Statics (11/13/2019)

Let the angle that the rod makes with the horizontal be $\theta$ .

The coordinates of the COM of the rod are:

$x_1 = \frac{L}{2}\cos{\theta}$ $y_1 = \frac{L}{2}\sin{\theta}$

The coordinates of the end of the rod away from the origin is:

$x_2 = 2x_1 \ ; \ y_2 = 2y_1$

The elongation of the spring at this instant is:

$L_e = \sqrt{(x_2 - 0)^2 + (y_2 - 2)^2} - Lo$

Finally, the total potential energy of the rod-spring system is:

$V = mgy_1 + \frac{1}{2}kL_e^2 = PE_{Gravitational } + PE_{Spring }$

This is a function of $\theta$ . At equilibrium, the system has minimum potential energy. Differentiating the above with respect to $\theta$ and equating to zero, and simplifying gives:

$\frac{dV}{d\theta} = \frac{36\,\cos\left(\mathrm{\theta}\right)}{\sqrt{13-12\,\sin\left(\mathrm{\theta}\right)}}-21\,\cos\left(\mathrm{\theta}\right) = 0$

Solving for $\theta$ gives:

$\sin{\theta} = \frac{493}{588}$

In degrees, this is $\boxed{\theta \approx 56.9753 ^o}$

Of course, one has to carry out the second derivative test to be sure of a minimum, but that is left out along out the number crunching.

One can also notice that $\theta = 90 ^o$ is also a solution to the above equation. In that configuration, the total potential energy is maximum.

I like this problem, let's call the angle between the horizontal and the rod $\theta$ . In order for this system to be at rest, we must have equilibrium of torque on the rod. Firstly, let's consider gravity, which will act on the centre of mass, perpendicular to the rod. If the rod's length is $\frac{\ell}{2}$ , then torque supplied by gravity is given by

$\tau_g \ = \ mg \frac{\ell}{2} \cos \theta$

This must be equal to the torque supplied by the spring. However, we must only consider the component of the spring that acts perpendicular to the rod at length $\ell$. Doing a bit of basic geometry, we find that the torque supplied by the spring is:

$\tau_s \ = \ k \Delta x \ell \sin \alpha$

Where the angle $\alpha$ is the angle between the spring and the rod. We know that the length from the end of the rod at the pivot point to the part of the spring attached to the wall is $2$ . In addition, the length of the rod is $3$ , and the length of the spring will be given by $\Delta x \ + \ 1$ . We can now use the Law of Sines to find the angle $\alpha$ . Since we have the side length of all the sides of the triangle formed by the wall, the rod, and the spring (we will call the length of the wall $W$ . The length of the spring is given by $1 \ + \ \Delta x$ ):

$\frac{W}{\sin \alpha} \ = \ \frac{\Delta x \ + \ 1}{\sin (\frac{\pi}{2} \ - \ \theta)} \ = \ \frac{\Delta x \ + \ 1}{\cos \theta} \ \Rightarrow \ \sin \alpha \ = \ \frac{W \cos \theta}{\Delta x \ + \ 1}$

Like we said before, we must set these torques in equilibrium:

$\tau_s \ = \ \tau_g$ $k \Delta x \ell \frac{W \cos \theta}{\Delta x \ + \ 1} \ = \ mg \frac{\ell}{2} \cos \theta \ \Rightarrow \ k \Delta x W \ = \ \frac{(\Delta x \ + \ 1) mg }{2} \ \Rightarrow \ \Big( kW \ - \ \frac{mg}{2} \Big)\Delta x \ = \ \frac{mg}{2}$ $\Rightarrow \ \Delta x \ = \ \frac{mg}{2 \Big( kW \ - \ \frac{mg}{2} \Big)} \ = \ 0.7142857$

Finally, we can use Cosine Law to find the angle $\alpha$ . We have:

$(\Delta x \ + \ 1)^2 \ = \ \ell^2 \ + \ W^2 \ - \ 2\ell W \cos (90^{\circ} \ - \ \theta)$ $\Rightarrow \cos(90^{\circ} \ - \ \theta) \ = \ \frac{\ell^2 \ + \ W^2 \ - \ (\Delta x \ + \ 1)^2}{2 \ell W} \ = \ \frac{3^2 \ + \ 2^2 \ - \ (0.7142857 \ + \ 1)^2}{2 (3) (2)} \ = \ 0.83843537823$ $\theta \ = \ 56.975^{\circ}$