[Physics] Will a falling rod stay in contact with the frictionless floor

classical-mechanicsnewtonian-mechanicsrotational-dynamicsrotational-kinematics

Question

A uniform rod of mass $M$ is placed almost vertically on a frictionless floor. Since it is not perfectly vertical, it will begin to fall down when released from rest.

I have seen solutions online for this problem and while solving this problem, it is assumed that the end point of the rod that is in contact with the floor will continue to stay in contact with the floor till the rod, in its entirety, hits the floor horizontally. It is this assumption that lets us determine the normal force from the floor. However, how does one show that this assumption is true? Or is it taken to be an additional constraint of the problem?

Check the figure in D1 to verify if you've got the right setup in mind.

Duplicates in SE:

I believe the OP in D1 has asked the same question (along with other questions) but it has been closed as off-topic. Simon Robinson, one of the answerers in D2, has also expressed concerns about this. I ask this question because it hasn't been addressed properly on SE. I don't feel that the answer to this question is specific only to this vertical rod problem. Instead, I feel that this question is onto something basic that I don't yet understand regarding the necessary constraints that need to be specified in a physics problem.

My Attempt

The problem with this question is that I feel like I have given all the information that's necessary to predict the entire dynamics of the rod's motion after its release. I'm unable to accept the idea that "rod-cannot-lose-contact" constraint must be specified as an additional piece of information to solve this problem. If we accept that it's not an additional constraint, then we should be able to show that the rod's end point cannot lose contact. But, that's the problem. I've been thinking about it for days and I can't seem to find a way to show that.

I'm unable to see anything "violated" if it loses contact at some point during its fall. After it loses contact, it simply rotates about the center of mass with constant angular velocity [See $(1)$] and the rod's COM falls down with acceleration $\mathbf{g}$.
$$\frac{d\mathbf{L}_{CM}}{dt} = \boldsymbol{\tau}_{CM} \Rightarrow \text{$L_{CM}=I_{CM}\omega\;$ is constant} \tag{1}$$

Thanks for taking the time to read this question. I apologize if I have violated any code of conduct.

Any insight that addresses my question would be greatly appreciated.

Further Clarification, If Needed

Clarifications which will hopefully help PhySE users to better understand my question are made here. Reading the following information is not necessary to answer my question.

  1. It is important to note that even if the rod's bottom end point loses contact with the floor at some point during the fall, the centre of mass of the rod will continue to fall vertically straight down just as before (but now with acceleration $\mathbf{g}$). So, the fact that the COM falls vertically straight down cannot be used to prove that the rod's bottom end point doesn't lose contact with the floor.

    COM falls vertically straight down $\not\Rightarrow$ the rod's bottom end point doesn't lose contact with the floor

Best Answer

The technique to use in problems like this is to assume that the rod remains in contact with the table, and to then try to figure out whether the normal force ever switches sign for some angle $\theta$ as the rod falls. If it does, then the rod's lower tip will have to leave the table, as a "frictionless table" cannot pull the rod downward; it can only push it upwards. Similar techniques are used in the solution to the classic "puck slides down a frictionless hemisphere" problem, as well as the "toppling ruler" problem.

Actually doing this is something of a mess, but here's a rough sketch. Let $L$ be the length of the rod and $m$ be its mass. Let $I = \frac{1}{4} \beta m L^2$ be the rod's moment of inertia about its center of mass; note that $\beta = \frac{1}{3}$ for a rod of uniform density, while $\beta = 1$ if the mass is concentrated at the tips. This is done to provide a bit more generality; I will assume, however, that the mass distribution is symmetric, so that the center of mass is at the geometric center of the rod.

The ingredients you'll need are:

  • Geometric constraints: The vertical position of the center of mass of the rod will be $z = \frac{1}{2} L \cos \theta$ (taking positive $z$ to be upwards.) Differentiating this twice, we obtain for the velocity and acceleration of the center of mass $$ v = - \frac{L}{2} \omega \sin \theta, \\ a = - \frac{L}{2} ( \alpha \sin \theta + \omega^2 \cos \theta), $$ where $\alpha$ is the angular acceleration of the rod.

  • Conservation of energy: Since the table does no work on the tip of the rod, the mechanical energy of the rod is conserved. This gives a relationship between $v$ and $\omega$.

  • Newton's Second Law (translational): Using Newton's second law, you can relate $a$ and $N$.

  • Newton's Second Law (rotational): Calculating the torque about the center of mass of the rod, you can find a relationship between $N$ and $\alpha$.

This gives us a system of five equations and five unknowns $\{N, v, a, \omega, \alpha \}$ which can be solved. After going through it, I find that the normal force as a function of $\theta$ is $$ N = \frac{mg \beta (\beta + (1- \cos \theta)^2)}{(\beta + \sin^2 \theta)^2} $$ which is manifestly positive for any value of $\theta$. Thus, the tip of the rod does not leave the table; the table continually maintains an upward normal force as it falls.