# Yo-Yo Physics

You are given the yo-yo in the figure that has an axle of radius $R_0$, an outer radius $R$, and a radius of gyration $k$. There is a very thin string wound around its axle having negligible mass. The yo-yo is released with one end of the string is fixed, What is the acceleration of the yo-yo on the vertical? (use Lagrangian formulation)

$L=T-U$ where $T$ is kinetic energy and $U$ is potential energy

If we chose x the general coordinate (distance of falling) then

$T=T_{rotation}+T_{translation}=(Iω^2)/2+(m v^2)/2=I/(2(R_0)^2 )*(x^2 ) ̇+m/2*(x^2) ̇$

where

$x ̇=dx/dt(=v=ωR_0)$

$U=+mgh=-mgx$

In this equation $x$ is the falling distance. Thus (up until a constant) the Lagrangian is

$L(x,x ̇,t)=(I/(2R_0^2 )+m/2)*(x^2 ) ̇+mgx$

The associated Euler-Lagrange equation (said differently the equation of motion) is

$(d/dt) (∂L/(∂x ̇ )=∂L/∂x$

$(d/dt) [(I/(R_0^2 )+m) x ̇ ]=mg$

$a=x ̈=frac{mg}{m+I/(R_0^2 )}=frac{mg}{(m+(k^2 mR^2)/(R_0^2 )}=frac{g}{1+(kR/R_0 )^2} $

Because the definition of radius of gyration k is: $I=m(kR)^2$