A thin, smooth rod of length 1.0 m lies in a horizontal plane…

Step 1. Along the rod. The rod is smooth, so it can only push the bead at right angles to itself. In the rod's frame, the bead moves only along the rod, and the only force along the rod is the centrifugal force $m\omega^2 r$, outwards. Forces at right angles to the rod do no work in this frame. So

Step 2. Over the ground. The bead also goes round with the rod, at $\omega L = 2.0\ \text{m/s}$, at right angles to $v_r$. So $v = \sqrt{(1.936\ \text{m/s})^2 + (2.0\ \text{m/s})^2} = 2.78\ \text{m/s}$.

Step 3. The sideways force. The bead's angular momentum about the axis is $L_z = m r^2\omega$ (only the sideways part of its velocity, $\omega r$, counts). $\omega$ is fixed, but $r$ grows, so $\dfrac{dL_z}{dt} = 2mr\omega\dfrac{dr}{dt}$. This needs a torque, and the only force that gives one is the rod's horizontal sideways push $N$, at distance $r$. (The rod also holds the bead up against its weight, but that vertical force has no torque about the vertical axis.) $Nr = 2mr\omega v_r$, so

Step 4. Work. In the ground frame, the rod's push is the only force that does work on the bead. So the work equals the gain in kinetic energy. At the start the bead moves at $\omega r_0 = 0.5\ \text{m/s}$:

This is $m\omega^2(L^2 - r_0^2)$: half of it, $0.1875\ \text{J}$, went into the motion along the rod and half into the motion round the axis.