Question

find the moment of inertia of the ring shown in the figure​

Answer 1

Answer:

Mass of the Ring=M

Mass per unit length of ring λ=

2πR

m

Consider a small segment dl

Mass of dl=λdl

Moment of inertia of this segment about axis of rotation.

dI=(λdl)R

2

Moment of inertia of the total ring is

I=∫dI

⇒=∫λdlR

2

=λR

2

∫dl

⇒I=2πR.λR

2

⇒I=2πR.

2πR

M

R

2

⇒I=MR

2

Answer 2

Answer:

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