8/20/2023

 Today I did a few past F=MA problems, and I encountered a lot of new ideas! My favorite problem was the following:

Problem:

A system of cylinders and plates is set up as shown. The cylinders all have radius $r$, and roll without slipping to the right with angular velocity $\omega$. What is the speed of the top plate?

Source: 2020 F=MA #19

Solution:

In this solution, let plates 1, 2, and 3 denote each plate, numbered 1 to 3 from bottom to top.

Consider the bottom row of cylinders. Since each cylinder is rolling without slipping, the velocity of the cylinders' topmost point, which is the velocity of the 1st plate, is $2\omega R$. Now, note that the cylinders in the second row are also rolling without slipping. In the reference frame of their bottom point, plate 2 moves with velocity $2 \omega R$. However, since the bottom point has no relative velocity with respect to plate 1 (the plate it's rolling on), the instantaneous velocity of the bottom point with respect to the floor is the velocity of plate 1, which is $2 \omega R$. Hence the velocity of plate 2 with respect to the floor is $2 \omega R + 2 \omega R = 4 \omega R$. Using the same logic, the velocity of plate 3 with respect to the ground is $2 \omega R + 4 \omega R = \boxed{6 \omega R}$. 

This was a really cool problem!