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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...I'm not sure if they're completely correct, but they'll do for now.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)?I'm not sure. I assume it would be similar in some ways,

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oxnume said: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?

Conservation of energy will usually do it.

Show us how far you get, and where you're stuck, and then we'll know how to help!

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This is what my partner has came up with so far. But I can't seem to understand it and there are some parts that just appear from nowhere. I would've approached it a bit differently but I'm not sure if I'm right. So can you please look at this and tell me if it makes any sense?

The attachment is .zip because the original .doc file is too big (358kb), no virus, promise!

Also, in the file we used an example with a block on a ramp, would the effect be the same if we used a ball instead of a block?

The attachment is .zip because the original .doc file is too big (358kb), no virus, promise!

Also, in the file we used an example with a block on a ramp, would the effect be the same if we used a ball instead of a block?

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Since this involves kinematics and calculus, would the physics section be better or is this fine?

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Can someone please take a look at this?

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sorry, oxnume, i don't like .doc files

can't you type it out for us?

can't you type it out for us?

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oxnume said:There's a lot of equation editor stuff that I can't reproduce here. I printed them into images. That's fine right?

Yes, that's fine …

What equation editor stuff do you need?

With a few copy-and-paste symbols like θ, and the X

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http://oxnume.comuv.com/temp/2009.05.24%20-%20curve.docx01.jpg

http://oxnume.comuv.com/temp/2009.05.24%20-%20curve.docx02.jpg

http://oxnume.comuv.com/temp/2009.05.24%20-%20curve.docx03.jpg

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Ok the attachments are approved.

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oxnume said:Ok the attachments are approved.

ok, I can se them now (

but what are you asking us about?

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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?

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Hi oxnume!

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

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)

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)

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.

oxnume said:Q1.The second normal force (F_{N,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 F_{N,2}= mv^{2}/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

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

btw, this is a matter of

I lose understanding right where it starts talking about the second portion of the normal force.

Yes, as I said, it

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)

(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.

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tiny-tim said:However, I think most members of PF would strongly disagree with calling it aforce(your book calls it the "second normal force") … it's really the mass times the centripetal acceleration, and comes on theRHSof F = ma, not the LHS …

Is there another more appropriate name for it?

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oxnume said:Is there another more appropriate name for it?

Well, it's not a force, so just the mass times the centripetal acceleration.

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Heh, we weren't actually taught this. This is just an "explore by yourself" project that we chose.

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?

tiny-tim said: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)

Is there

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However, there is air resistance, and also something called "rolling resistance", which essentially is loss of energy through deformation of the ball … see wikipedia.

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