Least Time ... This One!

quantyst · July 16th, 2008, 9:34 am

Consider a frictionless incline S that is a straight line with slope m and that passes through the point (0,1). An object is placed on the incline at point (0,1) and let slide. Determine the value of m that minimizes the time before the object hits the semicircle y=-(1-x^2)^(1/2)=-sqrt(1-x^2).

MCarreira · July 16th, 2008, 12:31 pm

m=+1 or m=-1

quantyst · July 16th, 2008, 1:07 pm

QuoteOriginally posted by: MCarreiram=+1 or m=-1Hi MCarreira,The problem is more interesting than your answer suggests. Give it a try one more time.

vixen · July 16th, 2008, 1:31 pm

Is it not independent of the slope, as long as it intersects the semi-circle?

MCarreira · July 16th, 2008, 1:35 pm

I found that it is almost independent of the slope ... there should be a curve where this happens, I'm not sure it is the semicircle.I have a chart at the end of my file with the time, it does not seem to be constant.

quantyst · July 16th, 2008, 2:01 pm

QuoteOriginally posted by: vixenIs it not independent of the slope, as long as it intersects the semi-circle?Yes, it is independent of the slope. That is, all inclines (as long as they intersect the semi-circle) give the same time of descent.Question: Is the semi-circle y=-sqrt(1-x^2) the only curve that has this property for the given point (0,1)?

MCarreira · July 16th, 2008, 2:28 pm

So what is the formula then ? I did not find it to be independent. Edit: Sorry, I'm stupid. I had an error when simplifying the distance ... t=2/Sqrt[g]

phuebu · July 16th, 2008, 6:21 pm

QuoteOriginally posted by: quantystQuoteOriginally posted by: vixenIs it not independent of the slope, as long as it intersects the semi-circle?Yes, it is independent of the slope. That is, all inclines (as long as they intersect the semi-circle) give the same time of descent.Question: Is the semi-circle y=-sqrt(1-x^2) the only curve that has this property for the given point (0,1)?Well if other functions do exists they must satisfy the conditionwhere x_c is the value of x at which the incline and curve cross and alpha is a constant.

phdquant · July 17th, 2008, 8:12 am

yes it is independent of m and time of intersection= \frac{2}{\sqrt{g}}

daveangel · July 18th, 2008, 9:50 am

agreed with phdquant - 2/sqrt(g). the longer the distance it travels the faster the acceleration .

phdquant · July 18th, 2008, 10:00 am

BTW, I have LaTeXed both of least time problems, can eventually upload them

Paul · July 18th, 2008, 10:25 am

Supplementary question, what if the line is not straight but allowed to be any curve. What is the shape of the curve that gets you from (0,1) to any point on the semicircle fastest?P

quantyst · July 18th, 2008, 4:40 pm

QuoteOriginally posted by: PaulSupplementary question, what if the line is not straight but allowed to be any curve. What is the shape of the curve that gets you from (0,1) to any point on the semicircle fastest?PExcellent Question!Along the same lines, we can ask the following:1. Given points (0,b) and (a,0), with b>0, find a frictionless curve y=f(x) with f(0)=b, f(a)=0 such that y=f(x) minimizes the time of descent between the two points (0,b) and (a,0) for an object released at point (0,b).2. Given a point (0,b) and a curve y=g(x), find a frictionless curve y=f(x) with f(0)=b such that y=f(x) minimizes the time of descent between the point (0,b) and a point of intersection of y=f(x) and y=g(x) for an object released at point (0,b).Of course, when y=g(x)=the semi-circle, we get Paul's question.

khpyc · July 18th, 2008, 7:42 pm

Isn't it the Brachistochrone curve?

Olescro · July 23rd, 2008, 8:30 pm

instead of , let s use theta as the angle with the y axis distance = 2 * cos (theta)gravity = g * cos (theta)time = 2*sqrt ( distance/gravity) = 2/sqrt(g)