IIT JEE , AIEEE Forum - Physics , Mechanics

computer001

hey 2day v had a q in fiitjee test:
il jus describe it:
u have a ring
initially on the inner circumfrence @ topmost pt u have a small particle of mass m...
EDIT
oh sorry i 4 got to mention :
it is attached to the ring
mass of ring also m
ring follows pure rolling.. assume sufficient friction n all
wat is vel of the particle wrt person on ground when the ring turns thru 90^o

my logic is tht chng in pot energy = mgr
chng in KE=1/2mv^2 for particle + 1/2 I w^2 for rot of the ring
+ 1/2 m (v')^2 for linear motion of ring
v'=rw as pure rolling
v=rw as particle m is fixed to inner surface..
hence v get 3/2mv^2=mgr so v=rt{2/3*gr}

so vertical vel of particle=rt{2/3*gr}
hor vel also = rt{2/3*gr} due to translation of ring
so net v=rt{4/3gr}

some1 pls temme wat my mistake is...
n ofcourse elegant solution r always welcome :)

arpan1

i did the same as u did except for the fact that i took

I = MR² + MR² = 2MR²

anandghegde

why? is your ans wrong?

srini

Hey,

Gokul:chng in KE=1/2mv^2 for particle + 1/2 I w^2 for rot of the ring
+ 1/2 m (v')^2 for linear motion of ring
at that point the velocity will be v in i direction and rw in the j direction.so v²+v².

correcta?

computer001

@ arpan n anand:
nope my ans is not even an option!!!
@ srini:
u have prpendicular components u have to do lik rt(x^2 + y^2)

srini

Hey,

errr...you are asking or telling?

computer001

i dunno ..
if ur prev post was a questuion then my post was an answer...if ur prev post was just a statement then my post was confirmation :P

anandghegde

gokul how can you write change in pot enrgy =mgr ?

srini

Hey,

0Romba kolapara)

ya after turning 90,there will be the velocity component in both the directions...soroot and den square

.

computer001

see the ring is only rolling its com does not move so no chng in Pe for it alone
but for the particle ini it is @ the top @ a dist 2R from the ground...now when the ring rotates by 90 the particle comes to the same level as the centre of ring...so it comes down by a ht mgr

computer001

@ srini:
THAT IS EXACTLY WAT I HAVE SAID BUT I THINK THIS IS THE IRRELEVANT PART>>>THE MAIN Q IS ABT THE LOGIC Y R V DISCUSSIN ABT HOW TO CALC NET VEL WID X AND Y COMPONENTS???????

anandghegde

cant the particle slide inside the ring?

computer001

oh sorry i 4 got to mention :
it is attached to the ring...like welded to the ring or wateva

anandghegde

ok.....

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