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Finding a position of equilibrium by considering energy
By Edward Toman (P2478) on Sunday, December 17, 2000 - 05:38 pm:

This question comes from book "M4", Ex5A, Q3:

A uniform heavy rod AB, of mass m and length 4a, can turn in a vertical plane about one end A which is fixed. To the other end B is attached a light elastic string of natural length 3a and modulus (1/2)mg. The other end of the string is attached to a light ring which can slide on a smooth horizontal bar which is fixed at a height of 8a above A and in the vertical plane AB. Find the equilibrium positions of the rod and determine their nature.

I can find the two positions of unstable equilibrium (0 and p- if measured clockwise from the upward vertical).
However, the answer book states that the other position is when this angle (Q) is p/3, but I get it as cosQ=0.89.
Could anyone tell me which answer is correct, and could you also give the working.
By Kerwin Hui (Kwkh2) on Sunday, December 17, 2000 - 08:58 pm:

From what I can remember of the M4 exercises, the book answer is correct for the last chapter. I'll try to dig out my answer from 2 years ago.

Kerwin

By Kerwin Hui (Kwkh2) on Sunday, December 17, 2000 - 09:19 pm:

OK, consider the rod AB making an angle q with the upward vertical. Thus,

GPE of rod wrt A = 2mga sin q
extension of string = 8a - 3a - 4a cos q
EPE of string = mga (5 - 4 cos q)2 / 12

Thus, V=mga[(1/12)(5-4 cos q)2+2 cos q]

Now differentiate both side wrt q gives

dV/dq=mga sin q [(2/3)(5 - 4 cos q) - 2]

so dV/dq=0 gives sin q=0 or cos q=1/2.

Kerwin

By Edward Toman (P2478) on Thursday, December 21, 2000 - 09:20 pm:

Thanks a lot Kerwin.
Edward.