Physics Friday 103

Suppose we have a long, uniform plank of thickness T balanced on a cylindrical support of radius R.

Now, let the plank tilt slightly, with no slipping or sliding along the pivot (sufficient friction). There are two possibilities: first, the plank could continue to tilt until it falls off the support (unstable); or, secondly, the plank could tip back, in a small rocking oscillation on the pivot (stable); what is the condition needed for the latter situation?


Let the plank be tilted by a small angle θ from the horizontal; then the distance l along the bottom of the plank from the point of contact with the cylinder and the middle of the bottom of the plank is equal to the length of the arc along the support from the point of contact to the top of the support; thus l=.


Now, the only potential energy involved in our situation is the gravitational potential energy, which depends only on the height of the plank’s center of gravity. Thus, it will be stable if a small angle θ raises the center of gravity above the height it has at θ=0. Choosing our origin as the center of the cylinder, the height of the center of mass at θ=0 is just .

At an angle θ>0, we see the height of the center of mass yCM(θ) is
.
Now, yCM(θ) increases above yCM(0) if .
Differentiating,
,
,
and so
,
which is positive when T<2R; the thickness of the plank must be less than the diameter of the support; otherwise there are no stable oscillations, no matter how small.

The maximum size of the stable oscillations is limited by the angles where yCM(θ) no longer increases;
setting , we see that our oscillations must have a maximum angle less than θmax, where θmax is the first positive solution to the equation
, which can only be solved numerically; the closer the ratio gets to 1 (the thicker the board), the smaller this value, and thus the smaller the stable oscillations must be.

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