InterviewSolution
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InterviewSolution
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Demonstrate that in the case of a thin plate of arbitrary shape there is the following relationship 1, 2, and 3 define three mutually perpendicular axes passing through one point, with axes 1 and 2 lying in the plane of the plate. Using this relationship, find the moment of inertia of a thin uniform round disc of radius R and mass m relative to the axis coinciding with one of its diameters. |
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Answer» Solution :(a) Let us consider a lamina of an arbitrary shape and indicate by 1,2 and 3, three axes coinciding with X, y and z-axes and the PLANE of lamina as x-y plane. Now, moment of INERTIA of a point mass about x-axis, `dI_x=dmy^2` Thus moment of inertia of the lamina about this axis, `I_x=int dmy^2` Similary, `I_y=intdmx^2` and `I_z=int dmr^2` `=intdm(x^2+y^2)` as `R=sqrt(x^2+y^2)` Thus, `I_z=I_x+I_y` or, `I_3=I_1+I_2` (b) Let us TAKE the plane of the disc as x-y plane and origin to the centre of the disc (figure). From the symmetry `I_x=I_y`. Let us consider a ring element of radius r and thickness `dr`, then the moment of inertia of the ring element about the y-axis.  `dI_x=dmr^2=(m)/(piR^2)(2pirdr)r^2` Thus the moment of inertia of the disc about z-axis `I_x=(2m)/(R^2)underset(0)overset(R)intr^3dr=(mR^2)/(2)` But we have `I_x=I_x+I_y=2I_x` Thus `I_x=I_x/2=(mR^2)/(4)` 
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