Hi oxnume!
oxnume said:
Q1. The second normal force (FN,2) is what throws me off (and the rest of the sin and cos stuff with derivatives). The equation of that normal force is given as FN,2 = mv2/R
That looks awfully like the centripetal force equation for a circle with radius R. But in this case where is the circle? I originally thought that other force would consist of the inertia force from the previous "piece" of the curve, but I have no idea how to express that in terms of math. Is this correct at all?
Yes, F
N,2 is the "centripetal force", mv
2/R.
Here, R is the radius of curvature of the curve at that point … the radius of the circle that most closely fits the curve.
However, I think most members of PF would strongly disagree with calling it a
force (your book calls it the "
second normal force") … it's really the mass times the
centripetal acceleration, and comes on the
RHS of F = ma, not the LHS …
btw, this is a matter of
geometry, not physics … if an object goes at speed v along a curve with radius of curvature R, then its acceleration is automatically v
2/R
I lose understanding right where it starts talking about the second portion of the normal force.
Yes, as I said, it
isn't part of the normal force (except possibly for a frame of reference
moving with the object, for which fictitious forces such as this have to be invented so that Newton's first law still works)
Q2. The "trig identities" on page 3, I have never seen those identities in my life and my partner refuses to explain where she got them from. Are they correct at all? I just get completely lost from that point on...
The tangent of the slope of the curve, tanθ, is dy/dx, ie f'(x).
The line-element is ds = dx√(1 + (dy/dx)
2) = dx√(1 + (f'(x))
2) (so distance along the curve is ∫ds = ∫dx√(1 + (f'(x))
2))
(there's some "squareds" missing in your formulas
)
(and so sinθ = dx/ds and cosθ = dy/ds)
Q3. If we were to do an experiment of this. Would having a ball roll down a curve be the same as having a piece of block slide down the curve (like the diagram)?
No, because
i] there's no friction force impeding a rolling object (the point of contact is stationary, so there is no https://www.physicsforums.com/library.php?do=view_item&itemid=75" by the reaction force)
ii] you'd have to add angular momentum and energy to the equations.