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.
In the following diagram, is work done by static friction 0 ?, since the point of application is also moving with speed v w.r.t. ground here and is only stationary w.r.t. the block on which sphere is rolling w.r.t. ground here.
Static friction itself is 0. The formula $f_s=\mu N$ defines the maximum possible magnitude of the static friction force, not the true static friction force. In this case, there is no other acceleration, so there is no need for static friction. Static friction only comes into play when the two bodies are attempting to be in relative motion with each other. This is not the case here, at the point of contact the velocities of the corresponding points on the wheel and platform are equal and there is no force trying to stop this.
When you're standing on the ground, you're not mysteriously being pushed by friction. It's the same thing here, the wheel is "standing" with respect to the point of contact, though the points of contact are changing over time.
Best Answer
There is more than one approach one can use to calculate the work done by a contact force on a rigid object. All approaches are valid so long as you're careful about how you actually use your calculated work.
One way is to use the infinitesimal displacement of the point of application $d\vec l_\text{app}$:
$$W = \int \vec F \cdot d\vec l_\text{app} \tag{1}$$
In the absence of other forces (and forms of potential energy), this will correctly give you the quantity $1/2 \int v^2 dm$, i.e., the change in total kinetic energy.
In your object-down-the-ramp example, the static frictional force would do zero work.
Another way to approach the same problem is to use the center of mass displacement $d\vec l_\text{CM}$:
$$W = \int \vec F \cdot d\vec l_\text{CM} \tag{2}$$
In the absence of other forces (and forms of potential energy), this will give you the change in center-of-mass kinetic energy $\Delta K_\text{CM} \equiv \Delta\left(\frac{1}{2}mv_\text{CM}^2\right)$.
Yet another way is an extension of method #2, in which one considers the rotational aspects of the motion:
$$W = W_\text{CM} + W_\text{rot} = \int \vec F \cdot d\vec l_\text{CM} + \int \vec\tau \cdot d\vec\theta \tag{3}$$
This latter approach lends itself well to separating out work that causes changes in the center-of-mass kinetic energy and changes in the rotational energy of a rigid body:
$$\Delta K_\text{tot} = \Delta K_\text{CM} + \Delta K_\text{rot}$$
More info on the different definitions of work: Article by Sherwood