Question

Consider a fixed hemispherical bowl of radius $R$ with its opening horizontal. One end of a light rod of length $l = R$ is hinged at a point A on the brim of the bowl. A small bead of mass $m$ is attached at the other end the rod. Initially the bead is held at a point B also on the brim and then released. Find the tensile force in the rod when the bead passes the lowest position. Acceleration of free fall is $g$.

Answer 1

The problem as stated leads to a contradiction. A likely intended solution, though physically questionable, is $\sqrt{3}mg$.

Answer 2

The bead is constrained to move on the surface of the hemisphere and maintain a fixed distance $R$ from point A. This implies the bead moves on a circle defined by the intersection of the sphere $x^2+y^2+z^2=R^2$ (hemisphere) and $(x-R)^2+y^2+z^2=R^2$ (rod constraint, with A at $(R,0,0)$). This intersection is the circle $x=R/2$, $y^2+z^2 = 3R^2/4$. The center of this circular path is $C(R/2, 0, 0)$ and its radius is $r_c = (\sqrt{3}/2)R$.

The lowest position for the bead on this path is $P(R/2, 0, -(\sqrt{3}/2)R)$.

By conservation of energy, the speed $v$ at this lowest point is found:

• Initial energy (at B, on brim, at rest) = 0 (taking brim as reference potential energy).
• Final energy (at P) = $(1/2)mv^2 + mg(-\sqrt{3}/2 R)$. $0 = (1/2)mv^2 - mg(\sqrt{3}/2)R \implies v^2 = \sqrt{3}gR$.

At the lowest point P, the centripetal acceleration is directed vertically upwards (towards C). $a_c = v^2/r_c = (\sqrt{3}gR) / ((\sqrt{3}/2)R) = 2g$.

Forces acting on the bead at P:

1. Gravity: $mg$ downwards.

2. Normal force (N): From the bowl, acting along the radius from O to P. Vector $\vec{OP} = (R/2, 0, -\sqrt{3}/2 R)$. So $\vec{N} = N(1/2 \hat{i} - \sqrt{3}/2 \hat{k})$.

3. Tensile force (T): From the rod, acting along the rod from P to A. Vector $\vec{PA} = (R/2, 0, \sqrt{3}/2 R)$. So $\vec{T} = T(1/2 \hat{i} + \sqrt{3}/2 \hat{k})$.

Applying Newton's second law:

$\sum F_x = ma_x$: The motion is confined to the plane $x=R/2$, so $a_x=0$. $N(1/2) + T(1/2) = 0 \implies N+T=0$.

$\sum F_z = ma_z$: The acceleration is $a_z = a_c = 2g$ (upwards). $T(\sqrt{3}/2) - N(\sqrt{3}/2) - mg = m(2g)$ $(T-N)\sqrt{3}/2 = 3mg \implies T-N = 2\sqrt{3}mg$.

From $N+T=0$, we have $N=-T$. Substitute into the second equation: $T - (-T) = 2\sqrt{3}mg$ $2T = 2\sqrt{3}mg$ $T = \sqrt{3}mg$.

This result implies $N = -\sqrt{3}mg$. A negative normal force means the bowl would have to pull the bead inwards, which is physically impossible for a smooth surface. This indicates a flaw in the problem statement or an implicit assumption that the normal force can be negative or that the bead leaves the surface. However, if a solution is expected, $T = \sqrt{3}mg$ is the mathematical result derived from the given constraints.