1. Geometry of the Path

The bead is constrained by the bowl surface (Sphere, radius $R$) and the rod (distance $R$ from hinge A). Let the bowl center be origin $O(0,0,0)$. Hinge $A$ is on the brim, say at $(-R, 0, 0)$. The bead starts at $B$ on the brim, such that $AB=R$ (length of rod). The bead moves on the intersection of two spheres: 1. $x^2 + y^2 + z^2 = R^2$ (Bowl) 2. $(x+R)^2 + y^2 + z^2 = R^2$ (Rod constraint)

Subtracting the equations: $(x+R)^2 – x^2 = 0 \implies 2Rx + R^2 = 0 \implies x = -R/2$. The path is a circle in the vertical plane $x = -R/2$. Radius of this path circle: Put $x = -R/2$ in eq 1: $(-R/2)^2 + y^2 + z^2 = R^2 \implies y^2 + z^2 = \frac{3}{4}R^2$. Radius $r = \frac{\sqrt{3}}{2}R$.

2. Velocity and Forces at Lowest Position

Lowest Position: This occurs at the bottom of the path circle, i.e., $z_{min} = -r = -\frac{\sqrt{3}}{2}R$. Vertical drop from brim ($z=0$): $h = \frac{\sqrt{3}}{2}R$. Velocity from Energy Conservation: $$ \frac{1}{2}mv^2 = mgh \implies v^2 = 2g\left(\frac{\sqrt{3}}{2}R\right) = \sqrt{3}gR $$

Forces: At the lowest point, the bead is undergoing circular motion in the vertical plane $x=-R/2$. Radius of curvature is $r = \frac{\sqrt{3}}{2}R$. Centripetal Force required (Upwards in diagram): $$ F_c = \frac{mv^2}{r} = \frac{m(\sqrt{3}gR)}{\frac{\sqrt{3}}{2}R} = 2mg $$

The vertical forces acting on the bead are: 1. Gravity ($mg$ down). 2. Vertical component of Normal force from bowl ($N_z$). 3. Vertical component of Tension from rod ($T_z$). By symmetry and geometry at $x=-R/2$: The rod vector $PA$ and Normal vector $PO$ are symmetric with respect to the vertical $z$-axis in the $x$-$z$ projection. Angle with vertical $\alpha$: $\cos\alpha = \frac{z}{R} = \frac{\sqrt{3}/2 R}{R} = \frac{\sqrt{3}}{2}$. So, $T_z = T \cos\alpha = T\frac{\sqrt{3}}{2}$ and $N_z = N\frac{\sqrt{3}}{2}$. Also, horizontal equilibrium in the x-direction implies $T=N$.

Equation of Motion (Vertical):

$$ (T_z + N_z) – mg = F_c $$ $$ (T + N)\frac{\sqrt{3}}{2} – mg = 2mg $$ Since $T=N$: $$ 2T\frac{\sqrt{3}}{2} = 3mg $$ $$ T\sqrt{3} = 3mg \implies T = \sqrt{3}mg $$