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.