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Moment Of Inertia Of A Solid Sphere - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Moment of inertia of a SOLID SPHERE is considered one the most difficult concept.

  • 6 Questions around this concept.

Concepts Covered - 1

Moment of inertia of a SOLID SPHERE

Let I=Moment of inertia of a SOLID SPHERE about an axis through its centre

To calculate I

Consider a sphere of mass M, radius R and centre O. Mass per unit volume of the sphere =  \rho =\frac{M}{\frac{4}{3}\pi R^3}

 

              

Take an elementary disc of mass dm, whose centre is C and which lies between two planes perpendicular to the axis at a distance x and x + dx from its centre

Disc has Radius=AC and thickness =dx.  As shown in figure

So,  AC = \sqrt{R^2 - x^2}

So,  dV=\pi (R^2-x^2)dx 

dm=\rho dV=\frac{3M}{4\pi R^3}*\pi (R^2-x^2)dx=\frac{3M}{4 R^3}*(R^2-x^2)dx

The moment of inertia of the elementary  disc about the axis

dI=\int \frac{(AC)^2dm}{2} \ \ \ \ \ \ \ \ \left (\because I_{disc}=\frac{mR^2}{2} \right )

 Now integrate this dI  between the limits x=-R to x=+R

 

\\ I=\int dI=\int \frac{(AC)^2dm}{2} \\ \\ \\ =\int_{-R}^{R}(R^2-x^2)\frac{3M}{8R^3}(R^2-x^2)dx \\ \\ \\ =\int_{-R}^{R} \frac{3M}{8R^3}(R^2-x^2)^2dx \\ \\ \\ =\frac{3M}{8R^3}\int_{-R}^{R}(R^4-2R^2x^2+x^4)dx \\ \\ \\ \Rightarrow \mathbf{I=\frac{2}{5}MR^2}

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Moment of inertia of a SOLID SPHERE

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