# Rod rotated by elastic string [closed]

A uniform rod $AB$ of length 2m and mass 1kg, has a mass of 1kg attached at $B$. It can rotate freely about a horizontal axis through $A$. The end $B$ is attached by means of a light elastic string natural length 1m to a point a distance of 2m above $A$. The system is held at rest with $AB$ horizontal and is then released. If the rod just reaches the vertical position find the modulus of elasticity of the string.

This is what I've done so far.

Work done for rotation (Couple $\times$ Angle): $$\theta(2T -\frac{3mg}{2})=\frac{ \theta}{2}(\frac{ 4 \lambda x}{l}-3mg)$$

Change in Energy: $$mgh-\frac{\lambda x^2}{2l}$$

Equate these and rearrange to get: $$\lambda=\frac{lmg(2h+3 \theta)}{x(4\theta + x)}$$

Values: $$x=2\sqrt{2} -1, \; l=1, \; m=2, \; \theta=\frac{\pi}{2}, \; h=\frac{3}{2}, \; g=9.8$$

My answer is 10.2, the right answer is meant to be 10.4.

Would it be possible to explain to me where I've gone astray? I'm self teaching rotational dynamics and am fairly new to the topic.

The issue I think is that the couple/force provided by the tension is not constant as the extension changes as the rod rotates. I am fairly sure I don't need to (and can't ) integrate the tension (with respect to the angle) so there must be another way, still via energetics. What is the best way to tackle these kind of problems?

Here is the diagram I drew, this may be wrong too ($G$ signifies centre of gravity).

• I'm not sure I understand. Your diagram seems to show a final position that isn't described in the text. Also, you could find the total energy in the string, but to convert that to a modulus, I think you'd need a cross section which isn't given. Jan 29, 2015 at 7:55
• My mistake, the question adds that the rod just reaches the vertical position. I've edited the question. Jan 29, 2015 at 11:30
• I'm not sure why this has been closed, What could I do to improve this question? Feb 13, 2015 at 16:02