gravitational illusion
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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?
Quote: ssho88Same 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.
Quote: unJonThat’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 !
Direct: https://www.youtube.com/watch?v=Q93jouWh1dc
By the way, I love these slides!
Quote: gordonm888There 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).
