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Author Topic: Who is up for a challenge?  (Read 37643 times)

Offline Chemistry4me

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« Reply #75 on: 19/06/2009 03:06:53 »
P = E/t
 

Offline erickejah

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« Reply #76 on: 19/06/2009 03:08:28 »
so,
7.84492=E/60
E = 470.695J
 

Offline Chemistry4me

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« Reply #77 on: 19/06/2009 03:10:08 »
Yes. That was what I got.
 

Offline erickejah

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« Reply #78 on: 19/06/2009 03:13:09 »
thanks for baring with me. :)
 

Offline erickejah

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« Reply #79 on: 20/06/2009 01:20:59 »
when are u going to put the next one, I will double check my answers this time.. :)
 

Offline Chemistry4me

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« Reply #80 on: 20/06/2009 01:29:02 »
Are these questions too easy for you?
 

Offline erickejah

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« Reply #81 on: 20/06/2009 02:00:22 »
they are okay I guess, I only need resolve them earlier in the day. because fatigue affects my reasoning sometimes...
 

Offline Chemistry4me

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« Reply #82 on: 20/06/2009 04:22:11 »
Here:

Tom was looking at the image of newsprint using a concave lens by holding it close to the page of the newspaper. The actual height of the words on the newspaper is 3.0 mm. The image produced is 1.0 mm high when the lens is held 3.0 cm from the print on the newspaper. Calculate the focal length of the lens.
 

Offline lightarrow

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« Reply #83 on: 20/06/2009 19:51:19 »
Here:

Tom was looking at the image of newsprint using a concave lens by holding it close to the page of the newspaper. The actual height of the words on the newspaper is 3.0 mm. The image produced is 1.0 mm high when the lens is held 3.0 cm from the print on the newspaper. Calculate the focal length of the lens.
-15 mm.

Enlargement: x/y = p/q --> x/y = -3/1 (enlargement is negative if object and image are on the same side).

-3 = p/q --> q = -p/3 = -10 mm (negative because the image is on the same side of the object).

1/p + 1/q = 1/f --> 1/30 - 1/10 = 1/f --> f = -15mm (negative because is a diverging lens).
http://en.wikipedia.org/wiki/Lens_(optics)

Edit: for a better explanation of the terms, x is the height of the object, y of the image; p is the object's distance from the lens' centre and q the image's distance from the lens' centre (negative if on the same side of the object).
« Last Edit: 21/06/2009 10:30:04 by lightarrow »
 

Offline Chemistry4me

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« Reply #84 on: 21/06/2009 00:30:27 »
Yes, correct, you're on to it lightarrow. :)
 

Offline erickejah

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« Reply #85 on: 21/06/2009 02:37:37 »
nice, I would have done the same exactly thing,,,  :D :D JK lightarrow is the man  :)
 

Offline erickejah

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« Reply #86 on: 21/06/2009 23:09:57 »
hit me up with other one
 

Offline lightarrow

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« Reply #87 on: 24/06/2009 17:07:55 »
hit me up with other one
From a height R falls (from zero speed) a mass point which follows a curved trajectory of 1/4 circumference (so the total lenght is π/2*R).
From the same height R falls, at the same instant of time, from zero speed, a mass point along a stright inclined plane which is long π/2*R as well.

Which mass point arrives first?

« Last Edit: 24/06/2009 21:45:38 by lightarrow »
 

Offline syhprum

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« Reply #88 on: 24/06/2009 18:00:16 »
For point masses they both arrive at the same time, for real life rolling balls or cylinders the whole situation is much more complex as some of the potential energy is stored in the rotation of the objects.
 

Offline lightarrow

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« Reply #89 on: 24/06/2009 21:43:16 »
For point masses they both arrive at the same time,
Why?
 

