The radius of gyration of a disc of mass 50,g | vedclass

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Question

(B) The moment of inertia $(I)$ of a disc about an axis passing through its centre of gravity and perpendicular to its plane is given by $I = \frac{1}

Vedclass · Accepted Answer

$1.76$

Vedclass · Answer

(B) The moment of inertia $(I)$ of a disc about an axis passing through its centre of gravity and perpendicular to its plane is given by $I = \frac{1}{2}MR^2$.By the definition of the radius of gyration $(K)$,the moment of inertia is also expressed as $I = MK^2$.Equating the two expressions: $MK^2 = \frac{1}{2}MR^2$.Canceling $M$ from both sides,we get $K^2 = \frac{R^2}{2}$,which simplifies to $K = \frac{R}{\sqrt{2}}$.Given the radius $R = 2.5\,cm$,we calculate $K = \frac{2.5}{\sqrt{2}} \approx \frac{2.5}{1.414} \approx 1.767\,cm$.Rounding to two decimal places,we get $K = 1.76\,cm$.

The radius of gyration of a disc of mass $50\,g$ and radius $2.5\,cm$,about an axis passing through its centre of gravity and perpendicular to the plane,is ....... $cm$.

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