Clock rotation (AMC 10 #14 2015)

Question

I'm having trouble understanding the solution for: What I did was ${ C }_{ S }=2\pi r=2\pi *10=20\pi \\ { C }_{ B }=2\pi r=2\pi *20=40\pi \\ \therefore ,\quad \frac { 20\pi }{ 40\pi } *360=180\Rightarrow 6:00$

However, Solution 3 on aops.com uses the radius of the Large circle+small circle for the circumference for some reason, which I don't understand. I used the circumference of the big circle as my denominator, because the smaller circle is rotating ON the larger one.

Answer 1

The error is that you assumed that the motion is similar to motion on a flat surface. In that case, the point of contact is always at the bottom. In this case, the point of contact moves with respect to the center of the small disk. That's an extra rotation that you did not take into account. If we use the angle $\theta$ in radians, the path that the small circle travels is $(2\pi -\theta)r$ which must be equal to $\theta R=\theta 2r$. From here $\theta=\frac{2\pi}{3}$, which correspond to $4:00$ on the clock

Answer 2

your answer doesn't take into account relative motion.

one rotation will come free

explanation:

the centre of disc (or smaller disc) REVOLVES around circle of radius (R+r) where R is radius of larger circle and r that of the disc

so for arrow to point again in vertical direction centre has to move distance

$D=2\pi(R+r)$

again note arrow itself is ROTATING inside only smaller circle so distance moved by it $d=r\theta$

so when disc completes it's one reolution the arrow moved by angle

$r\theta=2\pi(R+r)\implies \theta=2\pi\left(\dfrac{R}{r}+1\right)$ put $R=2r$ you get $\theta = 6\pi\, radians$

it means "when Centre of disc revolves once around bigger circle the arrow of disc has already completed three rotations inside disc itself (and it is of course again pointing in same vertical direction as it was initially pointing)

so for arrow to rotate only once(and for the first time to point in vertical direction) centre of disc only need to cover $\dfrac{1}{3}$rd of full REVOLUTION i.e, $\dfrac{2\pi}{3}$ radians that corresponds to $4$'o clock aka option (C)