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« Reply #90 on: 24/06/2009 22:49:39 »
I don't know about the specific two shapes in the original problem but,although you can say that they will arrive with the same speed, whatever the profile of the track (initial PE will always convert to the same KE), the time taken for the journey is not independent of the profile. Just imagine a track that starts off almost horizontal and then plunges down. It could take an hour to get to the edge of the first section.
From that, I conclude that it is unlikely that the two paths would take the same time. I'd have to do the integral for the quadrant journey - someone else can do it - or have they?
I imagine the fastest journey time would be for a vertical drop, followed by a very small radius curve and then a horizontal section to the end.
« Last Edit: 24/06/2009 22:51:59 by sophiecentaur »
 

Offline JP

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« Reply #91 on: 24/06/2009 23:13:05 »
I'd guess that the curved track mass will hit first.  My reasoning is that the curved track has most of its acceleration occurring at the beginning.  It starts from free-fall, and ends up rolling horizontally.  The straight track has constant acceleration all the way down.  Since the speed it has at any point depends on the integrated acceleration it has seen in the past, putting high acceleration at the beginning would give it a higher speed early on, which would make it cover the distance faster.
 

Offline erickejah

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« Reply #92 on: 24/06/2009 23:36:20 »
For point masses they both arrive at the same time,
Why?

I would like to say that the acceleration in both balls is the same in the Y direction, they both end the journey at the same time.  this is my assumption in a friction-free world and a flywheel effect free environment.

s
 

Offline syhprum

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« Reply #93 on: 25/06/2009 09:50:12 »
I would like in view of what Sopiecentaur said to withdraw my answer (that both would arrive at the same time) I now see it to be incorrect, it seems that despite the generous suppositions made a more detailed analysis is required.
A straight vertical drop would of course take the shortest time while on the inclined plane the effective value of G would be divided by the tangent of the angle of inclination relative to the vertical i.e if it was horizontal G would be zero.
I assume for the curved track we have to calculate the effective value of G for each infinitesimal point on the track, a job for those well versed in calculus.
« Last Edit: 25/06/2009 10:08:06 by syhprum »
 

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« Reply #94 on: 25/06/2009 09:53:21 »
For point masses they both arrive at the same time,
Why?

I would like to say that the acceleration in both balls is the same in the Y direction, they both end the journey at the same time.  this is my assumption in a friction-free world and a flywheel effect free environment.

s
No, the y component of acceleration is NOT the same.
 It is g/√2 on the straight, diagonal path but it starts as g and ends up as zero on the circular path.
jp's qualitative analysis has to be correct. It needs to be done with yer actual Maths if you really want the right answer but my two 'extreme' examples show the way things must be going.
Just read your post, syphrum. Thanks- I am just doing the sums - so far it seems that there is a factor of √2 !!! involved.
Edit -typo
« Last Edit: 25/06/2009 10:21:08 by sophiecentaur »
 

Offline syhprum

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« Reply #95 on: 25/06/2009 11:52:24 »
In many ways I see a similarity to this problem with that of calculating the time it takes for the gravity train to reach the centre of the earth (which also defeated me!)
 

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« Reply #96 on: 25/06/2009 12:17:31 »
The gravity train is 'just' simple harmonic motion - the attractive force is proportional to the distance from the centre and the 'constants involved just involve the density of the particular planet. You can either accept that the sums work or slog through it. What fascinated me about that problem was that it reveals that a piece of dust would have the same time period moving through a hole in a large granite rock as long as there were nothing else around and the rock was not tumbling (of course)

Anyway, I have had an interesting hour or so - which I should have spent doing something useful - working out some answers.

