The only external force with torque about the vertical Z-axis is gravity, which acts parallel to the axis (zero torque), and the reaction at the pivot (zero torque). Thus, angular momentum about the vertical axis, $L_z$, is conserved.

Let $v_\phi$ be the horizontal azimuthal velocity component.

$$L_{z, initial} = L_{z, final}$$ $$m(u) (l \sin \theta_0) = m(v_\phi) (l \sin \theta)$$ $$v_\phi = \frac{u \sin \theta_0}{\sin \theta} \quad \text{— (1)}$$

The problem states the rod exerts no force on the particle. The only force acting is gravity. For the particle to move on the spherical surface of radius $l$, the component of gravity along the rod (radial direction) must provide the necessary centripetal force.

$$F_{radial} = m g \cos \theta = \frac{m v^2}{l}$$ $$v^2 = gl \cos \theta \quad \text{— (2)}$$

Here $v^2 = v_\phi^2 + v_\theta^2$, where $v_\theta$ is the tangential velocity in the vertical plane.

Total mechanical energy is conserved.

$$\frac{1}{2} m u^2 + mgl \cos \theta_0 = \frac{1}{2} m v^2 + mgl \cos \theta$$ $$u^2 + 2gl \cos \theta_0 = v^2 + 2gl \cos \theta$$

Substitute $v^2 = gl \cos \theta$ from (2):

$$u^2 + 2gl \cos \theta_0 = gl \cos \theta + 2gl \cos \theta$$ $$u^2 + 2gl \cos \theta_0 = 3gl \cos \theta$$ $$\cos \theta = \frac{u^2}{3gl} + \frac{2 \cos \theta_0}{3}$$

Given: $u = \sqrt{2}$ m/s, $g = 10$ m/s$^2$, $\theta_0 = 37^\circ$ ($\cos 37^\circ = 0.8$).

$$\cos \theta = \frac{(\sqrt{2})^2}{3(10)(1)} + \frac{2(0.8)}{3}$$ $$\cos \theta = \frac{2}{30} + \frac{1.6}{3} = \frac{0.2}{3} + \frac{1.6}{3} = \frac{1.8}{3} = 0.6$$ $$\theta = \cos^{-1}(0.6) = 53^\circ$$

The question asks for the “horizontal component of velocity perpendicular to the rod”. This refers to the azimuthal velocity $v_\phi$, which lies in the horizontal plane and is perpendicular to the rod’s vertical plane.

From equation (1):

$$v_\phi = \frac{u \sin \theta_0}{\sin \theta}$$

Using $\sin 37^\circ = 0.6$ and $\sin 53^\circ = 0.8$:

$$v_\phi = \frac{\sqrt{2} \times 0.6}{0.8} = \frac{\sqrt{2} \times 3}{4} = \frac{3}{2\sqrt{2}} \text{ m/s}$$