Respuesta :
Answer:
[tex]I=8.8209\ kg.m^2[/tex]
[tex]\alpha=0.6348\ rad.s^{-2}[/tex]
Explanation:
Given:
- width of door, [tex]w=0.99\ m[/tex]
- height of the door, [tex]h=2.2\ m[/tex]
- thickness of the door, [tex]t=4.2\ cm[/tex]
- mass of the door, [tex]m=27\ kg[/tex]
- torque on the door, [tex]\tau=5.6\ N.m[/tex]
∵Since the thickness of the door is very less as compared to its other dimensions, therefore we treat it as a rectangular sheet.
- For a rectangular sheet we have the mass moment of inertia inertia as:
[tex]I=\frac{1}{3} m.w^2[/tex]
[tex]I=\frac{1}{3}\times 27\times 0.99^2[/tex]
[tex]I=8.8209\ kg.m^2[/tex]
We have a relation between mass moment of inertia, torque and angular acceleration as:
[tex]\alpha=\frac{\tau}{I}[/tex]
[tex]\alpha=\frac{5.6}{8.8209}[/tex]
[tex]\alpha=0.6348\ rad.s^{-2}[/tex]
The moment of inertia of the door is [tex]8.8209\ kg.m^2[/tex]
The angular acceleration immediately afterward is [tex]0.6348\ rad.s^{-2}[/tex]
Calculation of moment of inertia & the angular acceleration:
(a) The moment of inertia should be
[tex]= 1\div 3 m.w^2\\\\ = 1\div 3 \times 27 \times 0.99^2\\\\= 8.8209 \ kg.m^2[/tex]
(b) The angular acceleration should be
[tex]= 5.6 \div 8.8209\\\\= 0.6348\ rad.s^{-2}[/tex]
Learn more about the acceleration here: https://brainly.com/question/24815335
