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Stability physics problem

  • Thread starter mr bob
  • Start date
  • #1
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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:-
diagram.jpg


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.
 
Last edited:

Answers and Replies

  • #2
Pyrrhus
Homework Helper
2,178
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Hey Bob, how can you be sure the angle made by AB and OB is 90 degrees?
 
  • #3
38
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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 [tex]\cos\frac{\theta}{2}[/tex]?

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
38
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Just ran through it trying splitting the triangle and i got

[tex]OB = kmg(4a\cos\theta -4a\cos\frac{\theta}{2})[/tex]

As this is equal to the GPE of the rod.

[tex]mga\cos\theta = kmg(4a\cos\theta -4a\cos\frac{\theta}{2})[/tex]

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

Which is the correct answer.
 
  • #5
Pyrrhus
Homework Helper
2,178
1
Yea, you are right. Good work!.
 

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