Q. In the triangular sheet given PQ=QR=`l` . If M is the mass of the sheet.what is the moment of inertia about PR?A) `(Ml^2)/24` B) `(Ml^2)/12`…

Q. In the triangular sheet given PQ=QR=`l` . If M is the mass of the sheet.what is the moment of inertia about PR?A) `(Ml^2)/24` B) `(Ml^2)/12`…

Q. In the triangular sheet given PQ=QR=`l` . If M is the mass of the sheet.what is the moment of inertia about PR?A) `(Ml^2)/24` B) `(Ml^2)/12`…

Q. In the triangular sheet given PQ=QR=`l` . If M is the mass of the sheet.what is the moment of inertia about PR?A) `(Ml^2)/24` B) `(Ml^2)/12`…
For a any plane figure having x and y axis on the figure plane and z axis perpendicular to the figure plane (see figure attached) it can be shown that`I_x =I_y =(1/2)*I_z` regardless where is located the intersection of the x,y and z axis, and regardless how x and y axis are rotated in the plane of the figure (see the figure).For a thin square sheet having sides of length l, the momentum of inertia with respect to a perpendicular axis through its center is`I_z =(M*l^2)/6` Therefore, relative to the x and y axis parallel to square sides in the figure plane we have`I_x =I_y = (M*l^2)/12` .Since the diagonals in a square are also perpendicular to each other than we can rotate x and y axis to become x’ and y’ axis and the momentum above stay the same.`I_(x’) =I_(y’) =(M*l^2)/12` Now the momentum of the right triangle PQR in the figure with respect to the axis PR is just half of the above momentum (from symmetry considerations).`I_(“triangle”) =(1/2)*I_(y’) =(M*l^2)/24` The correct answer is A) `(M*l^2)/24` Images: This image has been Flagged as inappropriate Click to unflag