For the diagonal path.
The acceleration, a, is g/√2 and the distance to travel is l√2.
The equation of motion gives s = at2/2
So l√2 = gt2 /(2√2)
Giving
t = 2√(l/g)

For the circular path it is easiest to consider the angle from the centre of the circle – starting at θ = 0 and going up to 90o
At any point on the path, the tangential speed, v, is l dθ/dt and the 'imposed'  tangential acceleration is
g cos(θ) so you can write what is happening to the mass at that point
The acceleration will be dv/dt,
or  ld2θ/dt2
and you can write the equation as
g cos(θ) = ld2θ/dt2
I was messing around for ages before I realised that this is the diff equation for a simple PENDULUM –durrrr.
The period of a pendulum with small displacements (swings) is given by 2π√(l/g) and we are dealing with a quarter of a period so the time would be
t = (π/2)√(l/g) for a small displacement. (Edit: that's a pi - not an n - it isn't clear on my browser)
Or 1.57√(l/g)
BUT, for large displacements, the sums are much harder and I had to look it up. You can’t work it out simply and you have to get into “elliptic functions of the first kind – aaaargh!  Wiki tells us that, for a 90o swing, there is an 18% increase in time, giving
t = 1.85√(l/g)

Please check for accuracy but I think it's ok and shows that the straight path takes longer.

I then looked at what happens if the mass falls vertically, then, with the acquired speed (after a tight curve) it travels the horizontal distance.

The vertical drop takes time tv =  √2 √(l/g)  (same as the circular path, apparently - and a bit counter intuitive) and its final speed is √2gl, so it takes an extra time
th = l/√2gl  or √(l/2g)
Total time
t =  tv + th
t =  √(2l/g) +  √(l/2g)
or  √(l/g)  (√2 + √(1/2))
so
t = 2.12√(l/g)
Which, is greater than for the circular path.
Somewhere in there there must be an optimum.
There must be some Maths wizards out there who could find it?
« Last Edit: 25/06/2009 14:09:39 by sophiecentaur »
 

Offline lightarrow

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« Reply #97 on: 25/06/2009 14:29:21 »
The vertical drop takes time
...
t = 2.12√(l/g)
Which, is greater than for the circular path.
Of course you considered the fact that the total path in this case is greater than (π/2)R, isnt'it?
 

Offline lightarrow

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« Reply #98 on: 25/06/2009 15:13:55 »
Sophiecentaur and jpetruccelli have given the right answer: the point mass in the circular path arrives first.
The following is my solution.

Said:

m = mass of the point-mass
s = curved coordinate from the initial point
v = velocity = ds/dt
R = maximum height of the point mass = radius of the circumference of the curved path
h(s) = height of the mass point with respect to the lowest point of the path
E = total energy = mgR
V = gravitational potential energy = mgh(s)
Kinetic energy = E - V = mgR - mgh(s)

we have:

(1/2)mv2 = E - V --> v = ds/dt = Sqrt[2(E - V)/m]

--> dt =  ds*Sqrt[m/2(E - V)] = ds*Sqrt[1/2g(R - h(s))] = ds/Sqrt[2g(R - h(s))]

Let's compare h(s) in the two cases:

Circular path: h(s) = R[1 - sin(s/R)]   computed writing s = R*α, where α = angle between the instantaneous vector (from the circumference' centre and the point mass) and the horizontal

Straight path: h(s) = R - (2/π)s  (similar triangles)

You can see that h(s) for the circular path is always smaller than h(s) for the straight path:

R[1 - sin(s/R)]  < R - (2/π)s  --> sin(s/R) > (2/π)s/R

that is: sinx > (2/π)x   0 < x < π/2

This last inequality is true; it's simple to see it drawing a graph of the functions sinx and (2/π)x.

So h(s) is lower, for the circular path, then R - h(s) is greater and so dt = ds/Sqrt[2g(R - h(s))] is lower.

Integrating dt, it follows that the total time is lower for the circular path.
 

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« Reply #99 on: 25/06/2009 15:42:07 »
That's a novel way of looking at it.
But does it show that the third option I suggested takes longer? It would suggest that, as the height is lost quicker, it would get there faster so the inequality wouldn't always always apply.
So what is the shape of fastest path from top to bottom?
 

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« Reply #99 on: 25/06/2009 15:42:07 »

 

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