which is (36.7 ft/sec^2- average normal acceleration).

    If you want me to take pics of some of the trig that I tired let me know.
     
  5. Dec 30, 2013 #4
    I am not even sure how average tangential and normal accelerations are defined. An instantaneous acceleration can be decomposed into tangential and normal components because there are well defined tangential and normal directions at any point along the path, but what should they be "on average"? Do you integrate them along the path? Or just take the initial and final directions?
     
  6. Dec 30, 2013 #5
    In terms of the unit vectors in the coordinate directions, what is the equation for the average acceleration between your two points? In terms of the unit vectors in the coordinate directions, what is the equation for the displacement vector between your two points? In terms of the unit vectors in the coordinate directions, what is the equation for a unit vector in the direction of the displacement vector. What is the equation for the component of the average acceleration vector in the direction of the displacement vector? What is the equation for the component of the average acceleration vector in the direction normal to the displacement vector?

    Chet
     
  7. Dec 30, 2013 #6

    haruspex

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    That's an interesting interpretation of 'tangential and normal accelerations'. Is this a standard definition, or your own (very reasonable) interpretation?
     
  8. Dec 31, 2013 #7
    This is my own interpretation. The original question was clearly very ambiguous. I tried to figure out what the person who posed this question might have been looking for. Since this could also be regarded as a discrete approximation to the instantaneous tangential and normal components of the acceleration at some point, it tried to determine how I could make this approximation most accurate. My rationale was that it would be second order accurate only at the half-way point. That would entail resolving the average acceleration into components normal and parallel to the average velocity vector, or equivalently the displacement vector, between the two end points.

    Chet
     
  9. Apr 15, 2017 #8
    Use formula ∫vdv=∫[a][/t]ds to find the displacement covered from A to B. the change in angle is known just change it to radians and arc length formula to compute radius of curvature. once radius is know calculate normal acceleration for individual points and then calculate its average value.
     
  10. Apr 15, 2017 #9
    This thread has not had any activity for over three years, and the OP was last seen over three years ago. I am hereby closing this thread.
     
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