The radius of gyration of a disc of mass 100, | vedclass

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Question

(A) 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

$3.54$

Vedclass · Answer

(A) 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 definition,the radius of gyration $(K)$ is related to the moment of inertia by $I = MK^2$.Equating the two expressions: $MK^2 = \frac{1}{2}MR^2$.Canceling mass $(M)$ from both sides: $K^2 = \frac{R^2}{2}$.Taking the square root: $K = \frac{R}{\sqrt{2}}$.Given the radius $R = 5\,cm$,we have $K = \frac{5}{\sqrt{2}} \approx \frac{5}{1.414} \approx 3.535\,cm$.Rounding to two decimal places,we get $K = 3.54\,cm$.

The radius of gyration of a disc of mass $100\,g$ and radius $5\,cm$,about an axis passing through its centre of gravity and perpendicular to the plane,is ....... $cm$.

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