A metal rod with ends $A$ and $B$ is welded at its middle, $C$, to a cylindrical drum of diameter 12 . The rod touches the ground at $A$ making a $30^{\circ}$ angle. The drum starts to roll along $A D$ in the direction of $D$. How far along $A D$ must the drum roll for $B$ to touch the ground?

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-140.jpg?height=268&width=661&top_left_y=752&top_left_x=1147)
(A) $\pi$
(B) $2 \pi$
(C) $3 \pi$
(D) $4 \pi$
(E) $5 \pi$
A metal rod with ends $A$ and $B$ is welded at its middle, $C$, to a cylindrical drum of diameter 12 . The rod touches the ground at $A$ making a $30^{\circ}$ angle. The drum starts to roll along $A D$ in the direction of $D$. How far along $A D$ must the drum roll for $B$ to touch the ground?
(A) $\pi$
(B) $2 \pi$
(C) $3 \pi$
(D) $4 \pi$
(E) $5 \pi$

## Solution

The drum rolls so that point $C$ moves to $C^{\prime}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-260.jpg?height=344&width=1133&top_left_y=1129&top_left_x=493)

Since a line drawn from the centre of the circle makes angles of $90^{\circ}$ with tangents drawn to a circle, $\angle C O M=360^{\circ}-90^{\circ}-90^{\circ}-30^{\circ}=150^{\circ}$. By symmetry, $\angle C^{\prime} O M=150^{\circ}$ and thus $\angle C^{\prime} O C=360^{\circ}-150^{\circ}-150^{\circ}=60^{\circ}$. Since $\angle C^{\prime} O C=60^{\circ}$, this implies that point $C$ will have to travel $\frac{1}{6}$ th the circumference of the circle or $\frac{1}{6}\left(1^{2} \pi\right)=2 \pi$.

ANSWER: (B)