1. Set up the geometry

A uniform rod of length $L$ is attached at one end to a massless string of length $L$, whose other end is fixed at $O$.

For the rod to remain aligned with the string while rotating about $O$, the whole system must rotate like a rigid radial line about $O$.

So the rod must have angular velocity about $O$ equal to the angular velocity implied by the impulse.

2. Location of the rod's center of mass

The string has length $L$, and the rod is attached at its end. Hence the rod lies along the same line beyond the string.

Therefore the center of mass of the rod is at distance

from the pivot $O$.

3. Impulse given at a point on the rod

Let the rod mass be $M$.

The impulse $P$ is applied at a point whose distance from the rod's midpoint is

To produce rotation about $O$ with rod always aligned with string, the struck point must be such that immediately after impulse the motion is pure rotation about $O$.

That means:

• linear momentum gives CM speed,
• angular momentum about CM gives spin,
• and these must match the angular speed about $O$.

4. Linear velocity of center of mass

Since total impulse is $P$,

so

If the rod rotates rigidly about $O$ with angular speed $\omega$,

Hence

5. Angular momentum about the center of mass

The rod gets spin due to the impulse applied at distance $x$ from its midpoint.

So angular impulse about CM is

Thus,

For a thin rod,

Hence

Substitute $\omega=\frac{2P}{3ML}$:

Simplify:

Therefore

Since $x=\frac{L}{n}$,

6. Final answer

The derived answer matches the stored correct answer.