Question

5.
From a disc of radius R, a concentric circular portion of
radius r is cut out so as to leave an annular disc of mass M.
The moment of inertia of this annular disc about the axis
perpendicular to its plane and passing through its centre of
gravity is
[MP PET 2001]
(a) 1/2 M (RP+)
(b) 1/2 M (R2-)
( 12 M (R4+r4)
(d) 1/2 M (R4_r4)​

Answer 1

The moment of inertia is 1/2 M(R^ 2  +r^2 ).

Explanation:

• The formula of moment of inertia is given as 1/2 M(R^ 2  +r^2 ).

• The moment of inertia of this annular disc about the axis perpendicular to its plane will be  1/2 M(R^ 2  +r^2 ).

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The m.i of a disc is about 2 units. its m.i. about axis through a point on its rim and in the plane of the disc is?

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Answer 2

Answer:

THE MOMENT OF INERTIA (I)  IS 1/2 M(R^2+r^2)

Explanation:

I=MR^4/R^2-r^2/2 - M r^4/R^2-r^2/2

I=1/2M(R^4-r^4)/(R^2-r^2)

I =1/2 M (R^2+r^2)