The force on the bolt from the rod is the centripetal force, $F_c$, required to maintain the bolt's circular motion. The magnitude of this force is given by Newton's second law:

$$F_c = ma_c = m\omega^2 r$$

where $m$ is the mass of the bolt, $\omega$ is its angular velocity, and $r$ is the radius of its circular path.

The angular velocity $\omega$ can be determined from the stroboscopic observation. The strobe lamp flashes at a frequency $f_{strobe} = 2000 \text{ Hz}$. The time between flashes is $T_{strobe} = 1/f_{strobe}$. Since the bolt appears in $N=8$ distinct, equally spaced positions during one full rotation, the time for the rod to complete $1/N$ of a revolution must be equal to the time between flashes. The period of rotation, $T$, is therefore:

$$T = N T_{strobe} = \frac{N}{f_{strobe}}$$

The angular velocity is related to the period by $\omega = 2\pi/T$. Substituting the expression for $T$:

$$\omega = \frac{2\pi}{N/f_{strobe}} = \frac{2\pi f_{strobe}}{N}$$

Now, substitute this expression for $\omega$ into the centripetal force equation:

$$F_c = m \left( \frac{2\pi f_{strobe}}{N} \right)^2 r$$

Substituting the given values:

$m = 30 \text{ g} = 0.030 \text{ kg}$ $r = 3.5 \text{ cm} = 0.035 \text{ m}$ $f_{strobe} = 2000 \text{ s}^{-1}$ $N = 8$ $$F_c = (0.030 \text{ kg}) \left( \frac{2\pi (2000 \text{ s}^{-1})}{8} \right)^2 (0.035 \text{ m})$$ $$F_c = (0.030 \text{ kg}) (500\pi \text{ s}^{-1})^2 (0.035 \text{ m})$$ $$F_c = 262.5 \pi^2 \text{ N} \approx 2590.8 \text{ N}$$

Rounding to two significant figures, consistent with the given data:

$$F_c \approx 2600 \text{ N}$$