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

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There is a solid cylinder of radius R and mass M; now we take a disc of the same mass and same radius to compare them; Now a line is passing through the diameter of that disc; the ratio of the moment of inertia in between the solid cylinder and the disc about its diameter is:

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When a sphere of the moment of inertia $\mathrm{' I '}$ rolls down on an inclined plane then the percentage of rotational kinetic energy will be -

$40%$

$35%$

$20%$

$28%$

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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

$dI=\int (AC)^2dm$

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

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