MrV
Joined: Feb 13, 2010
• Posts: 7891
April 10th, 2021 at 4:35:38 PM permalink
Interesting; I guessed wrong ...

"What, me worry?"
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5856
April 10th, 2021 at 5:49:56 PM permalink
The fastest time from the top to the bottom is if it is a single cycloid curve (well, inverted cycloid curve).
ssho88
Joined: Oct 16, 2011
• Posts: 655
April 10th, 2021 at 6:25:02 PM permalink
Quote: MrV

Interesting; I guessed wrong ...

The "curve" path have longer distance(more surface resistance) and therefore more energy loss, the ball velocity should lower at the end.

Why the ball on "curve" path have higher velocity(more energy ?) at the end ? Energy = 1/2mv^2 ???
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5856
April 11th, 2021 at 9:57:00 AM permalink
Quote: ssho88

The "curve" path have longer distance(more surface resistance) and therefore more energy loss, the ball velocity should lower at the end.

Why the ball on "curve" path have higher velocity(more energy ?) at the end ? Energy = 1/2mv^2 ???

Energy doesn't really have anything to do with it, and I am assuming any surface resistance is negligible.

The only force on the ball is gravity, which is always straight down, but any modification to the velocity will be in the direction of the slope of the ground; the steeper the ground, the higher the increase in the ball's velocity will be. What effect particular shapes of the ground has is a little beyond what I learned in physics.
OnceDear

Joined: Jun 1, 2014
• Posts: 7054
April 11th, 2021 at 10:26:04 AM permalink
Quote: ThatDonGuy

Energy doesn't really have anything to do with it, and I am assuming any surface resistance is negligible.

The only force on the ball is gravity, which is always straight down, but any modification to the velocity will be in the direction of the slope of the ground; the steeper the ground, the higher the increase in the ball's velocity will be. What effect particular shapes of the ground has is a little beyond what I learned in physics.

I found the result surprising, mostly because the curved path would be have issues with the wildly changing rotational momentum of the ball.
Beware. The earth is NOT flat. Hit and run is NOT a winning strategy: Pressing into trends is NOT a winning strategy: Progressive betting systems are NOT a winning strategy: Don't Buy It! .Don't even take it for free.
rxwine
Joined: Feb 28, 2010
• Posts: 11714
April 11th, 2021 at 10:47:05 AM permalink
Hmm?

There's no secret. Just know what you're talking about before you open your mouth.
ssho88
Joined: Oct 16, 2011
• Posts: 655
April 11th, 2021 at 3:35:37 PM permalink
Quote: ThatDonGuy

Energy doesn't really have anything to do with it, and I am assuming any surface resistance is negligible.

The only force on the ball is gravity, which is always straight down, but any modification to the velocity will be in the direction of the slope of the ground; the steeper the ground, the higher the increase in the ball's velocity will be. What effect particular shapes of the ground has is a little beyond what I learned in physics.

"the steeper the ground, the higher the increase in the ball's velocity will be"
But the velocity of the ball will decrease when going UP HILL . . .

I thought TOTAL ENERGY before and after should be the same ?

TOTAL ENERGY before = TOTAL ENERGY after + energy loss due to resistance. Assumed energy loss due to resistance = 0,

Potential energy(mgH) + 1/2m(v1)^2 = Potential energy(mgh) + 1/2m(v2)^2

v1=0, so 1/2m(v2)^2 = mg(H-h), both cases should have SAME v2 at the end of the path, why the ball travel along "curve" path take shorter time to reach the end ?? Same v1 and v2(for both cases) does not mean same travel time ?
unJon
Joined: Jul 1, 2018
• Posts: 4189
April 11th, 2021 at 3:54:20 PM permalink
Quote: ssho88

"the steeper the ground, the higher the increase in the ball's velocity will be"
But the velocity of the ball will decrease when going UP HILL . . .

I thought TOTAL ENERGY before and after should be the same ?

TOTAL ENERGY before = TOTAL ENERGY after + energy loss due to resistance. Assumed energy loss due to resistance = 0,

Potential energy(mgH) + 1/2m(v1)^2 = Potential energy(mgh) + 1/2m(v2)^2

v1=0, so 1/2m(v2)^2 = mg(H-h), both cases should have SAME v2 at the end of the path, why the ball travel along "curve" path take shorter time to reach the end ?? Same v1 and v2(for both cases) does not mean same travel time ?

No. The acceleration is different for the two cases.

