When the upper ball of mass $m$ is given a horizontal velocity $v_0$, it begins circular motion in a vertical plane with centre at lower ball ( as for limiting case lower ball is an inertial frame). The tension $T$ in the rod acts downwards on the upper ball to provide the necessary centripetal force, in addition to the component of gravity.

The equation of motion for the upper ball in the radial (vertical) direction is: $$ T + mg = \frac{m v_0^2}{l} $$ Solving for Tension $T$: $$ T = \frac{m v_0^2}{l} – mg $$

For the lower ball (also mass $m$) to remain on the floor, the Normal reaction $N$ from the floor must be non-negative ($N \ge 0$). The forces acting on the lower ball are:

• Gravity $mg$ acting downwards.
• Tension $T$ from the rod acting upwards (since the rod pulls the bottom mass up as the top mass tries to fly out).
• Normal force $N$ acting upwards.

The force balance equation is: $$ T + N = mg \implies N = mg – T $$

The lower ball immediately loses contact with the floor if $N \le 0$ at the instant of release. $$ mg – T \le 0 \implies T \ge mg $$ Substitute the expression for $T$ from Step 1: $$ \frac{m v_0^2}{l} – mg \ge mg $$ $$ \frac{m v_0^2}{l} \ge 2mg $$