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Rolling motion in a special ramp

  1. May 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Assuming the sphere roles down without sliding prove that the acceleration of it's center of mass is:

    [tex] a= \frac{g\cdot \sin(\theta)}{1+\frac{2}{5}\cdot \frac{1}{1-\frac{1}{4}\cdot \xi^2}}[/tex]

    [tex]Where \ \xi=\frac{L}{R}[/tex]

    Note that the moment of inertia of the sphere is:

    [tex]I_{sphere}=\frac{2}{5}\cdot M\cdot R^2[/tex]


    2. Relevant equations

    [tex]\tau= F\cdot r\cdot \sin(\varphi)[/tex]

    [tex]\alpha\cdot R=a[/tex]

    3. The attempt at a solution

    The forces acting on the sphere are: the normal force, the force of gravity and the frictional force. The sum of the y components of the normal force will be equal to [tex]M\cdot g[/tex] and the sum of the x components will be 0. So that the sum of torques due to the normal force is zero as well as torques due to the sphere's weight.

    [tex]\tau_{a}=F_{a}\cdot R\cdot \sin(\varphi)[/tex]

    Where we have by the figure:

    [tex]\sin(\varphi)= \frac{\sqrt{R^2-(\frac{L}{2}^2)}}{R}= \sqrt{1-\frac{1}{4}\cdot \xi^2}[/tex]

    So we have that:

    [tex] \left\{ \begin{array}{ccc} 2\cdot F_{a}\cdot R\cdot \sqrt{1-\frac{1}{4}\cdot \xi^2} & = & I\cdot \alpha \\ -2\cdot F_{a} + M\cdot g\cdot \sin(\theta) & = & M\cdot a \end{array} \right. [/tex]

    Which will yield:

    [tex]F_{a}= \frac{I\cdot a}{2\cdot R^2\cdot \sqrt{1-\frac{1}{4}\cdot \xi^2}}[/tex]

    After some manipulation you arive to

    [tex]a=\frac{M\cdot g\cdot \sin(\theta)}{M+\frac{I}{R^2\cdot \sqrt{1-\frac{1}{4}\cdot \xi^2}}}[/tex]

    Which after substituting I for it's value leads to:

    [tex]a= \frac{g\cdot \sin(\theta)}{1+\frac{2}{5}\cdot \sqrt{\frac{1}{1-\frac{1}{4}\cdot \xi^2}}}[/tex]

    Can somebody please tell me where I've gone wrong
    Last edited: May 16, 2010
  2. jcsd
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