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Normal component of the acceleration

  1. Aug 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Given y[tex]^{2}[/tex] = 8x[tex]^{3}[/tex] + 8x

    where x and y are in metres, and y is positive

    What is the normal component of the acceleration

    When x = 4m, x = 7m/s, x = 2m/s[tex]^{2}[/tex]

    Note : Wasn't sure if it was Calc or Physics, but I was leaning toward Calc, as it is basically Algebra to solve for the answer.

    2. Relevant equations

    Not sure

    3. The attempt at a solution

    I understand that "x = 4m, x = 7m/s, x = 2m/s[tex]^{2}[/tex]" is just "y[tex]^{2}[/tex] = 8x[tex]^{3}[/tex] + 8x" getting differentiated for each part, first of which doesn't need it.

    y[tex]^{2}[/tex] = 8x[tex]^{3}[/tex] + 8x => y = ([tex]\sqrt{8x^3 + 8x}[/tex])

    I am just a little lost at "What is the normal component of the acceleration"

    Any and all help is greatly appreciated.

  2. jcsd
  3. Aug 2, 2010 #2


    Staff: Mentor

    These should probably be "... when x = 4m, dx/dt = 7 m/sec, and d^2x/dt^2 = 2m/sec^2. You can't have x being equal to three different quantities with three different units.
    Are you sure that you are giving us the exact wording of this problem? It doesn't make much sense to me as you have written it.
  4. Aug 2, 2010 #3


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    If this is not a trajectory in the plane, then the problem makes no sense- there is no "normal component". If it is a trajectory then you need to find the acceleration as the vector [itex](d^2x/dt^2}\vec{i}+ (d^2y/dt^2)\vec{j}[/itex]. Of course, to do that, you have to know the dependence of x and y on t.

    Knowing that [itex]y^2= 8x^3+ 8x[/itex] tells you that [itex]2y(dy/dt)= (24x^2+ 8)(dx/dt)[/itex] but uyou still need to know or be able to calculate either dy/dt or dx/dt.

    I agree with Mark44- please give the exact wording of the problem. "When x = 4m, x = 7m/s, x = 2m/s2" is impossible.
  5. Aug 4, 2010 #4
    I came across the same question. Sheen has typed the entire question. Only mistake being the repeated x. Which was meant to be x, x' and x''.
    I derived the equation twice and subbed in the given x values as i derived to get the corresponding y values. i.e. i solved for y, y' and y". Is this right?
    If this is right i still get stuck on how to find the normal component.
  6. Aug 4, 2010 #5


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    Hint: Since the motion is confined to a plane, the acceleration can be resolved into two components, the tangential component parallel to the velocity and the normal component perpendicular to the velocity.
  7. Aug 5, 2010 #6
    ***Post removed***
    Last edited: Aug 5, 2010
  8. Aug 5, 2010 #7
    y² = 8x³ + 8x

    2y dy/dt = (24x² + 8) dx/dt

    Velocity = sqrt ( (dx/dt)² + (dy/dt)² )

    Acceleration = sqrt ( (d²x/dt²)² + (d²y/dt²)² ) = ???? :S (not sure)

    2y dy/dt = (24x² + 8) dx/dt

    2 d²y/dt² = 48x d²x/dt²

    d²y/dt² = 24x d²x/dt²


    Acceleration = sqrt ( (d²x/dt²)² + (d²y/dt²)² ) = sqrt( (d²x/dt²)² + (24x . d²x/dt²)² ) = sqrt ( 2² + (24 . 4 . 2)²) = sqrt ( 36868 ) = 192.01 m/s/s

    Which doesn't seem right. :S

    Can someone look through and see if I did anything wrong. Thank you

    PS: I need to find the normal component of the acceleration, so I have yet to start that
    Last edited: Aug 5, 2010
  9. Aug 5, 2010 #8


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    This is good. Come up with numerical answers for what y and y' equal given the information you have about x and x'.
  10. Aug 5, 2010 #9


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    That's not right. You have to use the product rule.
  11. Aug 5, 2010 #10
    Sorry I don't understand how, do I use the product rule in that??

    Thanks for all the help so far vela, I really appreciate it

    uhmm ohh wait

    dy/dx = (dy/dt) / (dx/dt)

    2y dy/dt = (24x² + 8) dx/dt

    dy/dx = (dy/dt) / (dx/dt) = (24x² + 8) / 2y

    d²y/dx² = ( (48x . 2y) - 2(24x² + 8)dy/dx ) / 4y²

    I am a little lost.
    Last edited: Aug 5, 2010
  12. Aug 5, 2010 #11
    y = sqrt (8x³ + 8x) = sqrt (8.4³ + 8.4) = sqrt (544) = 23.324

    y' = (24x² + 8) x' / 2y = 58.82

    now what?
  13. Aug 5, 2010 #12
    I am unsure on how to find the "Normal component of the Acceleration"

    Vela can you let me know what to do step by step, rather than just one step at a time. I would really apreciate that.

    I also got the numerical values of y and y', and am once again stuck.

    Also how do I use the product rule on : 2y dy/dt = (24x² + 8) dx/dt

    Have never been taught.
  14. Aug 5, 2010 #13
    I think I got it:

    Product rule

    2y dy/dt = (24x² + 8) dx/dt

    (2 dy/dt) . dy/dt + d²y/dt² = (48x . dx/dt) + ( (24x² + 8) . d²x/dt²)

    so that means:

    d²y/dt² = (48x . dx/dt) + ( (24x² + 8) . d²x/dt²) - (2 dy/dt) . dy/dt

    d²y/dt² = -4791.58

    That seems very wrong.

    Also after you get d²y/dt² and d²x/dt², what do you do? How do you find the normal component?
    Last edited: Aug 5, 2010
  15. Aug 5, 2010 #14


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    You forgot the 2y in front of y''. The RHS looks okay.
    You have the two vectors a=axx+ayy and v=vxx+vyy. What you want to do is resolve a into components parallel and perpendicular to v. So how do you find the component of a parallel to v? Hint: Use the dot product.
  16. Aug 5, 2010 #15
    Drawing a vector diagram for this question helps ALOT. It helped me solve it =]
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