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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 [tex]\theta[/tex] with the downward vertical, the potential energy, V, of the system when the string is stretched is given by

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

I drew the following diagram:-

I equated the GPE of the uniform rod to be

[tex]AB = -mg\cos\theta[/tex]

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

[tex]OC = \frac{\lambda x^2}{2a}[/tex]

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

[tex]x = (2a\sin\theta - a)[/tex]

So:

[tex]OB = \frac{kmg(2a\sin\theta - a)^2}{2a}[/tex]

[tex]OB = \frac{kmga(4\sin^2\theta -4\sin\theta + 1)}{2}[/tex]

[tex]OB = 2kmgasin^2\theta -2kmg\sin\theta + \frac{kmg}{2}[/tex]

But since:

[tex]OB = \frac{kmg}{2}[/tex] is a constant, we can take it out of the equation.

Using the identity: [tex]\sin^2\theta = 2\sin\theta\cos\theta[/tex]

[tex]4kmga\sin\theta\cos\theta -2kmga\sin\theta[/tex]

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

Any help would be really appreciated.

Thank you.

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 [tex]\theta[/tex] with the downward vertical, the potential energy, V, of the system when the string is stretched is given by

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

I drew the following diagram:-

I equated the GPE of the uniform rod to be

[tex]AB = -mg\cos\theta[/tex]

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

[tex]OC = \frac{\lambda x^2}{2a}[/tex]

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

[tex]x = (2a\sin\theta - a)[/tex]

So:

[tex]OB = \frac{kmg(2a\sin\theta - a)^2}{2a}[/tex]

[tex]OB = \frac{kmga(4\sin^2\theta -4\sin\theta + 1)}{2}[/tex]

[tex]OB = 2kmgasin^2\theta -2kmg\sin\theta + \frac{kmg}{2}[/tex]

But since:

[tex]OB = \frac{kmg}{2}[/tex] is a constant, we can take it out of the equation.

Using the identity: [tex]\sin^2\theta = 2\sin\theta\cos\theta[/tex]

[tex]4kmga\sin\theta\cos\theta -2kmga\sin\theta[/tex]

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

Any help would be really appreciated.

Thank you.

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