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Author Topic: Newtons cradle question  (Read 3222 times)

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Offline swashy

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Newtons cradle question
« on: 22:02:45, 01 March, 2010 »
Suppose you had a newtons cradle that was a mile long (or longer), and you could observe both ends simultaneously, would there be a measurable time delay in the 'end' ball movement, and what effects would be causing the delay?
Ade

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Offline royholl

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Re: Newtons cradle question
« Reply #1 on: 22:12:33, 01 March, 2010 »
A very good question Ade, I'm still thinking about it, but I will get back with some kind of answer...
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Offline stewartw

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Re: Newtons cradle question
« Reply #2 on: 08:20:15, 02 March, 2010 »
Here's my estimate:

When the ball at one end strikes the next one, a compression wave is transmitted through the second ball and on to adjacent balls. This wave will propogate at the speed of sound in that material (which for steel is about 4512 m/s). So for a line of balls that is 1 mile long (sorry for the mixed units) I'd expect a delay of approximately 0.36s

Similar principals are applied when doing seismology - when there is an earthquake, sound waves propogate through the earth - they are more commonly referred to as seismic waves. By monitoring the time that receiving stations scattered around the earth receive the sound / seismic wave, deductions can be made in relation to the interior of the earth.

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Offline Nomis Elfactem

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Re: Newtons cradle question
« Reply #3 on: 09:38:52, 02 March, 2010 »
Good question Ade....

Great answer Williams - sounds spot on to me... yes, there will be a delay due to the time it takes for the energy of the first ball (or any number of balls) in motion to be transferred to the other end vie the "shock wave" and the composition of the ball material dictates this speed.

S.
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Offline starf

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Re: Newtons cradle question
« Reply #4 on: 11:03:47, 02 March, 2010 »
Wouldnt it depend on the initial force, mass of the spheres and the medium in which the pendulum swings?

We know the greater the mass of an object the slower its acceleration. eg given the same force (kick), a bowling ball will accelerate more slowly than a beachball. In other words, the ratio of the masses of two bodies is equal to the inverse ratio of their accelerations when the same force is applied to both.

Question didn't stated how big (massive) your balls are and whether you are swinging them in the air (full of inudendo), or for that matter what force you have applied to them.

Newtonian mechanics:

Newton's second law: The net force on a body is equal to the product of the body's mass and the acceleration of the body. Net force is a vector quantity with both magnitude and direction as is acceleration whilst mass is a scalar quantity, so mathematically,

[tex]\vec{F}_{net}=m\vec{a}[/tex]

1N = force necessary to accelerate a mass by [tex]1 m/s^2[/tex]

Offline Nomis Elfactem

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Re: Newtons cradle question
« Reply #5 on: 11:28:21, 02 March, 2010 »
Yeap, you're totally right Starf.... but I guess the answer is still yes... all of the above (plus a lot more besides) just dictates the length of the delay, not whether there is one or not ?!? 

As ever with these things it can get rather complex so I guess we simplified it as a result.... e.g there is also torque and angular momentum to be taken into consideration due to the "strings" that hold the balls plus what the strings are made of and friction in all the couplings and not forgetting the temperature of the system etc etc...

Gets the mind gym working for sure, doesn't it - Newtonian physics at it's best  ;)

S.
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Offline stewartw

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Re: Newtons cradle question
« Reply #6 on: 14:08:39, 02 March, 2010 »
Hi Starf,

Yes, I'd agree that what you mention is applicable but only in respect of the energy that is transferred from one end of the line of balls to the other and the resultant acceleration of the final ball, not with regard to the time it would take for that energy to be transferred.

The speed at which the "kick" is transferred is dictated solely by the composition of the medium (in this case the balls). The exception to my answer above would be the case where the initial ball was moving at a speed greater than that of the speed of sound in the material. The analogy here would be an object travelling at supersonic / hypersonic speeds.

Real life examples of pieces of metal colliding at these kinds of speeds (of the order of a few miles / second) this include things like collisions between objects in orbit. For solids travelling at such speeds my understanding is that they behave more like liquids and it gets way to complicated for my brain   :hurt:

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William

Offline chris.bailey

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Re: Newtons cradle question
« Reply #7 on: 08:19:31, 03 March, 2010 »
Think someone needs to build it to test the theory. We could call it the LSC (Large Spherical Colider).
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Offline swashy

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Re: Newtons cradle question
« Reply #8 on: 20:20:31, 05 March, 2010 »
I hadn't considered all the variables, but thanks for the answers  :thumbup:

So as I understand it, the impact of the first ball creates a shock wave which travels at a speed which is variable depending upon the material of the balls in the line, I suppose you could actually consider the 'line' of balls as one continuous length of material, in which case you would need an increasing force of impact depending upon the length of the line to have any measurable change at the other end, (otherwise I guess the ball would just bounce back) and any loss would be because of compression?

So theoretically, if the 'line' was made of a material which was unable to be compressed, and the impact of the first ball was high enough, the answer to the question would all simply depend upon the speed at which the shock wave can travel through the subject material?

 :gum:
Ade

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Offline SneezingRabbit

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Re: Newtons cradle question
« Reply #9 on: 23:28:42, 14 April, 2010 »
This kinda makes me want to buy a newtons cradle now. Wouldn't it loose momentum before it reaches the end of the mile?
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Offline swashy

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Re: Newtons cradle question
« Reply #10 on: 07:28:46, 15 April, 2010 »
This kinda makes me want to buy a newtons cradle now. Wouldn't it loose momentum before it reaches the end of the mile?

Depends how big your balls are!
Ade

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Offline spaismunky

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Re: Newtons cradle question
« Reply #11 on: 07:49:48, 15 April, 2010 »
Ang on a minute. This is all very highbrow and therefore unintelligible to me, but wouldn't there be some energy released as heat on each ball striking each other, which I believe is why perpetual motion is so damned hard to prove, thereby slowing it all down. Would it even reach a mile? It always boils down to size doesn't it? ;)
Another thing...you'd be at the receiving end watching the ball strike then BAM, your eardrums would explode as the sound caught up ;)
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Offline stewartw

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Re: Newtons cradle question
« Reply #12 on: 08:05:11, 15 April, 2010 »
Heat would be generated and so yes, the energy transmitted along the length would decrease. This would have the effect of reducing the amplitude of the compression wave. But the original question asked what the time delay would be and so the answers above still hold true.

Waves propogate at a speed that is the product of their wavelength and frequency (v=f?)

Notice here that there is no mention of the amplitude - the speed of propogation is independant of the amplitude.

A good analogy here is sound waves in air - do loud noises travel faster or slower than quiet ones? The answer is no, they both travel at the same speed. Loud ones do however travel further because they have more energy and so it takes longer for their energy to dissipate as heat.

Best regards

William


Offline spaismunky

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Re: Newtons cradle question
« Reply #13 on: 08:36:22, 15 April, 2010 »
You see? You did maths there and I'm a girl. How many kittens are there in v=r?
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Offline stewartw

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Re: Newtons cradle question
« Reply #14 on: 08:45:43, 15 April, 2010 »
Oops ... looks like my greek lambda character (see http://en.wikipedia.org/wiki/Lambda) got converted to a "?" during the posting process.

Lambda is the traditional grrek figure used to denote wavelength.

The formula for the speed of propogation of a wave is velocity = frequency x wavelength

 

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