Stability physics problem

1. Feb 25, 2007

mr bob

Just come across a question and I'm at a point where i see no further.

A uniform rod AB, of mass m and length 2a, is free to rotate in a vertical plane, about the end A. A light elastic string of modulus kmg and natural length a, has one and attached to B and the other end to a fixed point O which is vertically above A with OA = 2a. Show that when AB makes an angle $$\theta$$ with the downward vertical, the potential energy, V, of the system when the string is stretched is given by

$$V = mga[(4k - 1)\cos\theta -4k\cos\frac{\theta}{2}] + constant$$

I drew the following diagram:-

I equated the GPE of the uniform rod to be
$$AB = -mg\cos\theta$$

And when it comes to calculating the energy of the elastic string i tried:

$$OC = \frac{\lambda x^2}{2a}$$

Where x is the extension in the string, which i calculated to be:

$$x = (2a\sin\theta - a)$$

So:
$$OB = \frac{kmg(2a\sin\theta - a)^2}{2a}$$

$$OB = \frac{kmga(4\sin^2\theta -4\sin\theta + 1)}{2}$$

$$OB = 2kmgasin^2\theta -2kmg\sin\theta + \frac{kmg}{2}$$

But since:
$$OB = \frac{kmg}{2}$$ is a constant, we can take it out of the equation.

Using the identity: $$\sin^2\theta = 2\sin\theta\cos\theta$$

$$4kmga\sin\theta\cos\theta -2kmga\sin\theta$$

However i can't get the 2nd term in the above equation to equal $$4kmga\cos\frac{\theta}{2}$$

Any help would be really appreciated.

Thank you.

Last edited: Feb 25, 2007
2. Feb 26, 2007

Pyrrhus

Hey Bob, how can you be sure the angle made by AB and OB is 90 degrees?

3. Feb 26, 2007

mr bob

Hmm. Since the triangle OAB is isosceles, could i split it into 2 right angled triangles to get the length of the extension in terms of $$\cos\frac{\theta}{2}$$?

I'm not too clear why the sine of the angle theta can't just be used. Is there a fundamental flaw in my understanding of mathematics?

Thank you.

4. Feb 26, 2007

mr bob

Just ran through it trying splitting the triangle and i got

$$OB = kmg(4a\cos\theta -4a\cos\frac{\theta}{2})$$

As this is equal to the GPE of the rod.

$$mga\cos\theta = kmg(4a\cos\theta -4a\cos\frac{\theta}{2})$$

$$V = mga[(4k - 1)\cos\theta - 4k\cos\frac{\theta}{2}]$$