Time falling on slopes of same length

AI Thread Summary
The discussion revolves around the analysis of time falling on slopes of equal length, with participants suggesting the use of variational calculus and graphical solutions. A key point is the relationship between tangential speed and vertical position, expressed through the equation dt=dl/v(l), where v(l) is derived from conservation of energy. The integral T=C ∫_0^L (dl/√-y(l)) is proposed to investigate the time taken on different paths. However, the thread was closed due to the original poster's lack of effort in presenting their work, leading to a permanent ban for repeated infractions. The conversation highlights the importance of demonstrating understanding in academic discussions.
feynman1
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Homework Statement
A ball falls from the top of 2 slopes of the same length as shown. No friction and no assumption of the convexity of either curve (they might be convex and concave at various locations, but the blue is always above the red). Which schematic solution is the best in showing that it falls from the red curve faster?
Relevant Equations
mechanical energy conserved
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What are your thoughts? We cannot help until you have showed some effort yourself.
 
jbriggs444 said:
What are your thoughts? We cannot help until you have showed some effort yourself.
One can use variational calculus, but I'm looking for a graphical solution.
 
feynman1 said:
One can use variational calculus, but I'm looking for a graphical solution.
You'd have to figure out what to graph, which may be easier once you have solved it algebraically.
 
Time passing the path interval [l,l+dl] is
dt=\frac{dl}{v(l)}
where l is length parameter of the path and v(l) is tangential speed as function of l.
By conservation of energy v(l) = v(y(l)) where y is vertical coordinate of the path and y=0 at start top l=0.
 
anuttarasammyak said:
Time passing the path interval [l,l+dl] is
dt=\frac{dl}{v(l)}
where l is length parameter of the path and v(l) is tangential speed as function of l.
By conservation of energy v(l) = v(y(l)) where y is vertical coordinate of the path and y=0 at start top l=0.
But it is not necessarily the case that ##v_{lower}(l)>v_{higher}(l)## for all l.
 
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We shall investigate integral
T=C \int_0^L \frac{dl}{\sqrt{-y(l)}}
where y(0)=0, y(L)=-H.

For path 1 and path 2, in the case ##y_2(l) > y_1(l)## for all l, obviously ##T_2 > T_1##.
 
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Thread closed for Moderation (no effort shown by OP).
 
OP has received a standard infraction for showing no work on a schoolwork-type question, and since this infraction adds to a number of previous infractions for the OP, this has resulted in a permanent ban. Thread will remain closed.
 
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