What's the problem? Independent enrollment? It is crazy to ask such a question in high school. Major events will not produce such professional mechanical examination questions. No wonder undergraduate education is a mess now.

First of all, it is necessary to judge that the ring is sliding, that is to say, the translation speed of the center of mass C includes not only rolling, but also sliding. Why? Because the speed of the ring (which is problematic, the estimation refers to the speed of the center of mass C) Vc=V0 is not equal to wR. According to the title, wR is equal to 3v0, and the direction is left, that is, the actual "sliding speed" is 4w0*R, and the direction is right.

Well, after the above analysis, we need to add a little mechanical knowledge: the theorem of velocity composition. The velocity of any point on a rigid body is equal to the composition of the centroid velocity vector and the relative velocity vector. Since displacement = velocity *t (actually integral, but not used here), it can also be synthesized in this way.

Everything is done, let's formally solve the problem.

The first question, the trajectory of A, can be seen from the above analysis that the displacement of A consists of two parts: the translation speed of the center of mass C and the rotation effect. Set the initial position of point C as the origin, the horizontal direction to the right as the positive direction of X axis, and the vertical direction as the Y axis, and establish a Cartesian coordinate system. For A and C, the horizontal displacement caused by translation is S 1=v0*t, while the displacement caused by rotation is

δx =-R * cos(w0 * t+pi/2)

δy =-R * sin(w0 * t+pi/2)

The final trajectory equation can be written as

x=v0*t-R*cos(w0*t+pi/2)

y=-R*sin(w0*t+pi/2)，

You can also choose to cancel the parameter t and write an equation, which is the helix.

The second question, we already know that it is still sliding, the sliding speed is 4w0*r, and the direction is to the right. Assuming that rolling friction does not do work, then only sliding friction does work, then the question becomes, how much does the point in contact with the ground slide relative to the ground? Analyze, there are vA=vC+vCA=v0-wR=-2v0. This is the speed that actually causes sliding friction to do work. In dt time, this contact point does work W=fS=u*mg*vA*dt. Because the speed is the same, there is no need to integrate, just change dt to T, but there is a big BUG in this problem. What does the whole exercise mean? Is it a circle? It seems so. If this is the case, then from the above analysis, w0*R=3v0, and the sliding speed of point A relative to the ground =-2v0, that is to say, in the same time of one revolution, the friction stroke is 2/3 of a circle, that is, 4/3 * pi * r. Multiply this value by f=umg to get the result.