Question 13

The rod revolves about the pivot point O. The distance from O to the center of mass (COM) of the rod is $R = L + L/2 = 3L/2$.

For the rod and string to remain aligned, the angular velocity $\omega$ of the rod about O must be the same as the angular velocity generated by the impulse about the COM.

1. Linear impulse: $P = M \Delta v_{cm}$. Since it rotates about O, $v_{cm} = \omega R = \omega (3L/2)$.

$P = M \omega \left( \frac{3L}{2} \right)$

2. Angular impulse about the COM: $P \cdot x = I_{cm} \omega$, where $I_{cm} = \frac{ML^2}{12}$.

Substituting the value of $P$:

$\left[ M \omega \left( \frac{3L}{2} \right) \right] \cdot x = \frac{ML^2}{12} \omega$

$\frac{3L}{2} x = \frac{L^2}{12}$

$x = \frac{L^2}{12} \times \frac{2}{3L} = \frac{2L^2}{36L} = \frac{L}{18}$

Given $x = L/n$, we have $n = 18$.

The final answer is 18.