Question

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 ???? .

Answer 1

Answer:

$I=2.6363\ kg.m^2$

Explanation:

Given:

dimension of uniform plate, $(0.673\times 0.535)\ m^2$

mass of plate, $m=10.7\ kg$

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

$I_{cm}=\frac{1}{12} \times m(L^2+B^2)$

where:

$L=$ length of the plate

$B=$ breadth of the plate

$I_{cm}=\frac{1}{12} \times 10.7\times(0.673^2+0.535^2)$

$I_{cm}=0.6591\ kg.m^2$

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:

$s=\frac{\sqrt{ (0.673^2+0.535^2)}}{2}$

$s=0.4299\ m$

Now using parallel axis theorem:

$I=I_{cm}+m.s^2$

$I=0.6591+10.7\times 0.4299^2$

$I=2.6363\ kg.m^2$