13hint...
integrate the expression for moment of inertia of a disc about an axis from its centre, wrt radius from 0 to r...
hope dat will solve d problem...!!!
oops... i'm late...
What is the moment of inertia of a solid cone about an axis passing through the centre of the base circle of radius r?
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8 Answers
62let the density be k
and the angle be theta (of the slant length from the vertical line)
at a distance x take a disc of thickness dx
radius of the disc will be xtanθ
inertia along the central axis will be
(dm).r2/2 = k.π(xtanθ)2 . (xtanθ)2
=k.π(xtanθ)4
=k.π x4 tanθ4
take the integral from 0 to l
we get
=k.π L5 tanθ4/5
where k is linear mass density... or k=M/{1/3Ï€r2L}
k=3M/{πL3tan θ2}
If there is a calculation error.. pls fix it :)
13
1but bhaiyya, how can we directly integrate from 0 to l, bcoz even the radius is variable?
62inertia is additive...
so the inertia will be sum of the inertias of the individual disks..
each disc has a variable radius that varies directly proportional to the distance from the vertex of the cone...
the proportionality is tan theta... (theta is half the angle of the vertex)
Now the area of the disk is
constant . pi . d^2
where d is the distance form the vertex of the plate (disc)
NOw the sum of these discs give the area...