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Al P
Lv 7
Al P 發問於 Science & MathematicsPhysics · 9 年前

How long does Joe's trip last?

Joe the flying ant loves to run the rim of a hollow

cylinder at 12 m/s. After running from point (A), he decides

to fly at 3i m/s parallel to diametrically opposite point (B), but

he doesn't realize that his flight path (plane) is falling at rate

(g). So he ends up banging his head into the wall at point (B') directly

below (B). What is the maximum amount of time it takes Joe

to go from (A) to (B) if Joe, now dizzy, walks from B' to B at

2 m/s due to the lump on his head, and what unsimplified *physical *

equation represents his actual path? Thank you.

更新:

Very good feedback az_.

1. First, this question is my own creation.

2. Ok, let the radius of the cylinder be 50m.

3. (i) is Joe's unit vector.

4. "why is there an and": At times, a mathematically simplified result

hides a physical result (path). This may seem like a good thing, but

many times I think otherwise.

Thank you Koshka.

更新 2:

Koshka, you can choose any radius you want.

The TU is due to the fact that you cared

to answer, and you are good! Extra good!

Yes, as per your drawing Joe bangs his head

at B'. Hint: I would drop the vector (R') and

keep (r). He starts at A and ends up at B.

更新 3:

Just imagine he's flying "on" an imaginary

falling circular plane. Ant ESP :)

更新 4:

Note:

There's are two additional reasons, unrelated to the answer, I have for

asking this question. I'll disclose them if this Q&A concludes.

更新 5:

Koshka, I don't know how to spell you new

name;) : Yes, the Pythagorean Theorem can

be worked into this, but just before Joe

bangs his head, he is moving along a curve.

更新 6:

I was really trying to paint a clear picture using words, and I had a very

good reason for doing so. If anyone still cares try this:

http://i985.photobucket.com/albums/ae336/sdopqwe82...

更新 7:

falzoon, pretty thick is not a word that comes to mind

regarding you and your work on YA!. I've read many of

your excellent answers.

Let the length of the red and green curves be equal

since the blue cord is always parallel to AB. Let any

change in momentum not be a factor. It may help

to visualize the cylinder falling "UP" at g, and Joe

"stays" put on the circular plane.

Joe moves as follows according to him:

1. He moves from point A to some random point C (red curve)

2. He then moves from point C following blue straight line

to point C' always parallel to AB=A'B'. He watches the red

arc rise above him.

C'. Since he instantly realizes he'll bang his head at C' he

wisely follows the green arc where he is again running on a rim

no longer parallel to AB but he's still in the same initial plane.

3. I should not have gotten so cute with the head banging thing,

but this what happened at point B': He instantly moves along

the nor

更新 8:

mal vector at B' (bangs his head) and "sticks" to the

cylinder wall.

4. He must now walk from from B' to B at 2 m/s

5. So we have 4 discrete paths and maximum time

of travel to find.

Note: Length red arc = Length green arc

更新 9:

Yes, the drawing looks in the lab frame.

更新 10:

lost a word:

Yes, the drawing looks correct.

更新 11:

T = T1+T2+T3+T4

T = s1(r,a)/12 + s2(r,a)/3 + s3(T2(r,a),r,a) + s4(r,a)/12

Unsimplified path:

S = ra + [2r*cos(a) + (2*r^2*cos(a)^2*g/9)] + ra

T(a) = ra/12 + [2r*cos(a)/3 + (2*r^2*cos(a)^2*g/9)/2] + ra/12

更新 12:

u = cos(a)

d(u^2)/da = = 2u*du/a = -2cos(a)sin(a)

∂T/ ∂a = r / 6 - 2 * r * Sin(a) / 3 - (2 * 2 * cos(a) * sin(a) * g * r ^ 2 / 9) / 2 = 0

Some iteration shows that for maximum time

a ≈ 0.001519759 rad (close enough)

Plugging into T(a):

T ≈ 2758.34 s

Good Job falzoon.

2 個解答

相關度
  • falzoon
    Lv 7
    9 年前
    最愛解答

    I'm not understanding this problem very well.

