The ant goes marching

From the first attempted crossing, we know the disc rotates at $.5 rpm = \frac{180^{\circ}}{minute} = \frac{3^{\circ}}{second}$ . For simplicity let the radius of the disc be $30 cm$ so the ant walks at $1 \frac{cm}{sec}$

The time for the ant to cross the chord must be the same as the time for the far end of the chord to rotate to the other side.

Let the central angle formed by the chord to be $\theta$ . The chord must rotate $180^{\circ}-\theta$ which it will do in $(60-\frac{\theta}{3}) sec$

The chord has length $60\sin{\frac{\theta}{2}} cm$ which will take the ant $60\sin{\frac{\theta}{2}} sec$

Equate these times and solve to get $\theta=72.976^{\circ}$ which equates to approximately $\boxed{36} seconds$