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Author Topic: Who is up for a challenge?  (Read 37607 times)

Offline Chemistry4me

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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 »


 

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?
 

Offline Chemistry4me

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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
 

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 ?
 

Offline Chemistry4me

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

Offline syhprum

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Who is up for a challenge?
« 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
 

Offline Chemistry4me

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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.



The string is parallel to the ground when it breaks.
 

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 »
 

Offline Chemistry4me

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Who is up for a challenge?
« Reply #8 on: 31/05/2009 09:43:47 »
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.
 

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 
 

Offline syhprum

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« 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.
 

Offline Chemistry4me

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« Reply #11 on: 01/06/2009 00:33:25 »
Looks like everyone is on holiday. :)
 

Offline erickejah

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« Reply #12 on: 01/06/2009 04:07:00 »
100m?
 

Offline Chemistry4me

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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. :)
 

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?
 

Offline Chemistry4me

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« Reply #15 on: 01/06/2009 04:24:04 »
Fc = mv2
        r
 

Offline erickejah

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

Offline Chemistry4me

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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?
 

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... [:-[]
 

Offline erickejah

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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.
 

Offline Chemistry4me

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« Reply #20 on: 01/06/2009 04:44:44 »
She can't. That embarrassed smiley is permanently broken.
 

Offline erickejah

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« Reply #21 on: 01/06/2009 04:52:11 »
can she modify it ????


here try with this one:



 

Offline Chemistry4me

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« Reply #22 on: 01/06/2009 04:53:08 »
Your calculations are correct except for the last one, 0.707 x 10 doesn't equal 70.7 m. :) Then there is another step... :)
 

Offline erickejah

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« Reply #23 on: 01/06/2009 04:56:34 »
lol,,
 

Offline erickejah

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« Reply #24 on: 01/06/2009 04:57:16 »
what is the next step?
 

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Who is up for a challenge?
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