Time falling on slopes of same length

[feynman1]

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

 

[jbriggs444]

What are your thoughts? We cannot help until you have showed some effort yourself.

 

[feynman1]

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.

 

[haruspex]

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.

 

[anuttarasammyak]

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.

 

[haruspex]

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.

 

[anuttarasammyak]

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 $T_2 > T_1$.

 

[berkeman]

Thread closed for Moderation (no effort shown by OP).

 

[berkeman]

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.