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  • What is the moment of inertia of a disc having inner radius  and outer radius  about the axis p

What is the moment of inertia of a disc having inner radius R_1 and outer radius R_2 about the axis passing through centre and perpendicular to the plane as shown in diagram 

Option: 1

\frac{M}{2} \left ( R_{2}^{2}- R_{1}^{2} \right )


Option: 2

\frac{M}{2}\pi \left ( R_{2}^{2}- R_{1}^{2} \right )


Option: 3

M \left ( R_{2}^{2}- R_{1}^{2} \right )


Option: 4

\frac{M}{2} \left ( R_{2}^{2}+ R_{1}^{2} \right )


Answers (1)

best_answer

As we have learned

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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