A thin uniform rectangular plate of mass 2 , | vedclass

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(A) Given: Mass of the plate,$M = 2 \, kg$.Moment of inertia about the $x$-axis,$I_{x} = 0.2 \, kg \cdot m^{2}$.Moment of inertia about the $y$-axis,$

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) Given: Mass of the plate,$M = 2 \, kg$.Moment of inertia about the $x$-axis,$I_{x} = 0.2 \, kg \cdot m^{2}$.Moment of inertia about the $y$-axis,$I_{y} = 0.3 \, kg \cdot m^{2}$.According to the perpendicular axis theorem,the moment of inertia $I_{z}$ of a planar body about an axis perpendicular to its plane (passing through the origin $O$) is given by:$I_{z} = I_{x} + I_{y}$$I_{z} = 0.2 + 0.3 = 0.5 \, kg \cdot m^{2}$.We know that the moment of inertia is also related to the radius of gyration $k$ by the formula:$I = M k^{2}$Substituting the values:$0.5 = 2 \cdot k^{2}$$k^{2} = \frac{0.5}{2} = 0.25 \, m^{2}$$k = \sqrt{0.25} = 0.5 \, m$.Converting to centimeters:$k = 0.5 	imes 100 \, cm = 50 \, cm$.Thus,the radius of gyration is $50 \, cm$.

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