Respuesta :
Answer:
The moment of inertia is [tex]I=\frac{M}{12} a^{2}[/tex]
Explanation:
The moment of inertia is equal:
[tex]I=\int\limits^a_b {r^{2} } \, dm[/tex]
If r is [tex]-\frac{a}{2} <r<\frac{a}{2}[/tex]
and [tex]dm=\frac{M}{a} dr[/tex]
[tex]I=\int\limits^a_b {r^{2}\frac{M}{a} } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}[/tex]
[tex]I=\frac{M}{a} \int\limits^a_b {r^{2} } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\ I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}[/tex]
The moment of inertia of rectangular sheet about an axis that lies in the plane of the plate, passes through the center of the plate, and is parallel to the side with length b is
[tex]I = Ma^2/12[/tex]
The Moment of Inertia [tex]I[/tex] is given by equation
[tex]I = \int r^2dm[/tex]
[tex]dm = mass\; of \; the\; elementary \; particle \\\\r = distance \; of \; mass\; dm \; from \; the \; axis \\[/tex]
The mass of the rectangular sheet = M
Length of Rectangular sheet = a
Width of Rectangular sheet = b
Area of Rectangle = a [tex]\times[/tex] b
So mass per unit area = [tex]M/ab[/tex]
A figure is attached for the given situation
Mass of the strip in the attached figure = [tex]dm[/tex] = [tex](M/ab) \times )(bdx) = M/a\times dx[/tex]
The Moment of Inertia [tex]I[/tex] is given by equation
[tex]I = \int r^2dm[/tex]
[tex]I = \int_{-a/2}^{a/2} {r^2} dm[/tex]
[tex]I = \int_{-a/2}^{a/2} {x^2} (M/a) \times dx[/tex]
on Integration we get
[tex]I = Ma^2/12[/tex]
For more information please refer to the link below
https://brainly.com/question/19222814
