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Homework Help: Circular motion problem

  1. Mar 17, 2012 #1
    1. The problem statement, all variables and given/known data
    I've attached the question.

    2. Relevant equations
    normal acceleration=(v^2)/r where radius i think is going to be 3-1.2=1.8?
    |a|= sqrt(a_t^2 + a_n^2)

    3. The attempt at a solution
    the equation given x^2/9 + y^2/16 = 1 is distance i believe, so when you differentiate it v=2x/9 + y/8 = 0, and differentiate again would mean a=2/9+1/8=25/72. And these three equations are tangential.
    not sure what to do from here...

    Attached Files:

  2. jcsd
  3. Mar 17, 2012 #2


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    On what did you base the idea that radius is 1.8?
  4. Mar 17, 2012 #3
    See the problem in a different way.

    What is the minimum condition for the man to just complete the ellipse?(in terms of velocity at top most point.Why??)

    Use Energy equation rather than force.

    (Also ask yourself Can we conserve energy of just the man? External force of Normal reaction acts on him.)
  5. Mar 17, 2012 #4
    well because the radius of the loop is 3m and the centre of gravity of the guy is 1.2m. So when he get to the top, he's only 1.8m away from the centre.
  6. Mar 18, 2012 #5

    the minimum condition is weight force=normal force?
  7. Mar 18, 2012 #6
    (v^2)/r= mg??
  8. Mar 18, 2012 #7


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    Either m(v^2)/r= mg or (v^2)/r= g
  9. Mar 18, 2012 #8


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    3m is the length of one semi-axis of the ellipse - the other is 4m. Not convinced either of them is the radius of curvature at the apex - but it might be? but I certainly do not expect R=3 at the top.
  10. Mar 18, 2012 #9
    Correct. v^2/r = g.

    What is the radius at this point?
    How will you apply conservation of energy?
  11. Mar 18, 2012 #10
    Hey, I read your last post now.

    See its not necessary radius of curvature will be equal to semi major or minor axis.
    Infact during motion radius of curvature is changing.

    The way you calculate radius of curvature at any point is by writing (assume m is theta .The parametric angle of ellipse.i cant type theta as Am online via cell)
    X=acos(m) Y=bsin(m) .

    At lowest point m has value 270 degrees.
    (since y is negative , x is 0)
    Now the way we calculate radius of curvature at any anypoint
    Is by writing
    r=v^2 at that point/centripetal acceleration at that point.

    You may be surprised as we were supppsed to calculate velocity at that point with this equation.
    How is one equation enabling us to soLve for two variables r and v.

    However, in essence we are actually using two equations the other being equation of ellipse.
    As you will see radius of curvature just depends on a,b and m and is independent of v and a.(as it should be logically.it should depend on path of particle and not on how the path is travelled)

    You will get r as
    [(a^2)(sin^2m) +(b^2)(cos^m)]/sqrootof[a^2cos^2(m) +(b^2)(sin^(m))]

    (How??)(you can write general equation of velocity by differentiating and of centripetal component of acceleration and divide)

    Now at topmost point
    M has value 90 degrees.
    (y is positive x is 0)

    Plug in r and use energy equation
    Last edited: Mar 18, 2012
  12. Mar 18, 2012 #11
    so v^2/r=g ; v^2/4=9.81 thus v= 6.26418.
    then conservation of energy principle:
    0.5mv^2+mgh = 0.5mv^2 + mgh
    0.5*85*v^2 + 85(9.81)1.2= 0.5*85*6.26418^2 + 85*9.81*(8-1.2) thus v=12.21 which is incorrect
  13. Mar 18, 2012 #12
    R is not 4
  14. Mar 18, 2012 #13
    im still not quite sure why r isn't 4.
  15. Mar 18, 2012 #14
    I just posted the big formula.

    At top mist point r is a^2/b.
    Plug in and check if answer is right .
    And then we will discuss why r isn't 4 :-)
  16. Mar 18, 2012 #15
    im not exactly sure what a and b is in that formula.
  17. Mar 18, 2012 #16
    Oops.my mistake.
    Actually general equation of ellipse is take as x^2/a^2 + y^2/b^2 = 1
    a=3, b=4
  18. Mar 18, 2012 #17
    so now i have to sub in a=3 b=4 and m=90 into that equation that you gave me?
  19. Mar 18, 2012 #18
    You will obtain r=a^2/b for top most point.

    Then substitue this new value of v into the energy equation(you have written it correctly).
    And find velocity at bottom
  20. Mar 18, 2012 #19
    okay i did that and i got a velocity of 11.48 m/s. the answer is 11.15 m/s
  21. Mar 18, 2012 #20
    Switch value of g from 9.8 m/s^2 to 10 m/s^2 or vice versa (whichever you used) and check once again if the answer matches exactly.
  22. Mar 18, 2012 #21
    substituting 10 gives me v=11.596
  23. Mar 18, 2012 #22
    I think my equation is wrong.
  24. Mar 18, 2012 #23
    You have usef the equation correctly.
    Maybe answer is approximated.
  25. Mar 18, 2012 #24
    so i was right to include potential energy at the ground where the height would be 1.2m and then when calculating potential energy at the top i have to reduce 1.2m from the height, making it 8-1.2m=6.8m???
  26. Mar 18, 2012 #25
    Yes.Change in potential emergy is due to net displacement of COM.

    I am prettu sure I have derived the radius of curvature correctly

    See, the answer should be 11.48 m/s according to me.
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