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Absane
08-17-03, 11:06 PM
Err... I cannot figure this out.. I am disappointed. Plus, I have no idea how to search for this...

Say you have a hill will the shape
y = 2x^2 - x^4/4
You place a point-object very close to x = -2 on the right side. You let go and there is a certain frictional force with any coef. of friction (that makes sense, of course). I want to find out the maximum height the object will make on the second hill.

I am stuck... and it is making me mad. Well, I do not have a background in calc-based physics other than what I do for fun.. this is one of the "fun" things I like to do.

As far as I know... theta = arctan(dy/dx) (instantanious angle)

I am going to sleep.. maybe I will wake up with the answer :bugeye:

Tom2
08-18-03, 12:07 AM
Originally posted by 4DHyperCubix
Say you have a hill will the shape
y = 2x^2 - x^4/4
You place a point-object very close to x = -2 on the right side.

OK

You let go and there is a certain frictional force with any coef. of friction (that makes sense, of course). I want to find out the maximum height the object will make on the second hill.

No problem: Just use the work-energy theorem. Since you have zero kinetic energy in both the initial and final states, the theorem takes the form:

|&Delta;PE|=|W|

where PE=mgy(x). I put in the absolute value signs to avoid confusion over differences in sign convention. You can figure out the initial PE by plugging in x=-2 into that. It is the final PE that you don't know, but you can get it by calculating the work.

Remember, W=&int;F<sup>.</sup>ds, where the integral is over the path taken by the particle. In this case, the force is friction and the path is y=2x<sup>2</sup>-x<sup>4</sup>/4.

Give that a shot.

edit: fixed superscript bracket

Absane
08-18-03, 02:08 PM
So.. if mu = 0.1, the mass is 1 g.. the curve is y = 2x^2 - x^4/4...

I just do

theta = arctan(d[2x^2 - x^4/4]/dx)
F = 0.1*1*9.8*cos(theta) for the intantanious force...

&int;F<sup>.</sup>ds on x = -2 to 0

Which will give 0.98446 J of work done by friction?

I think.. haha.

Absane
08-19-03, 02:54 PM
Slightly paranoid.. can anyone confirm this?

metacristi
08-22-03, 01:49 PM
The problem is not at all easy to solve in practice,involving also more than the basic knowledge of calculus.

Let H=the final height and X=the coordinate on the x axis (X must be positive) for which y(X)=H=2X<sup>2</sup>-(1/4)*X<sup>4</sup> (0)

The system is nonconservative (assuming that air friction is negligeable and that mg and the frictional force are the only forces acting upon the system) therefore the maximum (final) height should be less than y(-2)=4.

The kinetic energy being 0 for the initial and the final points --->

mg(H-4)=&#916Ep=- L=- &#8747<sub>y(x)</sub> Fds (1)

(the work L is smaller than 0)

In every point of the trajectory |F|=&#956*m*g*cos&#952 (2)

cos&#952=&#8730{1/[1+tg<sup>2</sup>&#952]}=&#8730{1/[1+y'<sup>2</sup>(x)]} (3)

and

ds=&#8730[1+y'<sup>2</sup>(x)]dx (4)

Introducing (2) (3) and (4) in (1) --->

4-H=&#956*&#8747<sub>x=-2 to x=0</sub> dx+&#956*&#8747<sub>x=0 to x=X</sub> dx (5)

(5) can be solved ---> it is obtained H as a function of X.From eq (0) we have another formula for H that must be fulfilled ---> X can be obtained numerically (using the trial and error method for example) ---> H can be computed.

[edit to add] The result seems odd ((5) should depend on the trajectory y(x)),probably F and ds should be written in a vectorial form before solving (1)...I haven't tried to use the vectorial method to confirm the above result (anyway it should give the same result if the above way to solve the integral is correct-it should be,mathematically at least I see no mistake) but I've solved the system (5)+(0) for &#956=0.1:X=~1.67 and H=~3.65.Looks acceptable.