Respuesta :
Answer
given,
mass of the solid door = 22 Kg
dimension of door = 220 cm x 91 cm
moment of inertia about the hinge
[tex]I = \dfrac{1}{3}Mr^2[/tex]
r is the distance from the one edge which is equal to 91 cm or 0.91 m
[tex]I = \dfrac{1}{3}\times 22 \times 0.91^2[/tex]
[tex]I = 6.073\ kg m^2[/tex]
Moment of inertia about center for rectangular gate is equal to
[tex]I_{CM} = \dfrac{1}{12}Mr^2[/tex]
moment of inertia for rotation about a vertical axis inside the door, 15 cm from one edge
[tex]I = I_{CM} + MR^2[/tex]
[tex]I = I_{CM} + M(\dfrac{91}{2}- 15)^2[/tex]
[tex]I = \dfrac{1}{12}Mr^2+ M(0.0930)[/tex]
[tex]I = \dfrac{1}{12}\times 22 \times 0.91^2+ 22 \times (0.093)[/tex]
[tex]I = 3.56\ Kg m^2[/tex]
The door's moment of inertia for rotation about a vertical axis inside the door, 15 cm from one edge is; I = 3.5648 kg.m²
We are given;
Mass of door; m = 22 kg
Height of door; h = 220 cm = 2.2 m
Width of door; L = 91 cm = 0.91 m
- Using parallel axis theorem, moment of inertia for rotation about a vertical axis inside the door, 15 cm from one edge is;
I = (1/12)ML² + M(½L - L')²
Where L' = 15 cm = 0.15 m
Thus;
I = (1/12)(22 × 0.91²) + 22(½*0.91 - 0.15)²
I = 1.5182 + 2.0466
I = 3.5648 kg.m²
- In conclusion, the door's moment of inertia for rotation about a vertical axis inside the door, 15 cm from one edge is
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