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

π

- if measured clockwise from the upward vertical).

However, the answer book states that the other position is when this angle (Q) is $π / 3$ , 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

θ

with the upward vertical. Thus,

GPE of rod wrt A $= 2 m g a \sin θ$

extension of string $= 8 a - 3 a - 4 a \cos θ$

EPE of string $= m g a (5 - 4 \cos θ)^{2} / 12$

Thus, $V = m g a [(1 / 12) (5 - 4 \cos θ)^{2} + 2 \cos θ]$

Now differentiate both sides wrt $θ$ gives

$dV / d θ = m g s \sin θ [(2 / 3) (5 - 4 \cos θ) - 2]$

so $dV / d θ = 0$ gives $\sin θ = 0$ or $\cos θ = 1 / 2$ .

Kerwin

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

Thanks a lot Kerwin.
Edward.