'Two-dimensional' lagrangian problem. [It is a problem for which we will use two generalized coordinates, but it is not genuinely a 2-D problem.] A frictionless slab of mass m is attached to a wall via a spring. Rolling w/o slipping on the frictionless slab is a circular cylinder of mass M and radius R. Determine the frequency of oscillation of this combination.

Use two generalized coordinates

• x, which runs from the spring unstretched position to the current position of the spring (which we can take to be the edge of the slab)
• z. which runs from the edge of the cart to the CM of the cylinder
• thus, the distance from the inertial reference point to the cylinder CM is x+z
• I is the rotational inertia of the circular cylinder wrt its CM
• theta is the rotation angle of the cylinder, and theta=z/R
• theta-dot = omega = z-dot/R

• T = 1/2 m x-dot^2 + 1/2 M(x-dot + z-dot)^2+ 1/2 I omega^2
• T = 1/2 m x-dot^2 + 1/2 M(x-dot + z-dot)^2+ 1/2 I z-dot^2 / R^2
• T = 1/2 [ (m+M) x-dot^2 + 2M x-dot z-dot + (M+I/R^2) z-dot^2 ]

• U = 1/2 k x^2
• L = T-U
• px = generalized x-momentum = dL/dx-dot = (m+M) x-dot + M z-dot
• pz = generalized z-momentum = dL/dz-dot = M x-dot + (M+I/R^2) z-dot

• Now for the two equations of motion
• d/dt(px) =dL/dx = (m+M) x-dot-dot + Mz dot-dot = -kx
• d/dt(pz) = dL/dz = d/dt(M x-dot + (M+I/R^2) z-dot) = 0

Notice that pz is a constant of the motion because the Lagrangian does not contain the coordinate z. (The other constant of the motion is energy)

From pz = constant we get (M+I/R^2) z-dot-dot = - M x-dot-dot

We put this in the x-equation of motion and get

• (m+M) x-dot-dot + M x-dot-dot (-M/(M+I/R^2))= -kx
• x-dot-dot ( (m+M)(M+I/R^2) - M^2)/c
• x dot-dot (mM + (m+M)I/R^2 )/(M+I/R^2) = -kx
• If I = 1/2 MR^2 ( a uniform cylinder) then we have
• x-dot-dot (mM +(m+M)M/2 )/(3/2 M) = -kx
• x-dot-dot (2/3)(m+m/2 + M/2) = x-dot-dot (m+M/3) = -kx
• For a uniform cylinder, omega is sqrt( k/(m+M/3))