Time falling on slopes of same length

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Homework Help Overview

The discussion revolves around the time taken for an object to fall along slopes of equal length, exploring the relationship between path shape and time of descent. The subject area includes concepts from physics, particularly dynamics and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using variational calculus and graphical methods to analyze the problem. There are considerations about how to represent the paths graphically and the implications of energy conservation on speed as a function of vertical position.

Discussion Status

The discussion includes various approaches to the problem, with some participants suggesting algebraic solutions may facilitate graphical representation. However, the thread was ultimately closed due to a lack of demonstrated effort from the original poster.

Contextual Notes

There are indications that assumptions about the relationship between speeds at different heights are being questioned, particularly regarding whether the speed at lower elevations is consistently greater than at higher elevations.

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
1650451256606.png
 
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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##.
 
Last edited:
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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