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Naddika [18.5K]
2 years ago
9

Here's some points for some hardworking people​

Physics
1 answer:
Natasha2012 [34]2 years ago
8 0
Thankyouuuuuuu........
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The moment of inertia of a thin ring of mass M and radius R about its symmetry axis is ICM = MR2.
monitta

Answer:

The moment of inertia of large ring is 2MR².

(A) is correct option.

Explanation:

Given that,

Mass of ring = M

Radius of ring = R

Moment of inertia of a thin ring = MR²

Moment of inertia :

Moment of inertia is the product of the mass of the ring and square of radius of the ring.

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Using formula of moment of inertia

I=I_{cm}+MR^2

Where, I_{cm} = moment of inertia at center of mass

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R = radius of ring

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I=MR^2+MR^2

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Hence, The moment of inertia of large ring is 2MR².

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