1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Mechanics - Acceleration on an elliptical path

  1. May 3, 2013 #1
    This is a mechanics question which has come up a number of times in end of year exams in my college, I hope this is the right forum to post it in.

    1. The problem statement, all variables and given/known data
    A particle travels around an elliptical path with a constant speed of 1 m/s. If the equation
    of the ellipse is (x^2)/3 + y^2 = 1
    Determine the maximum magnitude of the acceleration.

    A - (√3)/2
    B - 3/2
    C - 1/2
    D - 0
    E - 1

    2. Relevant equations
    If the equation = (x^2)/(a^2) + (y^2)/(b^2) = 1
    Then a = (v^2)/ρ, where ρ is the radius of curvature and is equal to (a^2)/b

    3. The attempt at a solution
    Using the above equations gives:
    ρ = 3/1 = 3
    a = 1/3 which is not an answer provided?


    I'm unsure whether the answer should just be zero? As there is no tangential acceleration? Or do I have the wrong formula for calculating the centripetal acceleration?

    Any help or explanation would be great :)
     
  2. jcsd
  3. May 3, 2013 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Is this the maximum or minimum radius of curvature of the ellipse?
     
  4. May 3, 2013 #3
    This is the maximum radius of curvature.
     
  5. May 3, 2013 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    But you want to find the maximum acceleration.
     
  6. May 3, 2013 #5
    Yes but isn't the maximum centripetal acceleration when the particle is travelling at the point of largest radius?
     
  7. May 3, 2013 #6

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No...
     
  8. May 3, 2013 #7

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Remember:
     
  9. May 3, 2013 #8
    Had an entire post written and then realised that it's the smaller radius but furthest from centre of the ellipse.

    Ok so taking that into account would the answer be 1?

    Minimum radius = 1

    a = (v^2)/r = 1/1 = 1?


    Edit
    Thank you, can't believe I didn't get that >.< That's a great help.
     
  10. May 3, 2013 #9

    TSny

    User Avatar
    Homework Helper
    Gold Member

    No, the minimum radius of curvature of the ellipse is not 1. a2/b is the curvature of the ellipse at a point on the y-axis. This happens to be the maximum radius of curvature for your ellipse because the ellipse is elongated along the x-axis. Think about where the minimum curvature will occur and see if you can guess how the formula a2/b would change to get the minimum curvature.
     
  11. May 3, 2013 #10
    Dang just when I thought I had it.

    Well I'm guessing the equation will change to b2/a?

    Although that gives me 1/√3 which when subbed into a doesn't result in an answer?


    Edit:

    I was going under the assumption that a and b are the two radii? So whichever is smaller I just sub directly into a=v2/r, this gives me an answer for all 3 years this question has come up
     
  12. May 3, 2013 #11

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I think 1/√3 is the right answer even though it's not listed as one of the choices.

    I hope whoever wrote the question wasn't assuming that the smallest radius of curvature would be the smaller of a and b!
     
  13. May 3, 2013 #12
    Written by a mechanics lecturer in my college but unfortunately I haven't had a reply to any email I've sent him. The questions the other two years were:
    1. A particle travels around an elliptical path with a constant speed of 1 m/s. If the equation
    of the ellipse is 3x2 + 2y2 = 2
    Determine the maximum magnitude of the acceleration.

    A - (√3)/2
    B - √3/√2 (Answer using the radius being the smaller of a and b)
    C - 1/2
    D - 0
    E - 1

    2. A particle travels around an elliptical path with a constant speed of 1 m/s. If the equation of the ellipse is x2/2 + y2 = 1
    Determine the maximum magnitude of the acceleration

    A - (√3)/2
    B - 3/2
    C - 1/2
    D - 0
    E - 1 (Answer using the radius being the smaller of a and b)


    As you can see neither of these has an answer which would work using the b2/a formula
     
  14. May 3, 2013 #13

    TSny

    User Avatar
    Homework Helper
    Gold Member

    For #1, a is smaller than b. So, the smallest radius of curvature is a2/b = 2/3. Max acceleration is then 3/2.

    For #2, b is smaller than a. So, the smallest radius of curvature is b2/a = 1/√2. Max acceleration is √2.

    As you noted, neither of these corresponds to any of the selections. Hope I'm not overlooking something.
     
  15. May 4, 2013 #14
    Unless he meant the magnitude of the tangential acceleration? Which would be zero each time? But he doesn't say that
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Mechanics - Acceleration on an elliptical path
  1. Elliptical Orbit (Replies: 4)

  2. Elliptic pendulum (Replies: 1)

Loading...