So the ball on the steeper drop is at a higher velocity for longer which offsets the longer track.

You could do the problem as an integral using acceleration formula and trigonometry for the gravitational force at each point of the track based on the angle.

You could also do it as an integration of potential energy translating into kinetic and rotational energy (last point is easy to forget about).
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ssho88
Joined: Oct 16, 2011
• Posts: 655
April 11th, 2021 at 4:39:38 PM permalink
Quote: unJon

No. The acceleration is different for the two cases.

So the ball on the steeper drop is at a higher velocity for longer which offsets the longer track.

You could do the problem as an integral using acceleration formula and trigonometry for the gravitational force at each point of the track based on the angle.

You could also do it as an integration of potential energy translating into kinetic and rotational energy (last point is easy to forget about).

Same v1 and v2(for both cases) does not mean same travel time ?

v1= initial velocity, v2 = end velocity
unJon
Joined: Jul 1, 2018
• Posts: 4189
April 11th, 2021 at 5:19:02 PM permalink
Quote: ssho88

Same v1 and v2(for both cases) does not mean same travel time ?

v1= initial velocity, v2 = end velocity

That’s correct.

Imagine a car that goes from 0 to 60 in three seconds vs a car that goes from 0 to 60 in 13 seconds. Which car crosses the finish line first?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
DRich
Joined: Jul 6, 2012
• Posts: 10743
April 11th, 2021 at 5:22:35 PM permalink
Quote: ssho88

Same v1 and v2(for both cases) does not mean same travel time ?

v1= initial velocity, v2 = end velocity

When I see V1 & V2 all I think about is that I can't abort the takeoff and that if we haven't hit V2 I hope an engine doesn't go out.
At my age, a "Life In Prison" sentence is not much of a deterrent.
ssho88
Joined: Oct 16, 2011
• Posts: 655
April 11th, 2021 at 5:58:29 PM permalink
Quote: unJon

That’s correct.

Imagine a car that goes from 0 to 60 in three seconds vs a car that goes from 0 to 60 in 13 seconds. Which car crosses the finish line first?

So I think it can be summarized as below :-

If we divide the path into 10 sections.

A) Case "a" (Straight Path)

Velocity along the straight path = V1a, V2a, . . .V10a = increase, increase , increase, increase. . . .increase

Total travelling time, Ta= S1a/V1a + S2a/V2a + . . .S10a/V10a = Integration( dSa/dVa)

B) Case "b" (Curve Path)

Velocity along the curve path = V1b, V2b, . . .V10b = increase, decrease , increase, decrease. . . .increase

Total travelling time, Tb= S1b/V1b + S2b/V2b + . . .S10b/V10b = Integration( dSb/dVb)

V1a = V1b = 0, V10a = V10b(if surface resistance is negligible and apply Energy Law), Sb > Sa, but Tb < Ta !
Wizard
Joined: Oct 14, 2009
• Posts: 25734
April 12th, 2021 at 6:16:09 AM permalink
I know this point has been made already, but think of "big drop" water slides. They are not on an inclined plane but start with a steep drop. The idea being to build up acceleration at the beginning, which more than offsets the longer distance traveled.

By the way, I love these slides!
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
gordonm888
Joined: Feb 18, 2015
• Posts: 4622
April 12th, 2021 at 10:57:03 AM permalink
There are some minor non-linear terms in this. The roller coaster ramp has more frictional drag, because the travel distance is longer and because frictional drag (per unit length of travel) is proportional to V^2 and the ball is at higher velocity for more of the path. Air drag is the other energy loss term and increases linearly with velocity (for laminar flow) for each unit of length that is traveled. So there are more energy losses on the left ramp and less on the right ramp.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
unJon
Joined: Jul 1, 2018
• Posts: 4189
April 12th, 2021 at 12:18:50 PM permalink
Quote: gordonm888

There are some minor non-linear terms in this. The roller coaster ramp has more frictional drag, because the travel distance is longer and because frictional drag (per unit length of travel) is proportional to V^2 and the ball is at higher velocity for more of the path. Air drag is the other energy loss term and increases linearly with velocity (for laminar flow) for each unit of length that is traveled. So there are more energy losses on the left ramp and less on the right ramp.

My physics is rusty (let me preface with that). But I thought for a rolling ball that doesn’t slip, all of the friction force gets converted into rotational energy. So isn’t an energy “loss” (meaning getting converted from kinetic to heat energy).
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
7up
Joined: May 22, 2010