# Problem in Classical Mechanics

Tags:
1. Jul 5, 2016

### </Henry>

1. The problem statement, all variables and given/known data
I am stuck over a classical mechanics problem. I tried to solve it, but after having solved the first point, I got stuck on the second one. Here is the problem:

>A mechanical structure is composed by 4 rigid thin bars of length $\ell = 8\ m$, mass $m = 5\ kg$ each one. Those bars are connected each other at their extremities through frictionless hinges, such that the whole structure forms a vertical rhombus.
The structure is hanged from the ceiling by means of the upper hinge.
In the lower hinge, a mass $M = 30\ kg$ is hanged.
Between the two horizontal hinges, a rectilinear spring (whose mass is negligible) acts with elastic constant $k = 40.5\ N/m$ and rest length $L = 9\ m$, such that the angle between the two superior bars is $\theta = 40^{\circ}$.

>Calculate:

>1) The Force the spring exerts over one of the hinges, and the total tension of the bars connected to the spring in that point;

>2) What should be the elastic constant if $\theta$ remains the same, when the whole structure rotates with constant angular velocity $\omega = 2$ rad/s, around the vertical axis?

>3) The moment of Inertia with respect to the rotation axis of the previous question, and the angular momentum when it rotates with a general $\omega$.

Here is the figure

For the first point, I calculated, though some trigonometry, the length of the spring in those conditions. Calling $l$ the semi length, I got

$$l = \ell\sin\frac{\theta}{2} = 2.736\ m$$

So the total length is $2l = 5.472\ m$. Now the elastic force is simply

$$F_k = k\Delta l = k (L - 2l) = 142.88\ N$$

And it's ok.

For the second point of the first request I used

$$T_{\text{tot}} = 2T\cos 70 ~~~~~~~ 2Tcos 70 - F_k = 0$$

$$T = \frac{F_k}{2\cos 70} = 209\ N$$

And it's ok.

**Now I'm stuck over the second request**

**Numerical results:**

2) $$\Delta k = -15.5\ N/m$$

3) $$I = 49.9\ kg\cdot m^2$$ and $$L = 99.8\ kg\cdot m^2/s$$

Any help?

2. Jul 5, 2016

### haruspex

I have a concern with the problem statement.
The set-up appears overspecified. There is enough information to calculate the angle between the bars, and it is somewhat unlikely that it will turn out to be exactly 40 degrees.

For the second part, please post your working; can't tell where you are going wrong if you don't.