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  1. Naked Science Forum
  2. Non Life Sciences
  3. Physics, Astronomy & Cosmology
  4. Who is up for a challenge?
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Who is up for a challenge?

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Offline Chemistry4me (OP)

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Who is up for a challenge?
« on: 31/05/2009 04:15:59 »
It's a bit quite at the moment so how about having a physics problem of the week?  [:-X]
Here's an easy one for you:
A man spins a 0.5 kg stone on the end of a 2 meter string which has a breaking strain of 25 N. He whirls the stone 2.45 m above the ground. Eventually the string breaks and the stone flies off horizontally. Ignoring gravity  [:I] air resistance, how far does the stone land from his feet? 
« Last Edit: 31/05/2009 08:27:21 by Chemistry4me »
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Offline Madidus_Scientia

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Re: Who is up for a challenge?
« Reply #1 on: 31/05/2009 07:54:43 »
Ignoring gravity? I guess it would hit a building or a tree or something, don't think it would have enough energy to leave the Earths atmosphere before getting slowed down and stopped by air resistance. So, it wouldn't land at all?
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Offline Chemistry4me (OP)

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Re: Who is up for a challenge?
« Reply #2 on: 31/05/2009 08:03:13 »
OPPPS [:I], I meant air resistance! [:D]
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Offline syhprum

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Re: Who is up for a challenge?
« Reply #3 on: 31/05/2009 08:11:21 »
May we assume the Earth is flat so that the path taken is parabolic not elliptical ?
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Offline Chemistry4me (OP)

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Re: Who is up for a challenge?
« Reply #4 on: 31/05/2009 08:20:11 »
Yes. Parabolic.
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Offline syhprum

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« Reply #5 on: 31/05/2009 09:09:52 »
Ignoring the correction that should be made for the fact that the string is not horizontal when it breaks I calculate that the distance will be 22.66 meters
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Offline Chemistry4me (OP)

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« Reply #6 on: 31/05/2009 09:19:19 »
That's not what I got  [:-\]. Maybe the question will be better illustrated with a diagram.

[diagram=461_0]

The string is parallel to the ground when it breaks.
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Offline syhprum

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Who is up for a challenge?
« Reply #7 on: 31/05/2009 09:38:01 »
My fault I worked with a one meter string, the way you posed the question was fine its just that I can't read!

PS I assume it is the stone that is 2.45m high when the string breaks not the hands of the operator.
« Last Edit: 31/05/2009 09:41:41 by syhprum »
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Offline Chemistry4me (OP)

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Who is up for a challenge?
« Reply #8 on: 31/05/2009 09:43:47 »
Quote from: syhprum on 31/05/2009 09:38:01
PS I assume it is the stone that is 2.45m high when the string breaks not the hands of the operator.
I am not sure what you mean.
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Offline syhprum

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Who is up for a challenge?
« Reply #9 on: 31/05/2009 10:04:15 »
The string will never be horizontal there will always be a drop from the operators hands to the height of the stone.
This will have a small effect on the breaking strain which I have decided can be ignored but the height of the stone when it is released is highly relevent.
I await some other answers after my wrong result, this is a very commen cause of failure in exams not reading the question 
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Offline syhprum

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Who is up for a challenge?
« Reply #10 on: 31/05/2009 17:29:50 »
Is there no one out there who understands simple physics and can do arithmetic ?.
I will post no further calculations until I see what other solutions are offered.
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Offline Chemistry4me (OP)

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Who is up for a challenge?
« Reply #11 on: 01/06/2009 00:33:25 »
Looks like everyone is on holiday. [:)]
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Offline erickejah

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« Reply #12 on: 01/06/2009 04:07:00 »
100m?
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Offline Chemistry4me (OP)

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« Reply #13 on: 01/06/2009 04:11:07 »
What? 100m! No. A hammer throw doesn't even go that far. [:)]
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Offline erickejah

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« Reply #14 on: 01/06/2009 04:14:21 »
ok, yes I am wrong. can u tell me how to calculate the speed of the object just before the string breaks?
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« Reply #15 on: 01/06/2009 04:24:04 »
Fc = mv2
        r
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Offline erickejah

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« Reply #16 on: 01/06/2009 04:31:28 »
thanks.. :) what about 70.7107m
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Offline Chemistry4me (OP)

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« Reply #17 on: 01/06/2009 04:34:31 »
Err...that's not what I got. How are you calculating this?
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Offline erickejah

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« Reply #18 on: 01/06/2009 04:40:23 »
i got the final velocity by:

√(Fc*r) = v
     m

giving me 10m/s

then i just calculated the time which the object takes to touch the floor by:

Δy=1/2(Vi+Vf)t
Which gave me: .707107s

at the end I just multiply those values... [:-[]
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« Reply #19 on: 01/06/2009 04:42:26 »
by the way, if u Karen see this post. please fix my smiley.
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