Question

Three rings each of mass M and radius R are arranged as shown. The moment of inertia of the system about YY′ will be; (a)3MR2(b)32MR2(c)5MR2(d)72MR2

Answer 1

Answer:

Explanation:

Here is the question

Answer 2

Answer:

Moment of inertia of ring 1 with respect to given axis be parallel axis theorem I.e.

MR^2/2

Moment of inertia for ring 2 and 3 with respect to the axis be parallel axis theorem I.e.

MR^2/2 + MR^2 = 3MR^2/2

Therefore moment of inertia of combined system I will be

I = I1 + I2 + I3

I = MR^2/2 + 3MR^2/2 + 3MR^2/2

I = 7MR^2/2

And the option is (d)

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