More on Mechanics?

We will be using the so called Euler Lagrange formulation

`d/dt((partialL)/(partial dot q_i))-(partial L)/(partial q_i)=Q_i`

where `L = T-V`. In this exercise we have `V=0` so `L = T`

Calling `x_a` the center of left cylinder coordinate and `x_b` the rigth one, we have

`x_b=x_a+R costheta+Lcosalpha`

Here `sinalpha=R/Lsintheta` so substituting for `alpha`

`x_b=x_a-R costheta + sqrt[L^2 - R^2 sin^2theta]`

now deriving

`dot x_b=dot x_a + Rsin(theta) dot theta-((R^2cos(theta)sin(theta))/sqrt(L^2-R^2sin^2(theta)))dot theta`

but

`T=1/2 J (omega_a^2+omega_b^2)+1/2m(v_a^2+v_b^2)`

Here `J` is the inertia momentum regarding the mass center. Also,

`v_a= dot x_a=R dot theta`
`omega_a = dot theta`

so, after substitutions and calling `xi(theta) = 1-(Rcos(theta))/sqrt(L^2-R^2sin^2(theta))` we have

`T=1/2(J+mR^2)(1+(1+sin(theta)xi(theta))^2)dot theta^2`

We choosed `theta` as the generalized coordinate. So we will reduce `F` actuating in the coordinate `x` to an equivalent force in `theta`. This coordinate acts rolling wise so we need a generalized momentum regarding the contact point in the floor, which is

`Q_(theta)=FR(1+ sintheta)`

The movement equations are obtained after

`(J+mR^2)((1+sin(theta)xi(theta))(cos(theta)xi(theta)+sin(theta)xi'(theta))dot theta^2+(1+(1+sin(theta)xi(theta))^2)ddot theta)=FR(1+sin(theta))` now solving for `ddot theta`

`ddottheta=(FR(1+sin(theta))-(J+mR^2)(1+sin(theta)xi(theta))(cos(theta)xi(theta)+sin(theta)xi'(theta))dottheta^2)/((J+mR^2)(1+(1+sin(theta)xi(theta))^2))`

Attached two plots. The first shows `theta` evolution and the second is for `dottheta`

Value of parameters:

`R=0.5,J=1,m=1,L=2` The applied force is shown in dased red.