1. Not finding help here? Sign up for a free 30min 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!

Acceleration on a curve

  1. Aug 10, 2004 #1
    a smooth wire is shaped in the form of a parabola y=ax^2 on a horizontal plane. a bead slides along the wire. as it passes the origin, velocity is Vo. what is the acceleration.

    ...............................................................................
    i see the parabola as part of a big circle and decided to use a=v^2/r and i know s=r(theta)... then i am lost... help!!! :cry:
     
    Last edited: Aug 10, 2004
  2. jcsd
  3. Aug 10, 2004 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    Sounds like you need to know r, which is the radius of curvature of the parabola at the origin.

    (...looks up radius of curvature...)

    Here it is:

    [tex]R=\frac{[1+\left(\frac{dy}{dx}\right)^2]^{3/2}}{\frac{d^2y}{dx^2}}[/tex]
     
  4. Aug 10, 2004 #3
    This is interesting. Would you show us how to derive this radius of curvature?
     
  5. Aug 10, 2004 #4
  6. Aug 10, 2004 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    But just knowing the velocity at one point tells you nothing about the acceleration, on a curve or straight line. Is there more information?
     
  7. Aug 11, 2004 #6

    ehild

    User Avatar
    Homework Helper
    Gold Member

    We can assume that the bead slides on the horizontal wire with constant speed. Otherwise the problem would be undetermined. The acceleration can be calculated without knowing about curvature, just from its definition: that it is the second derivative of the displacement, that is the componts of the acceleration are the second derivatives of the coordinates with respect to time. Choosing the plane of motion as the (x,y) plane, the components of the velocity are
    [tex]v_x=\dot x[/tex],
    [tex]v_y= \dot y = 2ax \dot x =2axv_x[/tex].

    The components of the acceleration are the derivatives of those of the velocity,

    [tex]a_x= \dot v_x [/tex],
    [tex]a_y=\dot v_y=2a\dot xv_x+2ax\dot v_y = 2a\left({v_x}^2+x\dot v_x\right)[/tex]

    Moreover,

    [tex]v_x^2+v_y^2=const \rightarrow v_x \dot v_x + v_y\dot v_y = 0 [/tex]

    The acceleration is asked at the origin, x=0, y=0. Here

    [tex]v_y=0\mbox{, so } v_x^2=v_0^2[/tex].

    Because of [tex]v_y=0[/tex] at x=0, the acceleration in x direction is also 0: [tex]\dot v_x=0[/tex].

    (The tangent of the curve at x=0 is the x axis, and the tangential acceleration is zero when the speed is constant)
    The result is :

    [tex]a_x=0[/tex].
    [tex]a_y=2av_0^2[/tex], the nagnitude of the acceleration is

    [tex] 2av_0^2[/tex].

    ehild
     
  8. Aug 11, 2004 #7
    thx for all the replies ... today my teacher gave me another way of solving this prob.

    s=r[tex]\theta[/tex]=2x (whereby x is the 2 points [tex]\theta[/tex] cuts at the x-axis)
    a=v^2/r
    y=ax^2
    [tex]\theta[/tex]/2= tan[tex]\theta[/tex]/2= dy/dx= 2ax
    therefore [tex]\theta[/tex]=4ax
    sub [tex]\theta[/tex]=4ax into s=r[tex]\theta[/tex]
    r=1/(2a)
    a=2av^2
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Acceleration on a curve
Loading...