02 Jul 20
@ponderable saidThe balls roll without slipping so you need to take angular momentum into account. In the case of particles the instantaneous speed is proportional to the square root of the depth. The total distance traveled depends on the shape of the track, since the horizontal distance is fixed. If the shallower track were perfectly flat then a particle would never reach the other end, which leads me to think that a particle on the deeper track would arrive first. But the question specifies a ball that rolls without slipping and that changes matters.
So this is probably an Energy balance Problem. I the beginning both balls have Zero velocit, but are accelerated by the difference in potential Energy:
E(pot)=m*g*dh
This Energy is converted into kinetic Energy
K(kin)= 1/2*m*v^2
At teh end of the track the kinetic Energy is converted back into potentila energy.
So we can consider the identical parts as being ide ...[text shortened]... the second step we can safely assume that track B is run through fast than case B (without friction)
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02 Jul 20
@ponderable saidYour order is correct. However, there is a unique solution, not two.
It is clear that B lies. So he has to have finished second.
C can only be true if he finished 1st.
Then we would have 1. C, 2. B, then D has to have lied and thus finished 3rd and A spoke truth and finished fourth.
But if D spoke truth it would be 1. D, 2. B. A speaking truth would still be fourth and C would have spoken untruth and finished third.
So I find that ...[text shortened]... e two Solutions.
Since Joe said that the second solution is wrong I go for 1. C, 2. B, 3. D, 4. A
If B's statement = TRUE:
"I am the 3rd place runner" contradicts 1st and 4th place statements TRUE and 2nd and 3rd place statements FALSE.
Thus it must be a FALSE statement.
So B must be 2nd or 3rd place.
It cant be 3rd, because B's statement again becomes TRUE.
Thus B is in 2nd Place.
Assume "C" has made a FALSE statement. That means they must be in 3rd place ( the only remaining place that can make a false statement ). But C's statement "I am not the 4th place runner" is a TRUE statement when they are in 3rd place. Contradiction.
Thus "C" has made a TRUE statement and must be in 1st place ( as 4th place finish for C would be a FALSE statement).
Order is currently C,B,X,X with one confirmed FALSE statement and TRUE statement.
Next, "D's" statement "I finished the race before B" must be FALSE, as the place finishing before B is 1st, which we just established is runner "C".
Thus, "D" finishes 3rd ( with a FALSE statement ).
And "A" finishes 4th with the remaining TRUE statement: "I finished the race after C"
Final Race Order:
C, B , D, A
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02 Jul 20
@ponderable said"Even if there is some additional way for the second step we can safely assume that track B is run through fast than case B (without friction)"
So this is probably an Energy balance Problem. I the beginning both balls have Zero velocit, but are accelerated by the difference in potential Energy:
E(pot)=m*g*dh
This Energy is converted into kinetic Energy
K(kin)= 1/2*m*v^2
At teh end of the track the kinetic Energy is converted back into potentila energy.
So we can consider the identical parts as being ide ...[text shortened]... the second step we can safely assume that track B is run through fast than case B (without friction)
Can you clarify that?
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02 Jul 20
@joe-shmo saidI think you are saying Track B will win the race, but that isn't exactly what you said.
"Even if there is some additional way for the second step we can safely assume that track B is run through fast than case B (without friction)"
Can you clarify that?
02 Jul 20
@joe-shmo saidTrack B wins (because the length in the depths is run through faster than in Track A)
I think you are saying Track B will win the race, but that isn't exactly what you said.
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02 Jul 20
1 edit
@ponderable saidOk. Correct! ( unless deepthoughts "second guessing" is justified in some subtle way )
Track B wins (because the length in the depths is run through faster than in Track A)
The tracks are identical before and after the second dip in track B. So focus on the common sections of track across the second dip.
Put a coordinate system across with respect to Track A.
x - direction parallel to track A at the flat length, y direction normal to it.
Where the two tracks deviate the initial velocity of both balls is "v" in the x-direction.
Track A maintains a constant velocity "v" over the sections in question.
The ball on track B once on the second slope experiences accelerations in both the y and x directions.
If we first analyze the forces acting on the ball in a coordinate system x' -y' aligned with the slope we find that the magnitude of the Normal Force ( the reaction force perpendicular to the slope ) is given by:
N = m*g*cosΦ
Where Φ is the angle of the slope with respect to the original coordinate x.
