Solving for the Potential Energy of a Uniform Rod Attached to an Elastic String

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The discussion centers on calculating the potential energy of a uniform rod attached to an elastic string in a vertical plane. The potential energy is expressed as V = mga[(4k - 1)cosθ - 4kcos(θ/2)] when the rod makes an angle θ with the downward vertical. Participants clarify the relationship between the angles and lengths in the system, leading to the correct formulation of the energy equation. A key point involves using the geometry of the triangle formed by the rod and the string to derive the necessary expressions. Ultimately, the correct potential energy equation is confirmed through mathematical manipulation and geometric insights.
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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 \theta with the downward vertical, the potential energy, V, of the system when the string is stretched is given byV = mga[(4k - 1)\cos\theta -4k\cos\frac{\theta}{2}] + constant

I drew the following diagram:-
diagram.jpg


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

Which is the correct answer.
 
Yea, you are right. Good work!.
 
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