B 16. Consider a fixed hemispherical bowl of radius R with its opening A EE <a horizontal. One end of a light rod of length I = 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.

1. Geometry first

• Bowl radius = $R$.
• Stick length $l = R$ is hinged at point A on the rim.
• Bead starts at point B on the rim. Since $AB = l = R$, the chord length gives the central angle $\theta\text{ such that }2R\sin\!\left(\tfrac{\theta}{2}\right)=R\;\Rightarrow\;\theta=60^\circ.$
• Both A and B lie in the horizontal rim plane that passes through the sphere’s centre, so they are at the same height.
• The lowest position (D) is reached when the stick hangs vertically below A, a distance $R$ lower.

Hence the bead descends a vertical height $h = R.$

2. Use conservation of mechanical energy

Take rim level as zero potential energy.

Initial energy (at B): only potential = $mgh = mgR$.

At the lowest point (D): potential is zero, kinetic is $\tfrac{1}{2}mv^2$.

Set them equal:

$\frac{1}{2}mv^2 = mgR \;\;\Rightarrow\;\; v = \sqrt{2gR}.$

3. Dynamics at the lowest point

A bead moving in a circle of radius $R$ with speed $v$ needs an inward (towards the hinge) centripetal force

$F_c = m\frac{v^2}{R}.$

At D two real forces act along the rod’s line:

1. Weight $mg$ (downward, away from hinge).
2. Tension $T$ (upward, toward hinge).

Taking inward direction as positive, $T - mg = m\frac{v^2}{R}.$

Plug $v^2 = 2gR$:

$T - mg = m\frac{2gR}{R} = 2mg \;\;\Rightarrow\;\; T = 3mg.$

So the stick pulls on the bead with three times its weight when it passes the lowest point.

What is happening?

Imagine a half-ball (a bowl) glued to the table. On the rim we pin one end of a very light stick that is exactly as long as the bowl’s radius. A tiny bead is tied to the other end.

1. We start with the bead resting on the rim, so the stick is slanted.
2. We let go. The bead swings down like a ride in a playground until the stick points straight downward (the lowest spot).
3. We want to know how hard the stick is pulling on the bead at that lowest point.

Two big ideas help us

• Energy never disappears (if we ignore friction). The bead’s height energy turns into speed.
• Going in a circle needs an inward pull (centripetal force). At the lowest point the stick must pull inward hard enough to keep the bead moving in a circle and still fight gravity pulling it down.

Put the two ideas together and you can find the speed, then the pull (tension). It turns out to be three times the bead’s weight.

Step-by-step solution

1. Vertical drop
   Bead descends from rim level to vertical line below hinge. Hence $h = R.$

2. Speed at lowest point
   $\frac12 m v^2 = mgR \;\;\Rightarrow\;\; v = \sqrt{2gR}.$

3. Inward (radial) force balance at lowest point
   Inward direction (toward hinge) positive: $T - mg = m\frac{v^2}{R}.$

4. Substitute $v^2$
   $T - mg = m\frac{2gR}{R} = 2mg \;\;\Rightarrow\;\; T = 3mg.$

5. Final answer
   $\boxed{T = 3mg}$