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The Proof is Trivial!

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Original post by DJMayes
That's the general convention in Mechanics questions I've seen although strictly speaking I don't think it's true.


OK - I've get the inequality, as long as μ<2\mu <2 . See my editted solution. The question's second part does seem to suggest that they are using the mechanics convention that mu is between 0 and 1... :smile:

Now for the next bit...
(edited 11 years ago)
Original post by Star-girl
OK - I've get the inequality, as long as μ<2\mu <2 . See my editted solution. The question's second part does seem to suggest that they are using the mechanics convention that mu is between 0 and 1... :smile:

Now for the next bit...


Your solution is incorrect. The GPE should be mgx, I don't know where the (l-x) is coming from. As for using the convention, I came up with the question so I know what it's expecting. :lol:
Original post by DJMayes
Your solution is incorrect. The GPE should be mgx, I don't know where the (l-x) is coming from. As for using the convention, I came up with the question so I know what it's expecting. :lol:


mgx^2/2 surely?
Reply 343
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Original post by metaltron
12absinθ \frac{1}{2}absin\theta we can see the area will be maximised by maximising two sides and making the angle between them a right-angle.

This needs to be justified.
Original post by DJMayes
Your solution is incorrect. The GPE should be mgx, I don't know where the (l-x) is coming from. As for using the convention, I came up with the question so I know what it's expecting. :lol:


Of course but it is actually mgx2mgx^2 (*) because the mass per unit length is m m and that bit of rope is x x units long... as for the (l−x) (l-x) , I was accidentally using the wrong section of rope! :rofl:

Time to modify that bit. XD

EDIT: (*) Actually bananarama is right about this and it is in fact half of this value.
(edited 11 years ago)
Original post by bananarama2
mgx^2/2 surely?


Where does the factor of 12\dfrac{1}{2} come in?
Reply 346
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Original post by Star-girl
the mass per unit length is m m

I think in the original question it was 1, but of course it doesn't really matter. It just might have avoided having to add m's where they will later cancel out.
Original post by Star-girl
Where does the factor of 12\dfrac{1}{2} come in?


Centre of mass of the dangly bit...
Original post by und
I think in the original question it was 1, but of course it doesn't really matter. It just might have avoided having to add m's where they will later cancel out.


Ah OK (in DJ's question, it doesn't specify, it just says that it's constant) - where's the original from?
Original post by bananarama2
mgx^2/2 surely?


I suppose that it would be different actually because of the rope, not simply mgh. That said, I used a different method altogether. Should I post it up?
Original post by bananarama2
Centre of mass of the dangly bit...


Oh, of course! :facepalm:

I may have overlooked that. :tongue:
Original post by DJMayes
I suppose that it would be different actually because of the rope, not simply mgh. That said, I used a different method altogether. Should I post it up?


Different to mine too?
Reply 352
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Original post by Star-girl
Ah OK (in DJ's question, it doesn't specify, it just says that it's constant) - where's the original from?


Oops, sorry, I read uniform as unit. :colondollar:
Original post by Star-girl
Ah OK (in DJ's question, it doesn't specify, it just says that it's constant) - where's the original from?


There isn't an original question. I was doing a STEP question similar to this (One of the Mechanics from STEP II, 1988) where you had a rope falling off of a smooth table and you had to find the time taken for it to fall off. I copied it over for someone on the Maths thread who was asking for interesting questions, but accidentally omitted the fact the table was smooth. They asked, and I decided it would be interesting to see how it would differ if friction was involved, so I added it and played around with it to see what it would do. My idea for the question initially involved finding the time taken to fall and the speed as it left the table, but I felt that that was too similar to the question it was adapted from and it missed a really interesting idea in whether it would fully fall off or not.

Original post by bananarama2
Different to mine too?


Yes.
(edited 11 years ago)
I'm very intrigued :tongue:
Original post by DJMayes
I suppose that it would be different actually because of the rope, not simply mgh. That said, I used a different method altogether. Should I post it up?


I think you, bananarama and I (once I've corrected it) should all post our solutions as they seem to be different. :cute:

Original post by bananarama2
Different to mine too?


See above.

Original post by und
Oops, sorry, I read uniform as unit. :colondollar:


No worries. :tongue:
Original post by Star-girl
The question's second part does seem to suggest that they are using the mechanics convention that mu is between 0 and 1... :smile:


There is no such convention.
Original post by und
What are the conditions for replacing everything with its modular counterpart? I'm not that familiar with modular arithmetic so I should probably read up on it a little.

Okay so this was what I was looking for (it's understandable that a few steps were skipped)
Since n7−n≡n(n6−1)n^7 - n \equiv n(n^6-1) and by Fermat's( little theorem for natural n non divisible by 7, n6≡1n^6 \equiv 1 (mod7)then

n6q+r≡(n6)q.nrn^{6q+r} \equiv (n^6)^q.n^r (mod7)

≡1qnr \equiv 1^q n^r (mod7)
i.e. we can reduce the power of n by 6q for any natural q.
2222≡3 2222 \equiv 3 (mod7)
and 5555≡−3 5555 \equiv -3 (mod7)
So we have
35555+(−3)22223^{5555} + (-3)^{2222} (mod7)
And as 6*925 = 5550 and 6*370 = 2220, by our power reduction;
≡35+(−3)2\equiv 3^5 + (-3)^2 (mod7)
≡32(33+1)\equiv 3^2(3^3 +1) (mod7)
≡32.28\equiv 3^2. 28 (mod7)
≡0 \equiv 0 (mod7)
Original post by DJMayes
...


You may be interested to know that now that I've corrected it and solved for v2 v^2 , it ends up as exactly lgμ21+μ\dfrac{lg\mu ^2}{1+\mu } , so the question may need some modification. :tongue:
Original post by Mr M
There is no such convention.


OK - thanks for clearing that up. The issue has been resolved now. :smile:

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