# Linear and angular velocity of a falling rod in the corner of a room [closed]

A uniform rod of length l and mass m is sitting vertically in a frictionless environment in the corner of a room, having a floor beneath it, and a wall to its left. It gets slightly off balance and so its top end starts falling towards the ground to the right; its bottom end may not move towards its left, but it may move further right; the bottom end may also not move downwards as it's sitting on the floor.

Would this rod gain both linear velocity as well as angular velocity? How would one go about computing the two velocities? I'm not asking for a full derivation since I don't want to be greedy (one will be much appreciated, of course!), just an outline for how to go about it and which formulas I'd need to use.

My intuition is telling me that the rod would gain both linear velocity as well as angular velocity, and so if the room were to completely disappear mid-fall, it would both keep spinning around its center of mass, as well as keep moving to the right and downwards. What I am unsure of is how conservation of energy would work in this case -- the initial potential energy should somehow have to split into both rotational and translational kinetic energy, a situation I haven't seen in my study of physics.

• @JohnAlexiou not really, I'm still trying to understand the concepts behind how kinetic energy is split between linear and angular; what I gather is that one can always choose an axis of rotation such that all motion is angular? But I'm still working through Michael's answer who was kind enough to help me
– Gabi
Nov 28, 2022 at 23:17
• Look at the accepted answer, and you will see how to solve these problems. First establish the kinematics (describe the position of the center of mass using the DOF variables) and then apply the laws of motion. Note that of net force is not zero then momentum will be affected, and if net torque is not zero then angular momentum is affected. Nov 28, 2022 at 23:20

It is not hard to see that the linear velocity of the center of mass $$v_{CM}$$ must be non-zero. If $$v_{CM}$$ was zero, but the angular velocity was not zero, then one of the ends of the rod would have to rotate through the wall.

To actually solve this problem, you'll need to use conservation of energy, and you'll also need to figure out a relationship between the linear velocity and the angular velocity of the rod. To find this latter relationship, start by figuring out the $$x$$ and $$y$$ positions of the CM of the rod as a function of the angle of the rod; then see what it implies about the path taken by the CM of the rod.

Finally:

What I am unsure of is how conservation of energy would work in this case -- the initial potential energy should somehow have to split into both rotational and translational kinetic energy, a situation I haven't seen in my study of physics.

If you've done a problem with a ball or a barrel or a hoop rolling down a hill, you've done a problem in which potential energy is converted into both translational and rotational kinetic energy. In that case, as well as in this one, there is a constraint between the object's translational velocity and its angular velocity that lets us write the total KE of the object in terms of only its linear velocity or its angular velocity.

• Thanks a lot! "start by figuring out the x and y positions of the CM of the rod as a function of the angle of the rod" -- I've done this under the assumption that the lower end will not move towards the right before the top end hits the floor -- is my assumption correct?
– Gabi
Nov 28, 2022 at 20:54
• @Gabi: Doesn't the bottom end have to move to the right as the top end falls? Otherwise it sounds like the bottom end is stationary while the top end is moving, meaning that the length of the rod is not constant. Nov 28, 2022 at 21:05
• Sorry, this one confused me -- I was thinking of the case where the bottom end stays fixed in the corner, taken to have coordinates $(0, 0)$, while the CM traces the path $(l/2) (\cos \alpha, \sin \alpha)$ from $\alpha = \pi / 2 \dots 0$ ; in this case the rod length would be constant. Am I missing something crucial perhaps? :) Thanks!
– Gabi
Nov 28, 2022 at 22:49
• @Gabi: Ah, I misunderstood. I thought you were doing the problem where the base of the rod slides out to the right while the top end slides down the wall. In either case, though, the easiest way to solve it is to assume that the end is fixed, and then look at the forces that are necessary to create that motion. If it happens, for example, that the wall must apply a force to the left to create this motion, then that's an indication that the rod loses contact with the wall at that point. Nov 29, 2022 at 12:09