MOMENT OF INERTIA

What is the moment of inertia of a solid cone about an axis passing through the centre of the base circle of radius r?

8 Answers

62
Lokesh Verma ·

let 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
MAK ·

hint...

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...

62
Lokesh Verma ·

sorry MAQ ...

1
Aditya ·

but bhaiyya, how can we directly integrate from 0 to l, bcoz even the radius is variable?

62
Lokesh Verma ·

i have taken r to be x tan theta.. which si varying.. isnt it?

1
Aditya ·

but i didnt get it exactly.... cud u pls explain once again

62
Lokesh Verma ·

inertia 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...

1
Aditya ·

thanks bhaiyya, i got it!

Your Answer

Close [X]