Here are some questions to ask before building equations:
What is the shape of the bowl?
What is the mathematical description of the shape of the bowl?
Is the bowl massless?
How does the bowl swing? Does it swing from a string? Is that string massless?
Does the bowl rotate? (in addition to its swinging and having a ball roll on its surface)
From a related question, "Consider a solid ball of radius $r$ and mass $m$ rolling without slipping in a hemispherical bowl of radius $R$ (simple back and forth motion). "
"The only torque acting on the ball at any point in its motion is the friction force $f$. So we can write
$\tau = I\alpha = fr$
again using the rolling condition $a = r\alpha$ and the moment of inertia for a solid sphere,
$\frac{2}{5}ma = f$
The net force acting on the system is the tangential component of gravity and the force of friction, so
$F = ma = mgsin\theta - f$ "
Since your bowl is swinging, $\theta $ changes with time.
(Imagine the bowl is like a swinging pendulum bob)
Now lets discuss the details about the swinging bowl.
Consider a swinging bowl on a massless string of length $L'$ with period of oscillation $T'$ and maximum angular displacement $\theta_{max} '$
We need to form equations describing the change of the angle on inclination of the bowl with respect to the rolling bowl as the bowl swings.
Therefore, we need some initial condition. Let's say the bowl is at its maximum angular displacement to our 'left', and the hemispherical bowl always 'points' towards the 'axis' of it's swinging. Our 'total' inclination angle must not sum up to $\frac{\pi}{2} rad$, otherwise the ball would fall vertically instead of rolling without slipping.
First off, let's deal with $\theta$, the angle of inclination. Angle of inclination=angle of ball in bowl + angle of bowl in pendulum system.
Secondly, we need the equation describing the change of $\theta $ with time. Assume the bowl is massless but rigid and doesn't rotate (due to the other torque, exerted by the ball on the bowl). However, for a short while, let's imagine that the bowl does rotate. We would have some critical case (or range of cases) such that the rotation of the bowl corresponds to the swinging of the bowl in such a way that we can have large maximum angular displacements for the swinging of the bowl (perhaps even $2 \pi$, corresponding to full revolutions!). In other words, the rotation of the bowl could help increase the stability of the system.
If the bowl rotates, we need to even information about the bowl.
So in the oscillation of the bowl, we only consider the mass of the ball.
To answer your questions in short:
- Why does friction stop exerting a force?
It doesn't!
Static friction still works. The wheel is not slipping; it is rolling. At the contact point the wheel and surface stay together because of static friction. This point of the wheel doesn't slide over the surface. So no kinetic friction, but certainly static friction. If not, then how would you start your car? Your tires need grip (static friction) on the surface, and must not slip.
The quote you give does though mention constant velocity. That is the same as saying no acceleration and of course also no angular acceleration. When that is the case, which it will be after some time, all torques must sum up to zero. So, if friction is the only force that causes a torque, then friction must be zero. Friction is only making the wheel start speeding up it's rotation - it makes the wheel start to turn, when your car speeds up. But when the rotation is constant, there is no more friction - just like when the block is sliding with constant speed on the ice, or when the space shuttle is drifting at constant speed, there is no friction that brakes it.
- Am I mistaken that friction exerts a force whenever an object and its surface might be distanced from each other?
Yes, you are mistaken. If I lift a bag of potatoes from the floor, there is no friction.
But I know what you mean: what if you move the object in parallel along the surface. Still, ideally you can have a no-friction icy surface, but apart from that I guess not.
Best Answer
Lets say that when your ball first contacts the ground, it has initial velocity $v_0$ and initial angular velocity $\omega_0 = 0$.
You have a constant torque being applied to the ball, so your differential equation is very easy to integrate to get:
$$\dot{\theta} = \omega = \frac{5g\mu}{2R} t + \omega_0$$
For the displacement, go directly with Newton's law, $\ddot{x}=-\mu g$, which also has a constant force and can be easily integrated once to get
$$\dot{x} = v = v_0 - \mu g t$$
From here you should be able to use your $v = \omega R$ condition to find out how long will it take the ball to start rolling without slipping, and once you have that time, integrate displacement once more to get
$$x = v_0 t - \frac{1}{2}\mu g t^2,$$
which will give you the distance traveled entering the time you calculated before.