This is similar to problems asking for the tension $T$ in the string attached to a falling mass $m$, the other end of which is attached to a flywheel or to a block which slides on a horizontal surface. Beginners are tempted to say either that $T=mg$ because that is the weight of the load, or that $T=0$ because the load is assumed to be in free fall. The truth lies between both extremes. The point at which the string is attached to the flywheel or the sliding block is also accelerating, at rate $a \lt g$. This changes the force in the string to $T=m(g-a)$.

Likewise here, if the disk were fixed in place, or very much more massive than the small particles - in both cases the disk does not (significantly) rotate. Then the points at which the strings are attached would not accelerate. The tension in them would then be the centripetal force required to keep the small particles moving instantaneously in a circle of radius $2R$, which is $T=ma_c$ where $a_c=\frac{v_0^2}{2R}$ is the centripetal acceleration, as you calculated.

However, the disk is relatively light so it can quite easily rotate, which means that the points of attachment accelerate vertically in the direction of the strings, reducing the tension in the strings below $T_c$.

What you need to do is draw separate FBDs for the disk and one mass, using the same unknown tension $T$ in each string. For the disk the 2 tensions $T$ form a couple which causes angular acceleration $\alpha$ about its centre. This results in a linear acceleration $a=R\alpha$ of the point at which the string is attached. The tension in the string is then $T=m(a_c-a)$.

You obtain 2 simultaneous equations relating $T$ and $\alpha$, enabling you to find both.

Your mistake is the same as the students who assume the tension in the string attached to a falling mass is $T=mg$ and then go on to calculate the acceleration of the flywheel or sliding block based on this value.

Note that although the particles may have obtained their velocity from an impulse force, this is not necessary, and it does not result in the tension in the string being an impulse force. The tension continues for a finite time.

If the impulse given to the particles had a component along the string, there would be an impulse tension as well as a continuing tension. But because the impulse is perpendicular to the string there is no impulse tension. An alternative, equivalent way of setting the system into motion is to accelerate the particles while holding the disk in place, then release the disk when the strings reach the vertical position. An impulse is not necessary to satisfy the given initial conditions.