24what happened to my question?
two particles of mass m and 2m are connected by rigid rod(mass≈0) and slide with μ=0 on circular path. of radius r inside of vertical circular ring.If system released at rest at θ=0°.
Determine
1)velocity v of particles when rod passes horz.position
2)max velocity of particles
3)max value of θ
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18 Answers
62oh not the one u said..
we dont need to find the potential energy after finding the CM..
you can do it by taking summation of mgH
24Answer is v2=gr(3-2√2/3√2)
plzzzz explain ""finally it will be 3mgr(1-1/√2)""
i am having confusion in soln becoz i am used to solve questions by first solving COM
62I edited my answer..
i think that should do..
if not i think i must be making some blunder which i am not able to see
24ok i agree with u....It will be easy for 2 or atmost 3 particles......But when particles increase should we follow CM method????
62well eureka that is not always necessary.. you can do without doing what u are saying :)
24But sir when a system of particles is there ,,, dont we first find CM and then proceed with Potential energy calculations????
62This i think you can do by simple work energy theoerm
Conservation of energy (PE+KE)
I dont think this is tough..
this only looks tough.. (Or else i am raelly missing something)
Eureka.. YOu have solved much tougher questions than this one..
give it a shot once... or else just tell me i will give the full soln :)
24sir howz initial potential energy 2mgr ?????????? Dont u think we first have to locate CM coordinates at t=0 and then get potential energy......
And then again when rod is horz?????
62second part..
Potential energy will be
2mgr(1-sinθ) + mgr(1-cosθ)
mgr(3-2sinθ-cosθ)
For kinetic energy to be max this has to be min...
so for min potential , 2sinθ+cosθ has to be max = √5
min potential will be mgr(3-√5)
Initially it is 2mgr
Change is mgr(√5-1)
This is equal to 1/2 (m+2m) v2
62haha.. ok :)
initially the potential energy is 2mgr
finally it will be 3mgr(1-1/√2)
both will have the same "speed"
So that will be
1/2(m+2m)v2=2mgr-3(1-1/√2)mgr
3/2 mv2=mgr(2-3+3/√2)
3/2 mv2=mgr(-1+3/√2)
v2=gr(√2-2/3)
Is this correct?
24Oh my GOD..........I missed something in figure........
here is the correct one
This came to light when I read this"" initially the potential energy is 2mgr""
24Sir I solved this type of question when masses were equal....In that case when rod became hrz angles made were 45°....I am not sure about the situation here........(Becoz we certainly need the position of COM)..............Plz help