Question

For the rectangle below, determine the CG and I xx using the 4 small rectangle sections and the coordinate system as a frame of reference.

1 1
↔ ↔
y ↑ _____________________
│ │ │ ↑
│Ractangle 1│ Ractangle 2│ 2
│_________ │__________│ ↓
│ │ │ ↑
│Ractangle 3│Ractangle 4 │ 2
│_________ │__________│ ↓ →x

Answer 1

The center of gravity for the symmetrical rectangle is at (0.5, 1). The moment of inertia
`(\(I_(xx)\))` about the x-axis, calculated from four small rectangles, is
`\((2)/(3)\)`.

To determine the center of gravity (CG) and the moment of inertia
`(\(I_(xx)\))` for the given rectangle, we can use the properties of symmetry.

1. Center of Gravity (CG):

Since the rectangle is symmetrical, the center of gravity lies at the intersection of its diagonals. The center coordinates are
`\((\bar{x}, \bar{y}) = (0.5, 1)\)` in the chosen coordinate system.

2. Moment of Inertia
`(\(I_(xx)\))`:

The moment of inertia for each small rectangle with respect to the x-axis is given by
`\((1)/(12)mb^2\)`, where m is the mass and b is the breadth.

`\[ I_(xx1) = (1)/(12) \cdot 2 \cdot 1^2 = (1)/(6) \] \[ I_(xx2) = (1)/(12) \cdot 2 \cdot 1^2 = (1)/(6) \] \[ I_(xx3) = (1)/(12) \cdot 2 \cdot 1^2 = (1)/(6) \] \[ I_(xx4) = (1)/(12) \cdot 2 \cdot 1^2 = (1)/(6) \]`

The total moment of inertia
`(\(I_(xx)\))` is the sum of the individual moments of inertia.

`\[ I_(xx) = I_(xx1) + I_(xx2) + I_(xx3) + I_(xx4) \] \[ I_(xx) = (1)/(6) + (1)/(6) + (1)/(6) + (1)/(6) \] \[ I_(xx) = (2)/(3) \]`

In summary, the center of gravity is at
`\((0.5, 1)\)`, and the moment of inertia
`(\(I_(xx)\))` about the x-axis is
`\((2)/(3)\)` for the given rectangle.

The complete question is:

(attached)