Answers (1)

Moment of inertia for continuous body -

I= \int r^{2}dm

- wherein

r is perpedicular distance of a particle of mass dm of rigid body from axis of rotation

 

 taking a strip of radius x  and thickness dx 

MI of ring dI = dmx^2

I = \int dI= \int_{R_1}^{R_2}\left [ \frac{ M}{\pi (R_{2}^{2}-R_{1}^{2})} 2 \pi x\times dx \right ]x^2\\\\I = \int dI = \frac{ M}{\pi (R_{2}^{2}-R_{1}^{2})} 2 \pi\left [ \frac{x^4}{4} \right ]_{R_1}^{R_2}\Rightarrow I = \frac{M}{2}(R_{1}^{2}+R_{2}^{2})

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