Now, I wish to compare velocities with respect to the original coordinate system in the "x" direction. The sum of the forces in x direction is as follows:
ΣF_x = N*sinΦ = m*a_x
m*g*cosΦ*sinΦ = m*a_x
a_x = g *cosΦ *sinΦ
Checking extremes to verify it makes sence let Φ = 0, we expect a_x = 0
a_x = g *cos0 *sin0 = g*1*0 = 0
And at Φ = π/2 we again expect the acceleration in x direction to be zero. ( the ball is in free fall )
a_x = g *cos(π/2) *sin(π/2) = g*0*1 = 0
what we have is that for:
0 ≤ Φ ≤ π/2
0 ≤ N ≤ mg
a_x ≥ 0
As a result, the velocity of the ball on track B in the "x-direction" grows from v to v' = v + δv as it continues down the slope. In otherwords for the simplest comparison the average velocity of the ball on B in the x-direction is greater than the velocity of the ball on track A over the same interval and same direction.
( 2v + δv )/2 > v
Next, the flat section the ball will again experience no acceleration in the x-direction, but it will maintain its velocity of v' = v + δv which is again greater than the velocity of the ball on track A over the same interval and direction.
v' = v + δv > v
Finally, on the ascent the ball will experience the mirror acceleration of the first section. the velocity of the ball on B will decrease from v' to v. Again, the average velocity of the ball on track B is greater than that of the ball on track A over the same interval and direction.
( 2v + δv )/2 > v
So in every section of the track B across the that second "dipped" region ball B experiences a higher average velocity than the ball on track A in the x-direction ( over the same distance in that direction ). So it arrives at the top of the trough with the same velocity as Track A, but it does so in a lesser amount of time.
Ball B wins the race.
I think that is a sufficient explanation, but if Deepthought justifies his "second guessing" with some subtle argument about angular momentum I'll be glad to hear it.
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02 Jul 20
@venda saidIt intuitively can seem like that when you realize whatever speed is gained on the downhill is lost on the uphill, but overall the speed is greater through the entire second dip as opposed to a speed just remaining what it is in track A as it crosses the same lateral distance as the second dip in track B.
Aren't we(or should I say Pondy and Deep thought -you've totally lost me with the energy thing)) overthinking this?
My answer is a tie -although I expect it'll be wrong!!
03 Jul 20
3 edits
@joe-shmo saidIt depends which ball has the further drop and rise. If B initially flattens at the same point as A, then B wins. B will travel faster than A, but never slower. This assumes of course that A and B are the same length.
Identical balls are released simultaneously along fixed tracks A and B from the left end. ( the beginning's and ends of the tracks are at the same position horizontally and vertically. The tracks are smooth and continuous. The illustration I have attempted does not perfectly represent that )
What is the result of the race? A wins, B wins, or A and B tie?
Assume that t ...[text shortened]... e frictional losses of energy.
Track A:
…\___________/
Track B:
…\__...………__/
………\____/
If the total drop is the same, just B has it broken up, then A wins because it will be traveling at B's greatest speed longer.
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03 Jul 20
2 edits
@eladar saidSorry..."Perhaps people around here were not smart enough to figure that out." - Eladar
It depends which ball has the further drop and rise. If B initially flattens at the same point as A, then B wins. B will travel faster than A, but never slower. This assumes of course that A and B are the same length.
If the total drop is the same, just B has it broken up, then A wins because it will be traveling at B's greatest speed longer.
I think most people correctly assumed from the accompanying illustration ( admittedly low quality ) that the total drop of Track B was greater than that of Track A as was intended to be demonstrated by said illustration.
05 Jul 20
@joe-shmo saidI had a bit of a think. The speed of the ball is v and if it rolls without slipping then the angular velocity w = v/a where a is the radius of the ball. This means that the kinetic energy has two contributions: T = mv^2/2 + Iw^2/2 = 1/2(m + I/a^2)v^2 = 1/2 M v^2. For a solid sphere I = 2/5 ma^2 [1], so M = 7m/5. So it just changes the effective mass of the ball. I don't think it'll change anything qualitatively.
Ok. Correct! ( unless deepthoughts "second guessing" is justified in some subtle way )
The tracks are identical before and after the second dip in track B. So focus on the common sections of track across the second dip.
Put a coordinate system across with respect to Track A.
x - direction parallel to track A at the flat length, y direction normal to it.
Where th ...[text shortened]... fies his "second guessing" with some subtle argument about angular momentum I'll be glad to hear it.
[1] https://en.wikipedia.org/wiki/List_of_moments_of_inertia