I solved this one a few weeks before the JEE...and it made quite an impact...took me approx. 20 mins to even get started...
Here'e the deal:
We have a particle and a ring, both smooth, on a horizontal plane. As shown, they are initially in contact and at rest, when the particle is given a velocity V (as shown).
Find the time in which the particle completes one full revolution inside the ring.
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ReplyDeletet = 2*π*r/v
ReplyDeleteis my answer correct???????also tell only answer bhaiya of SHM problem .i just wnt only the answer not the full solution.i am talking about the question---------- All the rods in figure are hinged smoothly as shown. The vertical red line is infact a spring of force const 'k'...
ReplyDeleteIn terms of the angle A, k, m find the period of small oscillations (along x) of the blocks.
Bhaiya...please give some hint to start the question...i started with analysing the motion in reference frame of the ring...but got ol messed up wid the equations..
ReplyDeleteif you have started by analysing in the ref. frame of the ring, then you are nearly done...
ReplyDeletehere's a hint: all the equations are useless..just see the forces which act on the particle in the frame of the ring.
is my answer correct??????????????????
ReplyDeleteyup...your answer is right!!
ReplyDeletegetting any hints people?
Can we do this by simple energy conversation in the frame of the ring, coz d energy of d system remains constant, nd ring is olways at rest in dat reference frame?
ReplyDeletei really don't know whether conservation of eng. is valid in frames...
ReplyDeletebut if you look at the forces acting on the particle in the frame, they are infact centripetal..
ie: normal reaction+pseudo force/
in the case of equal masses of the particle and the ring, this amounts to zero...
in short: the tangential velocity remains constant!!
@vaibhav
ReplyDeleteconservation of energy is not valid in frames...for that, you have a 'reduced mass system' equation....not in JEE syllabus
@anuj bhaiya....
plz explain this
"ie: normal reaction+pseudo force/
in the case of equal masses of the particle and the ring, this amounts to zero..."
Really sorry for that...i was wrong, the force in the frame cannot be zero..
ReplyDeleteLet us assume the normal reaction on the ring 'N', then the accn of the ring is N/m...radially outwards.
In the frame of reference, the ball experiences these force
1)'N', radially inwards..
2)the pseudo force which is m*(accn of the frame)...which comes out to be N, if both masses are assumed equal!
The mistake i did, was in the dirn. of the pseudo force.While it really is radially inwards, i took it outwards, hence getting the zero..
bhaiya then is the ans still that or something else??
ReplyDeleteby taking any general velocity at any point i am getting 6 eq in 6 variables so we can relate v with theta but they do not look easy to solve
@sambhav....if u noe abt d reduced mass concept...please try to solve dis with that concept
ReplyDelete@vaibhav
ReplyDeletethis is the energy conservation equation in frames...apply if you can....(i couldn't)
KE(mass 1) + KE(mass 2) - KE(center of mass)= KE(reduced mass)
1/2m1v1^2 + 1/2m2v2^2 - 1/2(m1+m2)vcm^2 = 1/2(m1m2/m1+m2)vrelative^2
hey!!! give up this reduced mass yaar...we have enough to do already./
ReplyDelete