Spinning Table

The problem author's solution is the most efficient one. Just for fun, I will avoid explicitly invoking a centrifugal force by using Lagrange analysis. Assume that the center of the table is the origin of a $2D$ coordinate system. The location of the mass on the table is given in polar coordinates as $(r,\theta)$ .

Kinetic energy:

$T = \frac{1}{2} M (r^2 \dot{\theta}^2 + \dot{r}^2)$

Spring potential energy (with springs at equilibrium length for $r = 0$ ):

$V = \frac{1}{2} \frac{3}{4} k r^2 + \frac{1}{2} \frac{1}{4} k r^2 + 2 \frac{1}{2} (6 k) \Big( \frac{r}{2} \Big)^2 = 2 k r ^2$

Lagrangian:

$L = T - V = \frac{1}{2} M (r^2 \dot{\theta}^2 + \dot{r}^2) - 2 k r ^2$

The equation of motion for the radial parameter is:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{r}}} = \frac{\partial{L}}{\partial{r}}$

Evaluating results in:

$M \ddot{r} = M r \dot{\theta}^2 - 4 k r$

Re-arranging a bit results in:

$\ddot{r} = - \Big(\frac{4 k}{M} - \dot{\theta}^2 \Big) r$

This corresponds to simple harmonic motion in $r$ with angular speed $\omega_{osc}$ .

$\omega_{osc} = \sqrt{\frac{4 k}{M} - \dot{\theta}^2}$

The spring system on the left simplifies to a single spring with spring constant $k$ while the system on the right simplifies to a single spring with spring constant $3k$ . Both of these springs then simplifies to a single spring with spring constant $4k$ , because they are in a parallel configuration.

Two forces will work on the object, the spring force and the centrifugal force.

The free body diagram of the object :

Newton's $2^{nd}$ law for the object :

$-4kx+M\omega^2x = M\ddot{x}$

Dividing both sides with $M$ ,

$-(\frac{4k}{M} - \omega^2)x = \ddot{x}$

This is a well-known differential equation, hence we have :

$\omega_{osc}=\sqrt{\frac{4k}{M} - \omega^2}$

and we can get the value of $\Omega$ by making $\omega_{osc} = 0$ , giving :

$\Omega=\sqrt{\frac{4k}{M}}$

Substitute the numerical values to get $\omega_{osc} = 9 \ rad/s$ and $\Omega = 10 \ rad/s$

Hence, $\mid\omega_{osc}-\Omega\mid = 1 \ rad/s$