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 Forum index » Diversions » Perplex City Puzzle Cards » PXC: Purple Puzzle Cards
[SOLVED] #185 - purple - spin cycle
Moderators: AnthraX101, bagsbee, BrianEnigma, cassandra, Giskard, lhall, Mikeyj, myf, poozle, RobMagus, xnbomb
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Author Message
Atrophied
Entrenched


Joined: 29 Aug 2004
Posts: 1133
Location: 53742E 4A6F686E27732C 4E4C00
zaeil wrote:
firebird wrote:
278 tether ... means radius of tether is 139


Nope. The tether goes from the point of rotation to the object being swung, meaning that its length IS the radius. Tie something to the end of a string and swing it around (carefully, please, don't want anyone losing an eye to science Wink )--you'll see what I mean.

EDIT: And your answer has to be in meters.


He blinded me with science!

sorry...
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PostPosted: Thu Nov 17, 2005 1:46 pm
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zaeil
Decorated


Joined: 31 Aug 2004
Posts: 233
Location: NC, US
Atrophied wrote:
zaeil wrote:
firebird wrote:
278 tether ... means radius of tether is 139


Nope. The tether goes from the point of rotation to the object being swung, meaning that its length IS the radius. Tie something to the end of a string and swing it around (carefully, please, don't want anyone losing an eye to science Wink )--you'll see what I mean.

EDIT: And your answer has to be in meters.


He blinded me with science!

sorry...


I knew somebody was going to say that! Razz
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PostPosted: Thu Nov 17, 2005 2:43 pm
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Escapist
Boot

Joined: 04 Dec 2005
Posts: 14
Location: Guernsey

 		This is really annoying, if you use g=3.7 which is the commonly accepted value (also by nasa) it says you got the wrong answer...

PostPosted: Tue Dec 06, 2005 5:39 pm
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Mark_1984
Greenhorn

Joined: 10 Dec 2005
Posts: 3
Location: London

 		I reckon they've got the answer wrong. The radius is 87 m, which means that the tether must be 174m. A free object will always rotate around it's centre of gravity. Assuming that the counter weight and craft are the same weight, the pair will spin around their centre of gravity, which is half way between the two. For the counter weigth to stay stationary, as suggested by zaeil, it would have to have an infinite mass.

PostPosted: Sun Dec 18, 2005 9:53 am
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Scamon01
Guest


 		Metres
I did the same thing first time. The tether measurement of 278 is in feet on that website. The answer is supposed to be in metres. Embarassed

Scamon

PostPosted: Fri Jan 27, 2006 11:59 am
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Phant
Veteran

Joined: 30 Nov 2005
Posts: 72
Location: mostly London
Mark_1984 wrote:
I reckon they've got the answer wrong. The radius is 87 m, which means that the tether must be 174m. A free object will always rotate around it's centre of gravity. Assuming that the counter weight and craft are the same weight, the pair will spin around their centre of gravity, which is half way between the two. For the counterweight to stay stationary, as suggested by zaeil, it would have to have an infinite mass.


I agree with everything you say, except: "they've got the answer wrong"

The question asks for the radius of the tether i.e. the distance from the hab to the point of rotation. This the only meaningful distance calculable from the information given.


Zaeil was mistaken in saying "the length is the radius", and the diagram could be interpreted as showing the counterweight moving at the same speed as the Hab (i.e. it has the same mass). But is not necessarily true that the rest of the tether is the same length, this would depend on the mass of the counterweight - a heavier CW requiring a shorter tether.

Picture the path the Hab follows, it is clearly a circle, with its centre at the point of rotation, and so it is the radius of this circle that is the answer.

While there is nothing wrong with Jeb's solution I believe it is worth me giving a slightly different method:
Spoiler (Rollover to View):
First it seems we must take acceleration due do gravity on mars to be: g mars = 3.8ms -2

and we are given: Period T = 30s (It rotates twice in 60s)

Knowing centripetal acceleration is: a = v 2 /r (v is velocity (speed) and r the radius we are trying to find)

and: v = 2πr/T (circumference / time taken to travel circumference)

and we want a = g mars

we can combine to get: g mars = (2πr/T) 2 /r

or: g mars = 4π 2 r/T 2

and rearranging get: r = g mars T 2 /(4π 2 )

which should give you ~87m after plugging in the values.


PostPosted: Wed Mar 08, 2006 11:55 am
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batgirl
Decorated


Joined: 03 May 2006
Posts: 261
Location: London

 		Just to note, as per the workings on the websites like the one posted by looosy

Spoiler (Rollover to View):
85


is also an accepted answer
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PostPosted: Tue May 16, 2006 6:22 pm
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griph
Greenhorn

Joined: 12 Jun 2006
Posts: 3

 		I was able to get away with an even lower value:

Spoiler (Rollover to View):
84


When I first read the card I wasn't sure what the axis of rotation was. I though it might be like the tether was a solid metal rod and the whole thing was turning such that the rod basically was the axis. (Babylon 5 style.) It took me a minute to figure out that if that was the case, the tether length wouldn't matter, so this thing must be going end over end.

PostPosted: Mon Jun 12, 2006 10:00 am
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Platinumflux
Boot


Joined: 03 Aug 2006
Posts: 43
Location: Ireland

PostPosted: Sun Aug 06, 2006 3:58 pm
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extrusion
Kilroy

Joined: 21 Jun 2006
Posts: 2

 		wikiD
Just to say if you look on wikipedia it says:

Spoiler (Rollover to View):
To produce 1g the radius of rotation would have to be 224 m (735 ft) or greater,


and then look a bit lower

Spoiler (Rollover to View):
An artificial gravity field of 0.38g (Mars gravity)


then multiply you get an accepted answer.

Pretty darn easy for a purple, just 1 web page and calc!

extrusion

PostPosted: Tue Aug 15, 2006 7:07 pm
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frog51
Kilroy

Joined: 13 Dec 2006
Posts: 2
Location: Livingston

 		Another accepted answer
I rounded up but
Spoiler (Rollover to View):
88
was still accepted

PostPosted: Fri Dec 22, 2006 9:51 am
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