The moment of inertia of a rectangular plate | vedclass

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(C) Let the mass density be $ho 	ext{ kg/m}^2$.Let the sides be $PQ = RS = a$ and $QR = PS = b$. Assume $a > b$.The mass of the rectangle is $M = \

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about $P > I/2$

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(C) Let the mass density be $ho 	ext{ kg/m}^2$.Let the sides be $PQ = RS = a$ and $QR = PS = b$. Assume $a > b$.The mass of the rectangle is $M = ho ab$.The moment of inertia of the rectangle about its center is $I_{cm} = M(a^2 + b^2)/12$.The distance of point $P$ from the center is $\sqrt{(a/2)^2 + (b/2)^2} = \sqrt{a^2 + b^2}/2$.Using the parallel axis theorem,the moment of inertia of the rectangle about $P$ is $I = I_{cm} + M(a^2 + b^2)/4 = M(a^2 + b^2)/3$.The mass of the triangle $PQR$ is $M' = M/2 = ho ab/2$.The moment of inertia of the triangle $PQR$ about an axis through its centroid $G$ perpendicular to the plate is $I_G = M'(a^2 + b^2)/18$.The distance of point $P$ from the centroid $G$ is $d^2 = (2a/3)^2 + (b/3)^2 = (4a^2 + b^2)/9$.Using the parallel axis theorem,the moment of inertia of the triangle $PQR$ about $P$ is $I_P = I_G + M'd^2 = (M/2)(a^2 + b^2)/18 + (M/2)(4a^2 + b^2)/9 = (M/2)(a^2 + b^2 + 8a^2 + 2b^2)/18 = (M/2)(9a^2 + 3b^2)/18 = (M/2)(3a^2 + b^2)/6$.Comparing $I_P$ with $I/2 = (M/2)(a^2 + b^2)/3 = (M/2)(2a^2 + 2b^2)/6$,since $a > b$,$3a^2 + b^2 > 2a^2 + 2b^2$,therefore $I_P > I/2$.

Column-$I$	Column-$II$
$(1)$ $SI$ unit of torque	$(a)$ $m$
$(2)$ $SI$ unit of radius of gyration	$(b)$ $N\,m$
	$(c)$ $Js^{-2}$

The moment of inertia of a rectangular plate about an axis passing through $P$ and perpendicular to the plate is $I$. Then,the moment of inertia of the triangular portion $PQR$ about an axis perpendicular to the plane of the plate and passing through $P$ is:

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Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ $SI$ unit of torque $(a)$ $m$

$(2)$ $SI$ unit of radius of gyration $(b)$ $N\,m$

$(c)$ $Js^{-2}$

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