IIT JEE , AIEEE Forum - Physics , Mechanics

prakharajabe

find the moment of inertia of disc abt an axis making an angle of "A"(90>A>0) WITH THE VERTICAL AXIS PASSING THROUGH THE CENTRE..... PLEASE SOLVE...

gorakavipraveen

hey, isnt it smae. I remember , I have read somewhere that it is same
Is it or not?

physics_surfer

Since Moment of Inertia is a VECTOR quantity so we can use the concept of projection of vectors and so the answer is

((M * (R^2) ) / 2) * cosA

gorakavipraveen

are yaar what happens when a=90* are , it would yield moment of inertia 0. Is it possible, Its actually M(r^2)/4 but not 0. How can you justify it man.

stareawe

if the axis is oriented at A to the vertical then the distance "r" of every particle increases by a factor of cosA . Hence.,
^{the moment of inertia increases by a factor of (cosA)}²

gorakavipraveen

but stareawe, If one particle increases on to one side, the other particles distance decreases by the same factor. This is obviously supporting the decrease of MI as axes is turned by Pi/2 radians.

shubham_sachdeva

Yeah.gorakivipravin is correct

bhupesh

Dear

There will be no effect until your axis is passing through the COM of the disc

edison

                   *    *             Weather disc is inclined or axis the case is similar.
             *            *                  Here, is an inclined disc with center C making an
         *                *                  angle A with the vertical.
     *                  *                   So any point at a distance 'r' from the center is at
*         * C     *                       a distance r sinA from the vertical axis.
*                *                            Now to find moment of inertia about this vertical
*            *                                axis
   *    *

let us consider a ring at a distance 'r' from the center and width dr,

if mass of the disc = M

then mass of this thin circular ring = M (2

r dr) /

R²

or dm = M (2

r dr) /

R²

so moment of inertia at vertical axis due to this portion is

dI = dm (r sin A )²or I = _{[0 ]}

^{[ R]} dm (r sin A )²or I = _{[0 ]}

^{[ R]} M (2

r dr) (r sin A )²/

R²

or I = _{[0 ]}

^{[ R]} 2M r³ sin²A / R² . dr

or I = 2Msin²A /R² ₀

^{[ R]} r³. dr

or I = MR²sin²A / 2

gorakavipraveen

edison, can Mi be 0. Please answere me sir.

edison

Dear gorakavipraveen,

For distributed mass systems like disc, cylinder, sphere, rod etc moment of inertia can never be ZERO.

shakirshafi12

in rod(thin) M.O.I can be zero about the axis parallel to the rod and passing through C.O.M

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