Oscillating disks on a step

I've been looking for opportunities to use Lagrangian Mechanics , and this is a good one. Call the position of the larger disk $x_1$ and the position of the smaller disk $x_2$ . The masses are similarly $M_1$ and $M_2$ . Note that the disks are rolling without slipping, which has two implications:

1) No energy losses - hence LM is applicable in a straightforward way
2) There is a definite relationship between rotation and translation

Accounting for both rotation and translation (left as an exercise for the reader), the kinetic energy is:

$\large{KE = \frac{3}{4} M_1 (\dot{x_1})^2 + \frac{3}{4} M_2 (\dot{x_2})^2}$

The spring potential energy is ( $L_0$ is the un-stretched spring length): $\large{U = \frac{1}{2} k (x_2 - x_1 - L_0)^2}$

The system Lagrangian is:

$\large{L = KE - U = \frac{3}{4} M_1 (\dot{x_1})^2 + \frac{3}{4} M_2 (\dot{x_2})^2 - \frac{1}{2} k (x_2 - x_1 - L_0)^2 }$

The Euler-Lagrange equations are:

$\large{\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x_1}}} = \frac{\partial{L}}{\partial{x_1}} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x_2}}} = \frac{\partial{L}}{\partial{x_2}}}$

Evaluating these yields:

$\large{\frac{3}{2} M_1 \ddot{x_1} = k(x_2 - x_1 - L_0) \\ \frac{3}{2} M_2 \ddot{x_2} = -k(x_2 - x_1 - L_0)}$

Plugging in specified parameters for the masses gives:

$\large{6 M \ddot{x_1} = k(x_2 - x_1 - L_0) \\ \frac{3}{2} M \ddot{x_2} = -k(x_2 - x_1 - L_0)}$

Multiplying the second equation by 4 gives:

$\large{6 M \ddot{x_1} = k(x_2 - x_1 - L_0) \\ 6M \ddot{x_2} = -4k(x_2 - x_1 - L_0)}$

After some consolidation:

$\large{6 M \ddot{(x_2 - x_1)} = -5k(x_2 - x_1) + \alpha \\ \ddot{(x_2 - x_1)} = \frac{-5k}{6M}(x_2 - x_1) + \beta}$

Thus the spring length, $x_2 - x_1$ , oscillates sinusoidally with the following angular frequency:

$\large{\omega = \sqrt{\frac{5k}{6M}}}$

Below is an alternative strategy, which uses forces and torques instead of the Lagrange approach.

Write Newton's 2nd Law for the horizontal movement of the larger disk:

$\large{F_{x1} = M_1 \ddot{x_1} \\ k(x_2 - x_1 - L_0) - F_{friction} = M_1 \ddot{x_1} }$

Now look at the rotational analog of Newton's 2nd law for the same body:

$\large{\tau = I_1 \ddot{\theta_1} \\ F_{friction} R_1 = \frac{1}{2} M_1 R_1^2 \frac{\ddot{x_1}}{R_1} \\ F_{friction} = \frac{1}{2} M_1 \ddot{x_1}}$

Plugging this into the previous equation gives:

$\large{F_{x1} = M_1 \ddot{x_1} \\ k(x_2 - x_1 - L_0) - \frac{1}{2} M_1 \ddot{x_1} = M_1 \ddot{x_1} \\ k(x_2 - x_1 - L_0) = \frac{3}{2} M_1 \ddot{x_1}}$

We could proceed similarly for $M_2$ and ultimately arrive at the same result as we got for the Lagrange approach.

@Steven Chase gave a beautiful solution using Lagrangians. I will post a more elementary one:

We can ignore the gravitational forces on the wheels since they are balanced by normal forces and don't generate torque. This leaves two forces on each wheel: the spring force $F_s$ toward the center and a friction force $F_i$ outward ( $i = 1,2$ ).

For each wheel we have the translational and rotational equations of motion $a_i = \frac{F_s - F_i}{m_i} \\ \alpha_i = \frac{F_i r_i}{\tfrac12 m_i r_i^2} = \frac{2 F_i}{m_i r_i}$ Rolling without slipping requires $a_i = r_i \alpha_i$ so that $\frac{F_s - F_i}{m_i} = \frac{2F_i}{m_i} \ \ \ \ \therefore\ \ \ \ F_i = \frac 1 3 F_s.$ In other words, the friction force on each wheel is one third of the spring's force, in the opposite direction. The acceleration of each wheel is $a_i = \frac{F_s - F_i}{m_i} = \frac{2 F_s}{3m_i}.$ Now the force $F_s = kx$ , where $x$ is the distance between the CMs of the wheels, and $a_1 + a_2$ is the acceleration of this distance (up to a negative sign). Thus $-\ddot x = a_1 + a_2 = \frac{2 F_s}{12 M} + \frac{2 F_s}{3M} = \frac{5F_s}{6 M} = \frac{5 k}{6M} x.$ Comparing this to the standard equation for a harmonic oscillator, $\ddot x = -\omega^2 x$ we see immediately that $\omega^2 = 5k/6M$ so $\omega = \sqrt{\frac{5k}{6M}}.$ We therefore submit the answer $5 + 6 = \boxed{11}$ .

Force (F) in the spring (x is the length relative to the rest length), +ve is further apart.
$F = -kx$

The torque on the large roller is balanced by the rotation inertia about the point of contact with the ground . Let a be the angle of rotation in the large roller (anti clock) and b in the smaller (clock wise) $\ddot a = \frac{2RF}{I_a}$ and $\ddot b = \frac{RF}{I_b}$

The moment of a roller about it centre axis is $\frac{mr^2}{2}$ but about the point of contact with the ground (using the parallel axis theorem) is $\frac{3mr^2}{2}$

Therefore $I_a = \frac{3} { 2 }\cdot 4M \cdot ( 2R )^2 = 24MR^2$ and $I_b = \frac{3} { 2 }MR^2$

Now the seperation of the rollers $\ddot x = 2R\ddot a + R\ddot b = FR^2 ( \frac{2 \cdot 2} {24 MR^2} + \frac{2} {3MR^2} )= \frac{5F} {6M}$

So $\ddot x = - \frac{5k}{6M} x$

This is solved for with a function of the form $K_1 \sin( \omega t + K_2 )$

where the angular velocity is $\sqrt{ \frac{5k}{6M}}$

Thus 5+6 = $\boxed{11}$ .