Answer to the Question 02/04
BALANCING PLANK
The question was:
A thick plank is placed on a log of semicircular cross section.
What is the thickest plank such that the plank on log
has stable oscillations?
(6/04) The problem has been solved
(17/4/04) by
Chetan Mandayam Nayakar, an undergraduate student at
Indian Institute of Technology, Madras
(e-mail mn_chetan@yahoo.com)
(his solution is presented below),
and by several other people.
The answer: The plank becomes unstable if its thickness
exceeds the diameter of the log.
The solution: Let the radius of the log be R and the thickness
of the plank be t. At equilibrium, the height of the
plank is
equal to R+(t/2). When the plank undergoes an
infinitesimal angular displacement of {theta} from
equilibrium, the height of the plank above the log axis
becomes
(R+(t/2))*cos({theta})+(R*{theta}*sin({theta})).
For stable
equilibrium, the new height should be greater than
R+(t/2). Applying the limit {theta} tending to zero and
solving the inequality, the result is t < 2R.
p.s. Actually the solution can be found in the answers to problem 03/03.
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