(B) The radius of gyration $k$ is given by the formula $I = Mk^2$, where $I$ is the moment of inertia and $M$ is the mass of the body.Case $1$: Axis p

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(B) The radius of gyration $k$ is given by the formula $I = Mk^2$, where $I$ is the moment of inertia and $M$ is the mass of the body.Case $1$: Axis passes through the centre and is perpendicular to the length.The moment of inertia is $I_1 = \frac{ML^2}{12}$.Thus, $Mk_1^2 = \frac{ML^2}{12} \implies k_1 = \frac{L}{\sqrt{12}} = \frac{L}{2\sqrt{3}}$.Case $2$: Axis passes through one end and is perpendicular to the length.The moment of inertia is $I_2 = \frac{ML^2}{3}$.Thus, $Mk_2^2 = \frac{ML^2}{3} \implies k_2 = \frac{L}{\sqrt{3}}$.The ratio of the radius of gyration in the first case to the second case is:$\frac{k_1}{k_2} = \frac{L / (2\sqrt{3})}{L / \sqrt{3}} = \frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}$.

Class 11 Physics System of Particles and Rotational Motion Moment of Inertia and Radius of gyration

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Let $M$ and $L$ be the mass and length of a thin uniform rod, respectively. In the first case, the axis of rotation passes through the centre and is perpendicular to the length of the rod. In the second case, the axis of rotation passes through one end and is perpendicular to the length of the rod. The ratio of the radius of gyration in the first case to the second case is:

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A
$1/4$
B
$1/2$
C
$1/2\sqrt{3}$
D
$1/6$

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System of Particles and Rotational Motion

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Moment of Inertia and Radius of gyration

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$A$ one-quarter sector is cut from a uniform circular disc of radius $R$. This sector has mass $M$. It is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

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