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Solid sphere vs. disk ramp test and time difference.

  1. Nov 27, 2006 #1
    A solid sphere and a solid disk are released from rest at the top of the ramp. The inclination angle is theta to the horizontal and the vertical height of the ramp is h. Determine the time difference between the objects for them to read the bottom of the ramp.

    This is what I've solved:

    I(disk)= 1/2mr^2 = I + md(parallel)^2 = 1/2 mr^2 + mr^2 = I = 3/2 mr^2

    Kinetic energy ---> Potential energy

    mgh = 1/2 I(omega)^2
    mgh = 1/2 (3/2 mr^2) (omega)^2
    Omega = square root [4mgh/3mr^2]

    Solving for the velocity by mulitplying by radius r:
    V= 2/r times sq. root [gh/3] times r = 2 sq.root [gh/3]

    Since it's a ramp, I used dsin theta for the hypotenuse and used a kinematic equation

    dsin theta = (v initial + v /2) t

    Solving for t (time)

    t= dsin theta / (2 sq. ro. [gh/3] ) / 2

    t = d sin theta / sq. r. [gh/3]

    The time for the sphere was also calculated the same way so no need for me to put it up here. My questios are: How am I supposed to find the time difference if there are no definate values for theta, and h. I would have to set up the final answers and subtract them from each other but that wouldn't get me any where. Also, is it valid to set d from the kinematic formula to equal d sin theta?

    Please help! Thanks.
  2. jcsd
  3. Nov 27, 2006 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Just do it symbolically, in terms of the given variables. Nothing wrong with that.

    No. If D is the distance along the ramp, and h is the height, then h = D sin(theta).
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