Rolling motion in a special ramp

[benf.stokes]

Homework Statement

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]

Homework Equations

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

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

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

 

[anathema]

?Your solution is correct, but you can simplify it further by using the given equations for the moment of inertia and \xi:

I = \frac{2}{5}MR^2

\xi = \frac{L}{R}

Substituting these values into your final equation, we get:

a = \frac{g\sin\theta}{1 + \frac{2}{5}\frac{1}{1 - \frac{1}{4}\left(\frac{L}{R}\right)^2}}

= \frac{g\sin\theta}{1 + \frac{2}{5}\frac{1}{1 - \frac{1}{4}\xi^2}}

which is the same as the given equation. So, your solution is correct.