66This and many more problems, you can find here:
http://kaymant.googlepages.com/101ProblemsInMechanics.pdf
(It was supposed to be a collection of 101 problems, but so far contains only 94 of them.)
The system above in the diagram iis released.
The mass of the black beads is m, while that of blue loop is M
What is teh minimum m/M ratio needed for the lower loop to take off from the ground..
(A variation of this question was in FIITJEE GMP long time back... I havent seen the latest one... )
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10 Answers
3i think asish u have done by applying force eqn.
N-2mg-Mg=0 if N=0 we get m/M=-1/2 which is not possible so i think nishant sir want us to think more.
66
62That is an awesome set of questions :) Each one seems to be a gem :)
I hope you are able to complete the 101 soon..
btw what;s with your book? found any publisher?
We have plans of publishing a book.. wanted to talk to you about our own progress... but we havent been able to meet.. last time rain Gods din want us to :D
66I had talk with a few publishers. And the talk didn't go the way I would have liked. They wanted to have full control over the presentation, but I wanted it my way. I am considering publishing it myself with the print-on-demand technology. I had plans of publishing it around december or so... but it seems that it will take a bit longer --- around march.
62Yeah the publishers want everything
even our magazine plan took a back seat because of that :(
they want too much controll....
1(at any instant at an angle θ )
for bead;
N+mgcosθ=mv2/r -(1)
and using COE =mgr(1-cosθ)=1/2 mv2
v=√2gr(1-cosθ)
now putting in (1)
N=mg(2-3cosθ) -(2)
now for 2 beads the upward force
F= 2Ncosθ=2mgcosθ(2-3cosθ) (from (1))
to get the θ for which this upward force is maximum ,we defferentiate the eq
d/dθ (2cosθ-3cos2θ)=0
we get cosθ=1/3
so upward force = 2mg/3
the hoop will rise if 2mg/3>Mg
m/M>3/2
is it correct sir