ML(R^2) / 2sqrt(L^2 + R^2)
do we have to consider the base of the cone...?
no!
just to clarify.. the axis shown is the natural axis..or the axis passing thru the apex and perpendicular to the base..
i asked that stupid question cuz i m getting the same as vibhav...plz see to it..areal density=[M/{pi*R*sqrt(L^2+R^2)}area of one elemental ring=perimeter*width=(2*pi*x*tanQ)*(dx)apply the formula for the moment of inertia of a ring of radius x*tanQthen integrate from 0 to Lat then end, substitute tanQ=R/L
i had a doubt with my answer as well.if L=0, it should give the MI of a ring which it doesnt.
mistake is in the 2nd step..area is NOT what you say it to be....keep on trying!also, at L=0, it gives the ans. of a disc, not a ring..
is it [MR^3/(2*sqrt(L^2+R^2)]..?
yeah. my bad.I made ditto same mistake.got the answer now though.sambhav, y dont u try converting height into radius and integrating from 0-Rull avoid the mistake then\and afterwards ponder wat u did wrong.
didn't get you....what is the answer?
THE ANSWER IS 3/10 MR^2
that's for a solid cone yaar...
is it 1/2MR^2
3/5 mr^2??
Sanchit got it right..Man! i thought it was simpler than Problem 5..
yess mr^2/2assume a small element dx, x height below the top.it will be a cylinder. dI=dm r^2find dm by taking S.A. ratio multiplied by total massintegrate. L^4 get cancelled
bhaiya mjhe meri galti kahan hai ye to pata lag gaya...par kya sahi lena tha yeh bata do..what would be the width of an elemental ring?
dx/cos(Q)
liquids ke sse1 me bhi to dx/cosQ liya hai
waise to its 2 late..but iv got it 2...
anuj bhaiya can u plz tell ur id wanna discuss something with u
shit man..!pehle dx lia...phir dx*secQ phir dx*tanQsecQ vale ko chodkar baki dono answers post kar die...
ML(R^2) / 2sqrt(L^2 + R^2)
ReplyDeletedo we have to consider the base of the cone...?
ReplyDeleteno!
ReplyDeletejust to clarify.. the axis shown is the natural axis..or the axis passing thru the apex and perpendicular to the base..
ReplyDeletei asked that stupid question cuz i m getting the same as vibhav...plz see to it..
ReplyDeleteareal density=[M/{pi*R*sqrt(L^2+R^2)}
area of one elemental ring=perimeter*width=(2*pi*x*tanQ)*(dx)
apply the formula for the moment of inertia of a ring of radius x*tanQ
then integrate from 0 to L
at then end, substitute tanQ=R/L
i had a doubt with my answer as well.
ReplyDeleteif L=0, it should give the MI of a ring which it doesnt.
mistake is in the 2nd step..area is NOT what you say it to be....keep on trying!
ReplyDeletealso, at L=0, it gives the ans. of a disc, not a ring..
is it [MR^3/(2*sqrt(L^2+R^2)]..?
ReplyDeleteyeah. my bad.
ReplyDeleteI made ditto same mistake.
got the answer now though.
sambhav, y dont u try converting height into radius and integrating from 0-R
ull avoid the mistake then\and afterwards ponder wat u did wrong.
didn't get you....
ReplyDeletewhat is the answer?
THE ANSWER IS 3/10 MR^2
ReplyDeletethat's for a solid cone yaar...
ReplyDeleteis it 1/2MR^2
ReplyDelete3/5 mr^2??
ReplyDeleteSanchit got it right..Man! i thought it was simpler than Problem 5..
ReplyDeleteyess mr^2/2
ReplyDeleteassume a small element dx, x height below the top.
it will be a cylinder. dI=dm r^2
find dm by taking S.A. ratio multiplied by total mass
integrate. L^4 get cancelled
bhaiya mjhe meri galti kahan hai ye to pata lag gaya...par kya sahi lena tha yeh bata do..
ReplyDeletewhat would be the width of an elemental ring?
dx/cos(Q)
ReplyDeleteliquids ke sse1 me bhi to dx/cosQ liya hai
ReplyDeletewaise to its 2 late..but iv got it 2...
ReplyDeleteanuj bhaiya can u plz tell ur id wanna discuss something with u
ReplyDeleteshit man..!
ReplyDeletepehle dx lia...phir dx*secQ phir dx*tanQ
secQ vale ko chodkar baki dono answers post kar die...