(C) The moment of inertia of a solid sphere about its diameter is given by $I = \frac{2}{5}MR^2$.When the sphere is recast into a disc of mass $M$ and

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(C) The moment of inertia of a solid sphere about its diameter is given by $I = \frac{2}{5}MR^2$.When the sphere is recast into a disc of mass $M$ and radius $r$,the moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is given by the parallel axis theorem: $I_{edge} = I_{cm} + Mr^2 = \frac{1}{2}Mr^2 + Mr^2 = \frac{3}{2}Mr^2$.Since the moment of inertia remains the same,we equate the two expressions:$\frac{3}{2}Mr^2 = \frac{2}{5}MR^2$.Dividing both sides by $M$ and solving for $r$:$r^2 = \frac{2}{5} \times \frac{2}{3} R^2 = \frac{4}{15} R^2$.Taking the square root of both sides:$r = \frac{2R}{\sqrt{15}}$.

Class 11 Physics System of Particles and Rotational Motion Mix Example - System of Particles and Rotational Motion

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$A$ solid sphere of mass $M$ and radius $R$ has a moment of inertia $I$ about its diameter. It is recast into a disc of thickness $t$ whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$. The radius of the disc will be:

MHT CET 2020 All MHT CET

A
$R/\sqrt{19}$
B
$R/\sqrt{15}$
C
$2R/\sqrt{15}$
D
$2R/\sqrt{19}$

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Mix Example - System of Particles and Rotational Motion

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$(a)$ $h = \frac{R}{2}$	$(i)$ Sphere rolls without slipping with a constant velocity and no loss of energy.
$(b)$ $h = R$	$(ii)$ Sphere spins clockwise,loses energy by friction.
$(c)$ $h = \frac{3R}{2}$	$(iii)$ Sphere spins anti-clockwise,loses energy by friction.
$(d)$ $h = \frac{7R}{5}$	$(iv)$ Sphere has only a translational motion,loses energy by friction.

$A$ solid sphere of mass $M$ and radius $R$ has a moment of inertia $I$ about its diameter. It is recast into a disc of thickness $t$ whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$. The radius of the disc will be:

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In the figure shown,the plank is being pulled to the right with a constant speed $v$. If the cylinder does not slip,then:

$A$ metal sphere of radius $r$ and specific heat $S$ is rotated about an axis passing through its centre at a speed of $f$ rotations per second. It is suddenly stopped,and $50\%$ of its rotational kinetic energy is used in increasing its temperature. The rise in temperature of the sphere is

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