Answer to the Question 02/04
BALANCING PLANKThe 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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