I have a project about analyzing motion down a curved ramp. I am quite confused about how to approach it. Can someone please point me in the right direction? (Is this the right place?)
Conservation of energy will usually do it.I have a project about analyzing motion down a curved ramp. I am quite confused about how to approach it. Can someone please point me in the right direction?
Yes, that's fine … but it'll take ages for the attachments to be approved.There's a lot of equation editor stuff that I can't reproduce here. I printed them into images. That's fine right?
Yes, FN,2 is the "centripetal force", mv2/R.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, 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)I lose understanding right where it starts talking about the second portion of the normal force.
The tangent of the slope of the curve, tanθ, is dy/dx, ie f'(x).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...
No, becauseQ3. 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)?
Is there another more appropriate name for it?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 …
Is there no friction force or is µ just dramatically reduced? Because there would still be a nonconservative force acting on the ball or else it would just go up and down forever right?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)