Falling pole lifting off the ground.

[Prologue]

I have heard (what I consider to be) a myth of a pole lifting off the ground before it hits. That is, set something like a telephone pole up on end with no initial falling velocity, then let it drop. The claim is that the bottom of the pole actually lifts from the ground before the whole thing crashes down. This is not a statement from experience (that I know of) but from a (very good) physics professor of mine, he says that it is true and I just don't believe him. There is a little ambiguity that I'll have to ask him about when I get the chance, whether or not the ground has friction, but I suppose it could be worked out both ways and it still would be crazy to me. He says the key it to look at the normal force, figure out when it is zero and then calculate the angle (measured from vertical) and it will be less than 90 degrees.

This isn't a homework problem, just something he mentioned a few months ago and I have some time to think about it now. I figured I would look at the frictionless case first and then take an energy approach, say

$$dE=0$$

And where M is the total mass of the pole, h is the height of the center of mass, the total length of the pole is given by 2l, I is the moment of inertia

$$Length=2l$$

$$h=l\sin{\phi}$$

$$\dot{h}=velocity \: of \: CM$$

$$I=\frac{1}{12}M(2l)^{2}$$

$$KE=\frac{1}{2}M\dot{h}^2+\frac{1}{2}I \omega ^{2}$$

$$U=Mgh$$$$E=KE+U$$

$$E=\frac{1}{2}M\dot{h}^2+\frac{1}{2I} L^{2}+Mgh$$

E has its structure because there will be rotation around the CM, there will be translation straight down of the CM, and there will be a potential energy associated with CM.

It is clear that h, L, and h dot are functions of time. I, M, g, E are not. My thought was to try and find the torque on the pole about the center of mass, then integrate over time to get the angular momentum. This looks like

$$\tau=Mg(h\sin{\phi})$$
$$L=\int_{0}^{t}\tau dt=\int_{0}^{t} Mg(h\sin{\phi})dt$$

But, this approach leaves me with a nasty equation

$$E=\frac{1}{2}M\dot{h}^2+\frac{1}{2I} (\int_{0}^{t} Mg(h\sin{\phi})dt)^{2}+Mgh$$I think that if I found h(t), then the rest would be tractable, but I don't see how to get it.

I would really like to get a handle on the method that one would use to solve this problem, as it seems mine are inadequate. My professor claims that it can be done with intro physics (calculus based) techniques. Got any ideas?

 

[kuruman]

You cannot find $h(t)$ analytically. Follow your professor's suggestion and use energy conservation to find the angular speed $\omega$ as a function of $\theta$ and hence the normal force as a function of $\theta$. Here is how.

We will assume that the pole is hinged at the point of contact and find the angle at which the vertical force on the hinge becomes zero. We take the zero of potential energy to be the vertical position of the pole. The change in potential energy is $\Delta U=-mgl(1-\cos \theta)~$ where $\theta$ is measured from the vertical. The pole rotates about the hinge so its change of kinetic energy starting from zero is $$\Delta K=\frac{1}{2}I \omega^2=\frac{1}{2}\times \frac{1}{3}m(2l)^2 \omega^2=\frac{2}{3}ml^2 \omega^2.$$Mechanical energy conservation says $\Delta U+\Delta K=0$ which gives the centripetal acceleration of the cm$$ a_C=\omega^2 l=\frac{3}{2}g (1-\cos \theta).$$

Now we draw the FBD for the rod

The torque equation about the hinge gives the tangential acceleration of the cm$$mgl \sin(\theta)=\frac{4}{3}ml^2\alpha=\frac{4}{3}ml^2\frac{a_{T}}{l}~\rightarrow~a_{T}=\frac{3}{4}g\sin \theta.$$We now write Newton's 2nd law in the vertical direction which is of interest here
$$F_{net,y}=F_y-mg=-ma_C \cos\theta-ma_T \sin\theta$$
Component $F_y$ is zero when
$$\frac{3}{2}g (1-\cos \theta)\cos\theta+\frac{3}{4}g\sin^2 \theta=g$$This equation simplifies to$$\cos^2 \theta-\frac{2}{3}\cos\theta+\frac{1}{9}=0$$Note that it's a perfect square, therefore the solution is $$\cos\theta=\frac{1}{3}~\rightarrow~\theta=70.5^o$$Answer: The rod's end will lift off the ground at angle $70.5^o$ with respect to the vertical.

Now that we have dispensed with the mythbusting, let's find the horizontal force at that angle
$$F_{net,x}=-ma_C \sin \theta+ma_T \cos\theta=-m\frac{3}{2}g (1-\cos \theta)\sin \theta+m\frac{3}{4}g\sin \theta \cos \theta$$ With $\cos\theta=\dfrac{1}{3};~\sin \theta=\sqrt{\dfrac{8}{9}}$, $$F_{net,x}=-\frac{1}{\sqrt{2}}mg.$$