MrV Joined: Feb 13, 2010
• Posts: 6897
April 10th, 2021 at 4:35:38 PM permalink
Interesting; I guessed wrong ...

"What, me worry?"
ThatDonGuy Joined: Jun 22, 2011
• Posts: 4684
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: 505
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: 4684
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: 5035
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.
Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..
rxwine Joined: Feb 28, 2010
• Posts: 10008
April 11th, 2021 at 10:47:05 AM permalink
Hmm? Quasimodo? Does that name ring a bell?
ssho88 Joined: Oct 16, 2011
• Posts: 505
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: 2431
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: 505
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: 2431
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.