A table featuring a smooth horizontal surface is situated in a...

In the rotating frame of the cabin, each mass experiences a centrifugal force directed radially outwards. Since the radius $R$ is large, the centrifugal force on a mass $M$ can be approximated as $M\omega^2 R$.

Let the two masses be $m_1 = 2m$ and $m_2 = m$. Both masses are placed such that the strings lie along the outward radius, meaning the centrifugal force acts outwards (away from the pulley) on both masses, while the tension $T$ acts inwards (towards the pulley).

The centrifugal forces on the masses are:
For the $2m$ mass: $F_1 = 2m\omega^2 R$ (outwards)
For the $m$ mass: $F_2 = m\omega^2 R$ (outwards)

Since $F_1 > F_2$, the heavier mass $2m$ will accelerate outwards, and the lighter mass $m$ will be pulled inwards. Let $a$ be the magnitude of their common acceleration relative to the cabin.

Writing the equations of motion for both masses in the frame of the cabin:
For the $2m$ mass (accelerating outwards):
$2m\omega^2 R - T = 2ma$

For the $m$ mass (accelerating inwards):
$T - m\omega^2 R = ma$

Adding the two equations to eliminate $T$:
$(2m\omega^2 R - T) + (T - m\omega^2 R) = 2ma + ma$
$m\omega^2 R = 3ma$
$a = \dfrac{\omega^2 R}{3}$

Now, substitute the value of $a$ into the equation for the $m$ mass to find the tension $T$:
$T = m\omega^2 R + ma$
$T = m\omega^2 R + m\left(\dfrac{\omega^2 R}{3}\right)$
$T = \dfrac{4}{3}m\omega^2 R$

Thus, the initial acceleration of the masses is $\dfrac{\omega^2 R}{3}$ and the tension in the string is $\dfrac{4}{3}m\omega^2 R$.