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Rubber ball Rotational Motion Question

  1. Nov 9, 2009 #1
    Note: I refrained from using any of the latex references, so bear with me.

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

    A rubber ball with a radius of 3.2 cm rolls along the horizontal surface of a table with a constant linear speed v. When the ball rolls off the edge of the table, it falls 0.66 m to the floor below. If the ball completes 0.37 revolutions during its fall, what was its linear speed, v?

    a) Variables extracted from the question

    r = 3.2 cm = 0.032 m
    h = 0.66 m
    theta = (2pi)(0.37 revolutions) = (2pi x 0.37) rad

    2. Relevant equations and physics principles/concepts

    a) Equations

    i. d = Vit + 0.5at2
    ii. v = rw

    b) Physics principles/concepts

    i. Law of conservation of energy (?)
    ii. Rotational kinetic energy and the moment of inertia (?)
    iii. Kinematics (?)

    3. The attempt at a solution

    a) Energy is conserved, therefore:

    Ki + Ui = Kf + Uf

    b) There is no potential energy as the bottom of the table. Furthermore, we are analyzing the rotational motion of a rubber ball (ie. I = 0.4mr2). Therefore:

    Ki + Ui = Kf
    0.5mvi2 + 0.5(0.4mr2)(vi/r)2 = mgh + 0.5mvf2 + 0.5(0.4mr2)(vf/r)2

    c) We can clean up the equation by eliminating the mass and radii parameters, and multiplying each term by 2.

    vi2 + 0.4vi2 = 2gh + vf2 + 0.4vf2

    d) Like terms can be collected and vi can be solved for.

    vi = sqrt ((5/7)[(7/5)vf2 - 2gh])

    e) We only need the final linear speed now. To determine that, we can use:

    vf = rwf

    i. As can be seen above, we need to determine wf. To do this, regular kinematics can be applied to find the total time the ball spends in the air.

    d = Vit + 0.5at2
    d = 0.5at2
    t = sqrt (2d/a)
    t = sqrt [(2 x 0.66 m) / (9.81 m/s2)]
    t = 0.37 s

    ii. At the very beginning, the amount of revolutions (in radians) was determined. By dividing the revolutions by the time obtained above, we can obtain wf, and ultimately vf.

    vf = rwf
    vf = (0.032 m)[(2pi x 0.37) / (0.37 s)]
    vf = 0.20 m/s

    f) Finally, we can solve for initial linear speed.

    vi = sqrt ((5/7)[(7/5)vf2 - 2gh])
    vi = sqrt ((5/7)[(7/5)(0.20 m/s)2 - 2(9.81 m/s2)(0.66 m)])

    So yeah, I do not know what I am doing wrong. Am I misinterpreting the question? Am I approaching the question in the wrong way? Does my logic make any sense?
    Last edited: Nov 9, 2009
  2. jcsd
  3. Nov 9, 2009 #2
    as the ball falls down, its rotational speed does not increase :)

    can u imagine why?

    and also, in the energy equation where should we put the potential energy, does the system lose or gain potential energy?

    hope that helps :)
  4. Nov 10, 2009 #3


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    Staff Emeritus
    Science Advisor
    Homework Helper

    Since ω does not change, the ball has this same ω when it leaves the table. Can you use that information to find vi=rω?
  5. Nov 10, 2009 #4
    Oh, wow, I think I figured it out.

    Thanks for the help you two.
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