Kevin S. Huang

2016: Problem 19

Topic: Energy
Concepts: Conservation of energy, Projectile motion

Solution:

By conservation of energy, when the bead travels to the bottom of the semicircle, it has speed

mg(2R)=\frac{1}{2}mv^2

v=2\sqrt{gR}

From kinematics, it takes time

H=\frac{1}{2}gt^2

t=\sqrt{\frac{2H}{g}}

to fall a distance $H$ . The range $D$ is given by

D=vt=2\sqrt{gR}\sqrt{\frac{2H}{g}} \propto \sqrt{RH}

D^2 \propto RH

Since $D^2$ and $RH$ are proportional to each other, their graph is a line through the origin. Thus, the answer is D.