33where does pi comes...
answer is clearly ..(15g/32R)1/2 whats the problem here.. :O
25 Answers
62after that find the potential energy of the sphere at slight displacement and then as all of u are suggesting ..
21LOL...... Yeah.......
Oh my.... i took the COMs for rign n disc!!!!!!!!!!
3yeah...tis b only ne...find the change in PE from unstable to stable equilibrium, equate it to change in roational KE about the axis of rotation (horizontal, passing through geometric centre of the compound sphere)...the answer is just waiting to be calculated !
33ooohh.... u have taken COM wrong...
its R/2 for hollow and 3R/8 for solid hemisphere
what u are taking 2R/pi
3R/4pi
are of half ring and disc... :P
3321HMMMM.....
IF MY COMs are correct i dont think ther cud b any mistake in the foll steps as i did them abt 3 times now....
21Sir, pl. notify me if ther is any mistake ( ther has 2b some otherwise answer does not match as i hav the factor Î on LHS which is not ther in the RHS so no question of getting cut and the answer does not hav a factr Î ) [2] [2] [2] [2] [7][7][7]
21COM of solid sphere = 4R/3Î -----> 1
COM of HOLO sphere = 2R/Î -----> 2
EQN 2 - EQN 1
2R/3Î ....
THER4 ∂(COM) = 2*(2R/3π)
THER4 ∂PE = (4R/3Π)*g*(2m)
equating da change in PE to gain in rot. KE => 8mRg/3Π= 0.5 * (16/15)(mR2)(ω2)
now we need to make ω the subject.....
62tapan i think you can take the average of the center of masses of the two part
one hollow hemisphere
second solid hemisphere
2/pi was the answer for one of the parts probably .. (not sure though!)
1yeah thanx i missed out on the fact that masses were equal so the COM would be above its lowest position
you know this habit of mine of taking some "obvious" assumptions [6]
2116/15MR2 as written in ur earlier post......
∂PE = ∂(COM) * g * 2m.....
sir can u help me find the xpression 4 ∂(COM) as masses of both spgheres as well as their volumes r SAME.
62yeah you have to.. so i want trying to find I (moment of inertia)
That because it will have pure rotation at the bottom most poitn..
21Yeah thnx PHILIP!!!!
and da ans 2 ur que....
"bcoz every body in Universe endeavours to attain minimum P.E. when free to do so"
so a slight displacement is enough for the sphere to be set into motion which definitely leads to complete inversion!!!!
So its ωmax can b found out....
[1]
33well ω is max when KE is max... when cahnge in PE is max... which is in case of complete inversion..
21I know to find the PE gained by using change in COM,
but the big que is wether in toally inverted postn wud the angular freq be max?????
How can u be sure of dat???
33yeah u have to do same PE =KE gained..
Just take there COMs...
21Sir,
I thot we wud hav to solve by equating the change in PE upon complete rotatn with the rotational energi gained.... [7]
Can u pl. solve it till the end ??? pl....
I cudnt carry on frm wer u left....
1sry ...
actually i din undersatnd the question...
i mean wat did it ask...
62this is a good question..
Find the CM of the compound sphere...
and then also the I along that axis.. that will be
2/3MR2+2/5MR2
16/15MR2