- #1

C. Darwin

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## Homework Statement

A cylinder with mass [itex]m[/itex] and radius [itex]a[/itex] rolls down a fixed cylinder with radius [itex]b[/itex]. Let [itex]\theta[/itex] be the angle between the plane containing the axis of each cylinder and the vertical. Find the normal component of the reaction force exerted on the rolling cylinder.

## Homework Equations

## The Attempt at a Solution

[tex]T = \frac{1}{2} m v^2 + \frac{1}{2}I\omega^2 = \frac{1}{2} m (R\dot{\theta})^2 + \frac{1}{2}(\frac{1}{2}ma^2)(\frac{b}{a}\dot{\theta})^2 = m[\frac{1}{2}(a+b)^2 + \frac{1}{4}b^2]\dot{\theta}^2[/tex][tex]V = mgh = mgRcos(\theta) = mg(a+b)cos(\theta)[/tex]

[tex] E = T + V \Rightarrow mg(a+b) = m[\frac{1}{2}(a+b)^2+\frac{1}{4}b^2}]\dot{\theta}^2 + mg(a+b)cos(\theta)[/tex]

[tex]\Rightarrow \dot{\theta}^2 = \frac{g(a+b)}{\frac{1}{2}(a+b)^2 + \frac{1}{4}b^2}(1-cos(\theta))[/tex]

Now here's where I'm not so sure (although my statements of energy may be off):

[tex] F_N = mgcos(\theta) - m\frac{v^2}{r} = mgcos(\theta) - mr\dot{\theta}^2 =mgcos(\theta) - m(a+b)\dot{\theta}^2[/tex]

[tex] F_N = mg[cos(\theta) - \frac{(a+b)^2}{\frac{1}{2}(a+b)^2 + \frac{1}{4}b^2}(1-cos(\theta))] [/tex]

I'm lost at this point, I think my statements for T and V are correct because I arrived at the same [tex]\ddot{\theta}[/tex] from both the energy analysis and the Langrangian.

According to the book, the answer is: [tex] F_N = mg(7cos(\theta) - 4)/3[/tex]

Thanks in advance.

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