A rod of mass $m$ and length $L$, pivoted at one of its ends, is hanging vertically. A bullet of the same mass moving at speed $v$ strikes the rod horizontally at a distance $x$ from its pivoted end and gets embedded in it. The combined system now rotates with angular speed $\omega$ about the pivot. The maximum angular speed $\omega_M$ is achieved for $x = x_M$. Then

Step 1: Apply conservation of angular momentum about the pivot.

The net external torque about the pivot during the collision is zero, so the angular momentum of the system is conserved.

$$L_i = mvx$$

Step 2: Find the moment of inertia of the combined system.

After the collision, the bullet is embedded in the rod at a distance $x$ from the pivot.

$$I = \frac{1}{3}mL^2 + mx^2$$ $$I = m\left(\frac{L^2}{3} + x^2\right)$$

Step 3: Determine the angular speed $\omega$ after the collision.

Equate the initial and final angular momentum.

$$mvx = m\left(\frac{L^2}{3} + x^2\right)\omega$$ $$\omega = \frac{vx}{\frac{L^2}{3} + x^2}$$ $$\omega = \frac{3vx}{L^2 + 3x^2}$$

Step 4: Find the position $x_M$ that maximizes the angular speed.

Differentiate $\omega$ with respect to $x$ and set the derivative to zero.

$$\frac{d\omega}{dx} = 0$$ $$3v \cdot \frac{(L^2 + 3x^2) - x(6x)}{(L^2 + 3x^2)^2} = 0$$ $$L^2 - 3x^2 = 0$$ $$x_M = \frac{L}{\sqrt{3}}$$

Step 5: Calculate the maximum angular speed $\omega_M$.

Substitute $x_M$ back into the expression for $\omega$.

$$\omega_M = \frac{3v\left(\frac{L}{\sqrt{3}}\right)}{L^2 + 3\left(\frac{L}{\sqrt{3}}\right)^2}$$ $$\omega_M = \frac{\sqrt{3}vL}{2L^2}$$ $$\omega_M = \frac{\sqrt{3}v}{2L}$$

Answer: The angular speed is $\omega = \dfrac{3vx}{L^2 + 3x^2}$, which is maximized at $x_M = \dfrac{L}{\sqrt{3}}$ with a maximum value of $\omega_M = \dfrac{\sqrt{3}v}{2L}$.