Displacement of a point on a circle There is a mark on a wheel of radius $30$ cm. The mark is in contact with a horizontal plane. The wheel rotates to a distance of $10\pi \approx 31.4$ cm. 
$1.$ What is the angle that the old position of the mark makes with the new position? 
$2.$ What is the distance between the new position of the mark with that of the ground?
[point B can be anywhere on the circle]

 A: The mark takes a trip on the dotted curve which is displayed on the animation below. The parametric equations of this curve, named as cycloid, are
$$\begin{align}
x_A&=R\phi-R\sin\phi \\
y_A&=R-R\cos \phi
\end{align} \qquad 0 \le \phi \le \pi/3$$
Where $R$ is the radius of the circle. 
Take a look at this animation. It is exactly made by your data included in the question. It may help to imagine better.
Can you derive the parametric equation?
Can you answer your questions based on the parametric equations?

A: Hint for your first question $r\theta = arc length$
Hint for your second question: you can make a triangle with the lines you've drawn, this triangle will be isoceles (remember the sum of angles in a triangle is $\pi$). I'm assuming the ground makes an angle of $\dfrac{\pi}{2}$ with the radius at A (correct me if I'm wrong). Try and make use of the sine rule to find a certain side of the triangle (I'll leave you to figure out which side). After doing all of this, you should be able to calculate the height of B from the ground by make use of $sin\alpha = \dfrac{opp}{hyp}$.
EDIT: I'm including the sine rule in case you don't know it. For a triangle ABC, with sides a,b,c opposite to their respective angles (i.e a is to the opposite side as A, etc) $\dfrac{sinA}{a}= \dfrac{sin B}{b} = \dfrac{sinC}{c}$
EDIT: I'm including the answer:
Let $\angle AOB = \theta$
$r\theta = 30\theta = 31.4$
$\therefore \theta = \dfrac{31.4}{30}$
Also Note that: 
$\angle OAB$ = $\angle OBA = \dfrac{\pi - \theta}{2}$
Let X be the point on the ground directly below B.  
$\therefore \angle BXA = \dfrac{\pi}{2}$
Note that $\angle OAX = \dfrac{\pi}{2}$ because the ground is a tangent to the circle at the point A.
$\therefore \angle BAX = \dfrac {\pi}{2}- \angle OAB$ 
Now, using the sine rule:
$\dfrac{sin\theta}{AB}$ = $\dfrac {sin\angle OAB}{r}$
Solve for AB by plugging in $\theta$ and $\angle OAB$ and $r$.
Now
$BX = AB \times sin\angle BAX$. BX is the height of B from the ground. Done. Hope this helped. :)

A: Angle subtended at center of by circle rolling without slipping
$$ x/r = \theta $$
$$ 31.4 /30 \approx 1.0467 \text{radians or } \approx 60 ^0$$
Displacement is $\frac16$ of circle circumference contacting on the ground.
