Falling Ladder

Suppose that the rod has length $2r$ and mass $m$ . While the rod is in contact with both the wall and the floor, its centre moves along a circle of radius $r$ with centre the point where the wall and floor meet. If the rod makes an angle $\theta$ with the downward vertical, then the velocity of the centre of mass is $r\dot{\theta}$ , and so the kinetic energy of the rod is $T \; = \; \tfrac12mr^2\dot{\theta}^2 + \tfrac12 \times \tfrac13mr^2 \times \dot{\theta}^2 \; = \; \tfrac23mr^2\dot{\theta}^2$ and so conservation of energy tells us that $\tfrac23mr^2\dot{\theta}^2 + mgr\cos\theta \; = \; mgr$ Thus the speed of the centre of mass is $v = r\dot{\theta}$ , where $v^2 \; = \; \tfrac32gr(1 - \cos\theta)$ , and so the horizontal component of the velocity of the centre of mass is $v_x \; = \; v\cos\theta \; = \; \sqrt{\tfrac32gr(1 - \cos\theta)}\cos\theta$ Then $\frac{d v_x}{d\theta} \; =\; \sqrt{\tfrac32gr}\frac{\sin\theta(3\cos\theta - 2)}{2\sqrt{1 - \cos\theta}}$ Since $\dot{\theta} > 0$ at all times and $m\dot{v}_x = R$ , we see that the rod remains in contact with the wall for as long as $\frac{d v_x}{d\theta} \ge 0$ , namely so long as $\cos\theta \ge \tfrac23$ .

The rod parts company with the wall when $\theta = \cos^{-1}\tfrac23$ , at which point $v = \sqrt{\tfrac12gr}$ and $v_x = V = \tfrac13\sqrt{2gr}$ . After this the rod is acted upon solely by gravity and the reaction force $N$ , and so we have the equations of motion $m\ddot{y} \; = \; N - mg \hspace{2cm} \tfrac13mr^2\ddot{\theta} \; = \; Nr\sin\theta$ where $y = r\cos\theta$ is the height of the centre of mass above the ground. From these equations we deduce (after eliminating $N$ ) that $r^2(1 + 3\sin^2\theta)\ddot{\theta} + 3r^2\sin\theta\cos\theta \dot{\theta}^2 \; = \; 3gr\sin\theta$ and so (solving this differential equation and matching the boundary conditions from the first stage of the motion) the second part of the motion is described by the formula $\dot{\theta}^2 \; = \; \frac{2g(8 - 9\cos\theta)}{3r(1 + 3\sin^2\theta)} \hspace{2cm} \cos^{-1}\tfrac23 < \theta < \tfrac12\pi$ Thus, when the rod hits the ground we have $\dot{\theta}^2 = \tfrac{4g}{3r}$ , and so the vertical component of the velocity of the tip of the rod is $2r\dot{\theta} = \sqrt{\tfrac{16}{3}gr}$ . Throughout the second stage of the motion the horizontal component of the velocity of the centre of mass of the rod remains as $V$ , and so the horizontal component of the velocity of the tip of the rod when it strikes the ground is also $V$ . Thus the speed of the tip of the rod at the point of impact is $\tfrac53\sqrt{2gr}$ .

With $2r =8.1$ and $g=10$ , we deduce that the speed of the tip of the rod at the moment of impact with the ground is $\boxed{15}$ m s ${}^{-1}$ .

It's a very small distance, so if we assume that distance is negligible, we'll get the height to be 8.1m.

$\frac {mv^2}{2}=mgh$

$\frac {v^2}{2}=gh$

$v^2=2gh$

$v=\sqrt {2 \times 10 \times 8.1}$

$v=\sqrt {2 \times 81}$

$v=9\sqrt {2}$

Which is approximately 15.

Since Mark Hennings posted the correct solution, I have no choice, but posting an incorrect one :-)

Initially, the rod has a potential energy of $mg \ell/2$ , where $\ell$ is the length and $m$ is the mass.

In the final state the rod will rotate around the end that touches the ground. We will assume that this end has no translational velocity. ( This plausible, but is wrong - see Mark's solution for the correct way to do this. ) The rotational kinetic energy is $\frac {1}{2} I \omega^2$ , where $I=\frac{1}{12}m \ell^2$ is the moment of inertia around the end point. The transalational kinetic energy of the center of mass is $\frac{1}{2} m v_c^2$ , where $v_c=\frac{1}{2} \ell \omega$ . From energy conservation we get

$\frac{1}{2} mg\ell= \frac{1}{2} \frac{1}{12} m\ell^2\omega^2+ \frac{1}{2} m \frac{\ell^2}{4}\omega^2$

We solve this for $\omega$ to get

$\omega^2=\frac{3g}{\ell}$

The velocity of the end point is $v=\omega \ell = 15.58$ m/s. This is not very different from the correct solution, BUT it is larger, because in the correct solution there is a residual horizontal velocity for the center of mass.

Note that in the correct solution (see Mark Hennings):

$\omega^2=\frac{8}{3} \frac{g}{\ell}$