Rolling a Circle Down a Function Let there be a circle $C$, with radius $r$. Also, there is an continuous curve on the Cartesian plane with coordinates $(x,y)$, defined by a natural exponential function: $y=e^{-x}$
The circle is held in place such that is touches the curve at the coordinates: $(a,e^{-a})$ for some $a \in \Bbb R :a<0$. That is, $a$ is a non-positive real number. Let $b$ be a positive integer, i.e $b \in \Bbb Z^+$.
I then have $3$ questions:
$(1)$ Let $a=-\log(10)$. What is the radius of the circle in terms of $x$ and $b$, given it takes exactly $b$ rotations until the circle touches coordinates $(0,1)$?
My Attempt:
I know the arc length formula is given by:
$$L=\int_a^b \sqrt{1+\left(\frac{dy}{dx} \right)^2}dx=\int_{- \log(10)}^0 \sqrt{1+e^{-2x}}dx=\int^{\log(10)}_0 \sqrt{1+e^{2x}}dx$$
I think $L$ can be found by using the substitution $u=\sqrt{1+e^{2x}}$. So we can simply equate b times the circumference with $L$ and solve for $r$, i.e. $L=2 \pi br \implies r=\frac L{2\pi b}$.
$(2)$ Let $a=- \log(10)$ and $b=4$. Also, assume the circle faces a constant acceleration due to gravity of $g=9.8ms^{-2}$ and there is no air resistance i.e. no drag. The initial angular velocity of the circle at $t=0$ is $v(0)=0$. What is the velocity at time $t\gt 0$, $v(t)$? Hence find the angular velocity of the circle when it is at coordinates $(0,1)$.
My Attempt:
First I plug in values for the radius in the previous question using $a$ and $b$. But I am slightly unsure as to how to calculate the angular velocity although I think that we would require the tangent at the point $(0,1)$ of the curve $y=e^{-x}$. I assume the question asks the the velocity of the circumference. So it is the rate of change of angle with respect to time. I am unsure as to where to start for formulating an equation.
(3) Find the angle at which to rotate the Cartesian plane such that if the circle is dropped from any point on the curve, it will almost surely come to rest at the coordinates (0,1). Also find the time $t$ such that if the circle passes the y axis for the second time, such that it initially at $t=0$, $x=- \log 10$.
My Attempt:
I know that we need to transform the Cartesian coordinates into Polar coordinates and use a matrix transform such that the derivative at our original coordinates $(0,1)$ is equal to $0$. That is find a $\theta$ such that the curve has a minimum at the coordinates the circle will end up. Part 2 is then a simple case of using results of the previous question and fitting in the values based on our new parameterisation of the plane.
I am not a Math/Physics major. I know this is possibly a long question, but is my though process correct? What are the possible ways I can improve my thinking towards this kind of problem? Are there are hints or simple ideas I am missing? Thank you.
 A: For question number 1: Note that the tangent to the curve at the point of contact is tangent to the circle. Since the curve has non-constant slope, it means that the point of contact is not at the same angle with respect to the center of the circle at the beginning and at the end. So while the circle makes a given number of turns, the point of contact travels more or less than the center of the circle, so you cannot write the path length as $2\pi rb$. You need to account for the extra angle.
For part 2: You need to make the assumption that the mass of the circle is equally distributed along the circumference, and that the circle rolls without slipping along the curve. If it can slip, there is no way to connect the angular velocity with the linear velocity of the center of the circle. If all this is true, then you can write that the velocity of the center $v$ is related to the angular velocity $\omega$ by $\omega=\frac{v}{r}$. Using conservation of energy, the potential energy difference is equal to the kinetic energy of the center of mass plus the rotation energy:
$$mg\Delta h=\frac{1}{2}mv^2+\frac{1}{2}I\omega^2$$
For the mass distribution assumed before, $I=mr^2$, so $$g\Delta h=r^2\omega^2$$
Note that you need to calculate the position of the center of the circle at the two contact points. The difference is slightly different than the difference between the heights of the two contact points, once again due to the fact that the position of the contact point varies with respect to the center, such that the curve and the circle are both tangent to the radius at the point of contact.
For part 3: calculate the tangent at the curve at the given point. You need to rotate the curve by an angle equal to the tangent's angle, so that in the new coordinate system the given point is a minimum (slope at that point will be $0$). 
