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A cylinder whose cross section is represented below is placed on an inclined plane. I would like to determine the maximum slope of the inclined plane so that the cylinder does not roll. The mass centre (CM) of the cylinder is at a distance r from the central axis. The cylinder consists of a cylindrical shell with mass $m_1$ and a smaller cylinder with mass $m_2$ placed away from the axis and rigidly attached to the larger cylinder. What is the influence of friction? Is it possible to establish the law of the movement? I think that the piece may roll upwards until it stops.

The figure was copied from Projecto Ciência na Bagagem -- Cilindro desobediente alt text

EDIT: Depending on the initial conditions is it possible to find the highest point the cylinder rolls to, before stopping?

EDIT2: From Institute and Museum of the History of Science -- Cylinder on inclined plane [another cylinder]

"When placed on the inclined plane, [another] cylinder tends to roll upward, coming to a halt at a well-determined position."

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Is the inner cylinder of mass $m_2$ rigidly attached to the larger cylinder? – David Z Nov 12 '10 at 0:36
@David Zaslavsky: Yes, it is. I edited the question. – Américo Tavares Nov 12 '10 at 0:58
up vote 6 down vote accepted

The effect of friction is to make the cylinder roll down the ramp rather than slide.

To find an equilibrium angle, use virtual work.

If $\phi$ changes by a small amount $d\phi$, as the cylinder rolls, then everything goes down a little (neglecting at first the small interior cylinder's upward movement) because you're moving down the ramp. You move $R d\phi$ down the ramp, and lose elevation $\sin \Phi R d\phi$. The total work done by gravity is $(m_1 + m_2) g \sin\Phi R d\phi$

On the other hand, the interior cylinder rises with respect to the center of the big cylinder by an amount $(R - R_2) d\phi$. The work done by gravity on the little cylinder is $g m_2 (R-R_2) d\phi$.

Equilibrium is achieved when these are equal, so

$m_2 (R - R_2) = (m_1 + m_2) R \sin\Phi$


$\sin\Phi = \frac{m_2(R-R_2)}{(m_1+m_2)R} = \frac{r}{R}$

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+1 I got the same answer by computing the potential energy. I would point out that $m_2(R - R_2)/(m_1 + m_2) = r$, which could make the answer look a little simpler. – David Z Nov 12 '10 at 2:06
Depending on the initial conditions is it possible to find how much the cylinder rolls upward, before stopping? – Américo Tavares Nov 12 '10 at 9:22
@Americo Do you want to know the equilibrium position or the highest point it rolls to? For the equilibrium you'd do the same as I did here, but allowing the small cylinder not to be on the far left of the big cylinder any more. For the highest height rolled you'd use conservation of energy. – Mark Eichenlaub Nov 12 '10 at 9:45
Thanks! I was thinking on the highest point. – Américo Tavares Nov 12 '10 at 9:55

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