Question

One quarter sector is cut from a uniform circular disc of radius r this sector has mass m it is made to rotate a water line perpendicular to its plane and passing through the centre of the original disc its moment of inertia about the axis of rotation is

Answer 1

Mass of the sector of the disc is "m"

now it is given that disc is one quarter so since the disc is considered to be uniform so the mass of whole disc will be four times more than the mass of sector

So total disc mass would be 4m

now we have moment of inertia of whole disc is given as

$I = \frac{1}{2}Mr^2$

here M = total mass of the disc

r = radius of the disc

so we have

$I = \frac{1}{2}*4m*r^2$

$I = 2mr^2$

now this is moment of inertia of total disc while we need to find the moment of inertia of quarter disc

So moment of inertia of quarter disc will be one fourth of total disc

$I' = \frac{1}{4}I$

$I' = \frac{1}{4}2mr^2$

$I' = \frac{1}{2}mr^2$

so moment of inertia of the quarter disc will be given by above equation

Answer 2

Answer: {A} (1/2)MR2

one quarter sector has mass M.

hence complete disc has mass 4m.

MI of disc rotating about line perpendicular to plane and passing

through centre = [MR2 / 2]

here MI of complete disc = 4m(R2/2) = 2MR2

hence by symmetry, for quarter of disc, MI = (1/4) (2MR2)

= [(MR2) / 2]