In a physics lab, Asha is given a 10.7 kg uniform rectangular plate with edge lengths 67.3 cm by 53.5 cm . Her lab instructor requires her to rotate the plate about an axis perpendicular to its plane and passing through one of its corners, and then prepare a report on the project. For her report, Asha needs the plate's moment of inertia ???? with respect to given rotation axis. Calculate ???? .

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rowboatontario rowboatontario · Answer 1 · 2020-04-17T09:32:29+02:00

Answer:

[tex]I=2.6363\ kg.m^2[/tex]

Explanation:

Given:

dimension of uniform plate, [tex](0.673\times 0.535)\ m^2[/tex]

mass of plate, [tex]m=10.7\ kg[/tex]

Now we find the moment of inertia about the center of mass of the rectangular plate is given as:

[tex]I_{cm}=\frac{1}{12} \times m(L^2+B^2)[/tex]

where:

[tex]L=[/tex] length of the plate

[tex]B=[/tex] breadth of the plate

[tex]I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)[/tex]

[tex]I_{cm}=0.6591\ kg.m^2[/tex]

We know that the center of mass of the rectangular plane is at its geometric center which is parallel to the desired axis XX' .

Now we find the distance between the center of mass and the corner:

[tex]s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}[/tex]

[tex]s=0.4299\ m[/tex]

Now using parallel axis theorem:

[tex]I=I_{cm}+m.s^2[/tex]

[tex]I=0.6591+10.7\times 0.4299^2[/tex]

[tex]I=2.6363\ kg.m^2[/tex]