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(D) As the ball $B$ moves from $A$ to $C$,it performs circular motion. The net force towards the center of the circular path is provided by the radial

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initially be in contact with the outer wall and later with the inner wall

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(D) As the ball $B$ moves from $A$ to $C$,it performs circular motion. The net force towards the center of the circular path is provided by the radial component of gravity and the normal force $N$ from the tube walls.Let $R$ be the radius of the tube and $	heta$ be the angle with the vertical. The radial equation of motion is $mg \cos 	heta - N = \frac{mv^2}{R}$.At the start (near $A$),$	heta = 0$,so $mg - N = \frac{mv^2}{R}$. Since $v$ is small,$N$ is positive,meaning the ball presses against the outer wall.As the ball descends,$v$ increases. The condition for contact with the outer wall is $N > 0$,i.e.,$mg \cos 	heta > \frac{mv^2}{R}$.Using conservation of energy,$\frac{1}{2}mv^2 = mgR(1 - \cos 	heta)$,so $v^2 = 2gR(1 - \cos 	heta)$.Substituting this,$mg \cos 	heta - N = \frac{m(2gR(1 - \cos 	heta))}{R} = 2mg(1 - \cos 	heta)$.$N = mg \cos 	heta - 2mg + 2mg \cos 	heta = mg(3 \cos 	heta - 2)$.For $N > 0$,we need $3 \cos 	heta > 2$,or $\cos 	heta > 2/3$. As $	heta$ increases beyond $\arccos(2/3)$,$N$ becomes negative,meaning the ball loses contact with the outer wall and hits the inner wall.

The tube $AC$ forms a quarter circle in a vertical plane. The ball $B$ has an area of cross-section slightly smaller than that of the tube,and can move without friction through it. $B$ is placed at $A$ and displaced slightly. It will

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