    Scenario 1 : Joe runs a certain distance from point A at 12 m/s, then stops, then flies

    along the blue line at 3 m/s, but is continually falling, so his path should be parabolic,

    but I can't see how he would hit point B' if he's going parallel to line AB. It looks like

    he should hit the bottom wall at a point (call it C') directly underneath the top of the

    green line (call it C).

    Scenario 2 : Same as scenario 1, except, as soon as Joe begins to fly, he's still got

    some momentum from the previous run, and although he flies parallel to AB, his final

    vector is going to carry him more towards the top of the circle in your diagram, in

    which case, there is still no way to crash into point B', as his first point of call.

    Scenario 3 : He flies in a direction more towards the centre of the circle, so that the

    sum of his flight vector and the red line vector will finally get him to reach point B', but

    then, he won't really be flying parallel to AB.

    So I'm flummoxed as to how he gets to B' without first hitting another part of the wall,

    or gets to B' by flying parallel to AB. I can see you've tried hard, and it's certainly true

    that I'm pretty thick sometimes when it comes to comprehension, so believe it or not,

    I feel some more clarification is required. It's an interesting question, but Joe is not the

    only one banging his head.

    EDIT: I'm still perplexed, so instead of many words, is my diagram correct?

    http://s205.photobucket.com/albums/bb192/falzoon/C...

    EDIT 2: Diagram - http://s205.photobucket.com/albums/bb192/falzoon/T...

    Given: r = radius of cylinder (m). I'm going to relate all times to θ = angle AOC.

    Joe's 1st and 3rd paths are equal. Their distances are each rθ metre, and at 12 m/s,

    the time for each is rθ/12 sec. The 4th path's distance is H metre, where H is the

    height of the cylinder, and at 2 m/s, the time is H/2 sec.

    Total time so far = T1 + T3 + T4 = rθ/12 + rθ/12 + H/2 = rθ/6 + H/2 sec.

    Joe's 2nd path is a parabola, with horizontal distance = d, and vertical distance = H.

    From the diagram, d = 2r*cos(θ).

    General distance equation : s = ut + (1/2)gt^2.

    Substituting values for horizontal displacement : d [= 2r*cos(θ)] = 3*cos(0º)*(T2).

    2r*cos(θ) = 3*(T2). Therefore, T2 = 2r*cos(θ)/3

    Substituting values for vertical displacement : H = 3*sin(0º)*(T2) + (1/2)g(T2)^2.

    H = g*(T2)^2/2. Therefore, T2 = √(2H/g). Equating the two T2's, we find that

    H = 2gr^2*cos^2(θ)/9, but this is more of an aside, as amazingly, knowing H is not

    required as it is dependent on θ. However, we do need this expression for total time.

    Now that we have T2, the total time of travel, T = rθ/6 + H/2 + 2r*cos(θ)/3 sec.,

    and on substituting for H,

    T = rθ/6 + gr^2*cos^2(θ)/9 + 2r*cos(θ)/3 sec.

    dT/dθ = r[3 - 12sin(θ) - 2gr*sin(2θ)]/18

    With g = 9.81 m/s^2, equating dT/dθ to zero to get the maximum, gives :

    19.62*r*sin(2θ) + 12*sin(θ) - 3 = 0

    This is not easy to solve, so I think we'll have to introduce r = 50 metre.

    Then, WolframAlpha (with a bit of tinkering), and knowing that 0 < θ < π, gives a

    maximum of T ≈ 2758.3396656583483976 sec. when θ ≈ 0.0015197591683283 rad.

    ≈ 0.0870757862215213º

    Which means that θ is not much more than zero degrees.

    If θ was zero, then T is a tad smaller at 2758+1/3 sec.

    Just to finish off,

    T1 = T3 ≈ 0.0063 sec., T2 ≈ 33.3333 sec., T4 ≈ 2724.9937 sec,

    H ≈ 5449.9874 metre, d ≈ 99.9998845166257415 metre.

    BTW, I don't know the answer to the"unsimplified physical equation" question.

  • az_lender
    Lv 7
    9 年前

    My questions:

    What do you mean by "3i m/s" ?

    Was there a diagram giving some dimensions of the cylinder?

    Why is there an "and" in the last question - does it mean that

    you omitted one of the questions in the textbook version?

    If we don't know the diameter of the cylinder,

    the maximum time from A to B could be

    much longer than Joe's life expectancy.

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