39tahts why i said i gove it from sources. i never said i make them.
do them if u feel to do .. syllabus blabla bla math lao yaha
1]Two beads of mass m are positioned at the top of a frictionless hoop of mass M andradius R, which stands vertically on the ground. The beads are given tiny kicks,and they slide down the hoop, one to the right and one to the left, as shown. Whatis the smallest value of m/M for which the hoop will rise up off the ground at some
time during the motion?
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25 Answers
3SO ZERO NA AS m>>M .....[OR IS IT DAT U SHUDNT TAKE DAT APPROX]
YAAR ITNA ACHHA KYSA TYPE KARTA HAI TU...................KOI SOFTWARE HAI KYA??????
1sorry everyone.
But I was frustrated with his questions.
I come here but dont post any questions .. but when i saw his question and that the is saying that he is typing all the question i really became angry.
62@chetan..
Dont get so emotional.
Sometimes you should oversee some of these things.
I think he was only trying to give some good questions.
62take it easy guys..
ltes forget this one..
I will try to get all the questions that he had posted.
Good atleast that we have the source of all those questions :)
3LUKS LIKE HES ALREADYT LEFT..... SAD I THINK HE MUST HAVE GOT OFFENDED .....
EVERY 1 MAKES SUM MISTAKES..... U SHUDA LEFT HIM ....
1er..... well lets be a little easy on him chetan
maybe he gave it from other sources..... but that does not imply he was a fool... btw the problems are really good na ? so wats the problem and he is not forcing everyone to solve 'em ..
3WAT ABT , 4TH 1 ????
AND THIRD 1 PLS SEE MAH METHOD ... I DUNNO HOW TO FIND LIMIT OF DAT 1......
3ar u sure we hav to assume M>>m .... then cum wat may the velocity of the big block will not change(wil be minute change) ...
and it will hit the wall..........
but if we apply COR/NLC and COM well see that the velocity of the ball after collision 1 is 2v0 .....
AFTER COLLISION 2 IT WILL REMAIN WITH THE BLOCK ITSEF AS IT WILL ASO HAVE VELOCITY V0 ...............SO NO MORE COLLISIONS AFTER THAT AND IT WILL HIT THE WALL.....
IVE GONE TERRIBLY RONG I TINK ......
OR IVE DONE TOO MUCH APPROX. WITH M>>m....
3and the cookie i feel will be semicircular in 1 half and elliptical[not exactly but sumwat elliptical] in the other!!!!!![3].............
106Q1. 
From FBD of m,.... let speed be v when it makes angle θ with vertical.
Using WE theorem,
mv2/2 = mgR(1-cosθ) .....
==> v2 = 2gR(1-cosθ) ...... (i)
Writing force equation along normal,
mgcosθ - N = mv2/R
==> N = mg(3cosθ-2) ...... (ii)
If the mass M just lifts up,
then resulting force on it must be equal to Mg upwards.
Writing force equation along vertical for mass M,
Mg = 2Ncosθ
==> Mg = 2mg(3cosθ-2)cosθ
==> m/M = 1/2(3cos2θ - 2cosθ) ..... (iii)
For m/M to be min. d(m/M)/dθ = 0
==> taking derivative and solving,
cosθ=1/3
==> from (iii), m/M = 3/2
3thanx dud for postin problem .........my bad luck i have to go now to study....hav comp. xam tommorow .... havnt started doin any thing!!!!!!
will attempt tommorow