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March 25, 2017

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A poorly designed playground slide begins with a straight section and ends with a circular arc as shown in the figure below.
img716.imageshack.us/img716/8381/25746193.jpg
A child starts at point P and slides down both sections of the slide. At some point on the circular arc, the normal force goes to zero and the child loses contact with the ramp.
Assuming the forces of friction are negligible, at what height from the ground will the child become airborne?
Answer in units of m.

  • Physics-Mechanics - ,

    relate the change in PE from the vertical fall to changes in KE

    KE change: 1/2 m v^2
    height change: 4.9-radius+ radius(1-CosTheta)
    PE change: mg*height change

    check that
    where theta is the angle measured from the point on the circle to the center and the x axis.

    Now, the object flies off when the normal component of weight (mgSinTheta) is equal to centripetal force (mv^2/r)
    set the equal.
    from setting PE change=KE change, you should be able to solve for v^2. Put that into the equation with the forces, and solve for angle.

  • Physics-Mechanics - ,

    Hi,

    I am confused on the height change.
    By 4.9 did you mean 4.8? and if so do I set it equal to zero to solve?

    is the v in the centripetal force different from the v in KE?

  • Physics-Mechanics - ,

    yes, 4.8 I was using my memory of the pic as 4.9

    KEchange=PE change
    1/2 m v^2=mg(4.8 - radius+ radius(1-cosTheta)

    solve for v^2
    then put that v^2 in this:
    mv^2/r=mgSinTheta
    and solve for Theta
    You may have some tricky algebra.

    It is the same v in KE, and force.

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