Quote:MrVInteresting; I guessed wrong ...

https://www.youtube.com/watch?v=MBSRPuUibxk

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 ???

Quote:ssho88The "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.

I found the result surprising, mostly because the curved path would be have issues with the wildly changing rotational momentum of the ball.Quote:ThatDonGuyEnergy 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.

Quote:ThatDonGuyEnergy 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 ?

No. The acceleration is different for the two cases.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 ?

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).

Quote:unJonNo. 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

Quote:ssho88Same 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?