Understanding Angular Velocity: Solving for Radius with Homework Statement

[mathman44]

Homework Statement

I think I've just majorly confused myself here...

In the above diagram, since the ball doesn't "slip" we know that
R*\frac{d\theta}{dt} = V

where theta is the angle of rotation on the disk itself.

I need to relate the angle phi to the angle theta. If I use the relation

R*\frac{d\theta}{dt} = p*\frac{d\phi}{dt}

then I get the "correct" answer for this problem; but shouldn't the relation really be

R*\frac{d\theta}{dt} = (p+R)*\frac{d\phi}{dt}

since the velocity vector is at the base of the moving disk? Major brainfart here!

 

[diazona]

Ah, but remember that as the disk rolls, the relative orientation of its center and its point of contact changes.

To see what's going on, think about what happens as the disk rolls through an angle of 2π (for example); that is, so that Δθ = 2π. Suppose it started out at the bottom of its "track," so that the contact point was at the bottom. Once it has rolled through an angle of 2π, it will be partway up one side of the track, and the contact point is no longer at the bottom. So the contact point will have effectively moved around the circumference of the disk by more than 2πR. If you think about that example carefully (you may want to draw diagrams for the initial and final states), you'll be able to come up with the equation you need.

 

[mathman44]

Actually your example pretty much made it clear, I take it "p" is just the average distance from the center to the initial "base" of the disk?

Thanks.

 

[diazona]

No, actually ρ (technically it's a rho, not a p, but that's not important) is the center-to-center distance. At least, that's what your diagram appears to show. It's a constant, not an average of anything.

 

[mathman44]

Doesn't p, the center-to-center distance, work out to be the average distance to a given point on the disk (that we define as corresponding to theta = 0) as it rotates around the circular well?

 

[diazona]

I don't think so. If you consider a circle of radius ρ concentric with the "track," it passes through the center of the disk and thus encloses less than half of the disk's circumference at any given time. So as a given point on the edge moves around that circumference, it will spend less than half its time within radius ρ of the center. (Then again, I haven't done the math in detail, so perhaps this argument is wrong)

 

[mathman44]

Well if you let p' be a vector from the center to an arbitrary point on the disc, it would have a length of p' = p + R*sin(theta), which would vary as the disk rolled, and the average of p' = p, no?

 

[diazona]

OK, you're right, although I had to do the math in detail to check it.

In any case, ρ is defined to be the center-to-center distance, and I don't think you need to use the fact that it is the average center-to-edge-point distance at all. So it threw me off a little when you brought it up, that's all ;)