7. A thin ring has mass M and radius R. Its moment of
inertia about the axis passing through its center and
perpendicular to its plane is
(A) MR2
(B) MPR
(C) MIRZ
(D) MAR
Answers
answer : option (a) MR²
A thin ring has mass M and radius R . we have to find moment of inertia of the ring about an axis passing through its centre and perpendicular to its plane.
cut an element of thickness dl as shown in figure.
dm = M/2πR × dl = Mdl/2πR
we know, dl = Rdθ
so, dm = Mdθ/2π
now moment of inertia of ring , I = (dm)R²
= MR²/2π ∫dθ
= MR²/2π [θ]
= MR²/2π [π - (-π)]
= MR²/2π × 2π
= MR²
hence moment of inertia of the ring about an axis Passing through its centre and perpendicular to its plane is MR².
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