1. Conservation of Angular Momentum

As the insects crawl towards the center, no external torque acts on the turntable-insect system about the vertical axis. Therefore, the total angular momentum is conserved.

Let $I$ be the moment of inertia of the turntable and $m$ be the mass of each insect. Let $r$ be their distance from the center.

$$L = (I + 2mr^2)\omega = \text{constant} = K$$

The angular velocity as a function of $r$ is:

$$\omega(r) = \frac{K}{I + 2mr^2}$$

2. Force on the Insects

The insects crawl “slowly,” implying we can neglect Coriolis and tangential acceleration forces compared to the centripetal force required to maintain their circular path at radius $r$. The maximum horizontal force required is the centripetal force provided by friction.

$$F(r) = m \omega^2 r$$

Substitute $\omega(r)$:

$$F(r) = m \left( \frac{K}{I + 2mr^2} \right)^2 r = m K^2 \frac{r}{(I + 2mr^2)^2}$$

3. Maximizing the Force

To find the distance $r$ where force is maximum, we calculate $dF/dr$ and set it to zero.

$$F(r) = C \cdot \frac{r}{(I + 2mr^2)^2}$$

Using the quotient rule:

$$\frac{dF}{dr} = C \left[ \frac{1 \cdot (I + 2mr^2)^2 – r \cdot 2(I + 2mr^2)(4mr)}{(I + 2mr^2)^4} \right] = 0$$

Simplify the numerator:

$$(I + 2mr^2) – r(8mr) = 0$$ $$I + 2mr^2 – 8mr^2 = 0$$ $$I – 6mr^2 = 0$$ $$r^2 = \frac{I}{6m}$$ $$r = \sqrt{\frac{I}{6m}}$$