ParthKohli
  • ParthKohli
What is this?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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ParthKohli
  • ParthKohli
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ParthKohli
  • ParthKohli
@ganeshie8 Help me out.
ParthKohli
  • ParthKohli
|dw:1449768649040:dw|

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ParthKohli
  • ParthKohli
First, I think there's some kinda problem with the question - it says \(\sqrt{2gL}\) in the statement but \(2\sqrt{gL}\) in the diagram. I'm going to suppose the former.
ParthKohli
  • ParthKohli
|dw:1449768850204:dw|
ParthKohli
  • ParthKohli
C, D definitely correct (I think)
ParthKohli
  • ParthKohli
Consequently B is incorrect. Let's check A though.
ParthKohli
  • ParthKohli
|dw:1449769130123:dw|
ganeshie8
  • ganeshie8
assuming \(u=\sqrt{2gL}\), would the block reach a height of \(L\) ?
ParthKohli
  • ParthKohli
yeah, see, that's the problem...
ParthKohli
  • ParthKohli
maybe the string will slacken (idk the right term) and it'd reach its highest point after that?
ParthKohli
  • ParthKohli
then that would redefine the meaning of the highest point of the trajectory, part of the reason I'm confused by this.
ParthKohli
  • ParthKohli
wait, why can that not happen?
ParthKohli
  • ParthKohli
maybe those conditions can simultaneously hold for a particular \(R\).
ParthKohli
  • ParthKohli
so what do you think? :\
anonymous
  • anonymous
just make the block to the height l .. rest is all correct you'll get CD and for the A part i am getting it 2R pi/3 will still hold
ParthKohli
  • ParthKohli
ok, so the problem was with the diagram...
ParthKohli
  • ParthKohli
|dw:1449769884441:dw|
ParthKohli
  • ParthKohli
Yes, should obviously be 2R... hmm.
anonymous
  • anonymous
yeo
ParthKohli
  • ParthKohli
Thanks guys :)
ganeshie8
  • ganeshie8
|dw:1449770397901:dw|
ganeshie8
  • ganeshie8
@ParthKohli
ParthKohli
  • ParthKohli
OK, that's a good question. The way I imagine it is that the string is always tangent to the point of contact.
ParthKohli
  • ParthKohli
Dunno if that makes much sense, but I've always treated that fact for granted. But it is pretty intuitive to think that the string should be tangent.
ParthKohli
  • ParthKohli
Otherwise, a nonelastic string cannot exert a tensile force.
ganeshie8
  • ganeshie8
Ahh that sounds legit..

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