# Equilibrium and movement of a cylinder with asymmetric mass centre on an inclined plane

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

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."

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

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

or

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

• +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. Nov 12, 2010 at 2:06
• Depending on the initial conditions is it possible to find how much the cylinder rolls upward, before stopping? Nov 12, 2010 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. Nov 12, 2010 at 9:45

The centre of mass (CM) is at distance $r=\frac{m_2 d}{m_1+m_2}$ from the geometrical centre of the cylinder.

In diagram (a), if the centre of mass lies within the red circle then it will be always on the downhill side of the point of contact P. It will always exert a clockwise torque about P, causing the cylinder to roll continuously down the incline. So the asymmetric cylinder cannot be placed in a position of stability unless
$r \ge R\sin A$.

In diagram (b) the cyinder rocks on the incline between P and Q. The cylinder is stationary at these two positions, and mechanical energy is conserved, so the PE must be the same - ie the CM must lie on the same horizontal line.

The distance PQ is $R(B+C)$ so the centroid has risen a vertical distance of
$R(B+C)\sin A=r\cos C-r\cos B$.

Given an initial orientation angle B, this transcendental equation gives the final orientation angle C. The distance PQ can then be found.

The following resources deal with the dynamics of the rocking/rolling motion :

Rolling motion of non-axisymmetric cylinders by Carnevali & May 2004
A jumping cylinder on an inclined plane by Gomez, Hernandez-Gomez & Marquina 2012
Librational motion of asymmetric rolling bodies and the role of friction force, Pavia University, undated
Rolling of asymmetric disks on an inclined plane by BYK Hu 2011