Moment of inertia for a rectangular plate

In summary, the moment of inertia for a rectangular plate, area of ab, is given by (1/12)(M)(a^2+b^2).
  • #1
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can anyone help me derive the moment of inertia for a rectangular plate, area of ab, (with the axis through the center)? i know it ends up being (1/12)(M)(a^2+b^2) but when i try it, my a's and b's end up canceling... its craziness. also, when i try the sphere, instead of (2/5)MR^2, i get(3/5)MR^2...
thanks...
 
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  • #2
It's hard to say where you went wrong since you don't tell us what you did but this is how I would do that problem:
The moment of inertia about a lamina (two-dimensional figure) with area density ρ is

[itex]\int\int\rho r^2 dA[/itex] where r is the distance from each point to the axis of rotation and dA is the differential of area.

In this problem, take the center of the rectangle to be at (0,0) so the vertices are (a/2, b/2), etc. Then the distance from (x,y) to the axis of rotation (passing through (0,0)perpendicular to the plate) is √(x2+ y2) and the moment of inertia is
[itex]\int_{-a/2}^{a/2}\int_{-b/2}^{b/2}\rho (x^2+ y^2)dydx [/itex]
= [itex]\int_{-a/2}^{a/2}\rho(bx^2+ \frac{1}{12}b^3)dx [/itex]
= [itex]\frac{1}{12}\rho(a^3b+ ab^3)[/itex].

Since M= ab ρ the moment of inertia is
[tex]\frac{M}{12}(a^2+b^2)[/tex].
 
  • #3
Similarly, for the sphere, set up spherical coordinates so that the sphere, of radius R, has center at (0,0,0) and the axis of rotation is the z-axis. Then the distance from each point to the axis is just the projection to the xy-plane, [tex]\rho sin\phi[/tex] and differential of volume is [tex]\rho^2sin\phi d\rho d\phi d\theta[/tex] so the moment of inertia, with density [tex]\lambda[/tex] is given by
[tex]\int_0^\pi\int_0^{2\pi}\int_0^R(\lambda\rho^4sin^3\phi)d\rho d\theta d\phi[/tex]
= [tex]\lambda (\int_0^{2\pi} d\theta )(\int_0^R\rho^4d\rho)(\int_0^\pi sin^3\phi d\phi)[/tex]

The first two of those can be integrated directly and give
[tex](2\pi)(\frac{1}{5}R^5)[/tex].
To integrate the third, let [tex]u= cos\phi[/tex] and we get
[tex]\frac{4}{3}[/tex]. The moment of inertia is given by
[tex]\frac{8}{15}\pi\lambda R^5[/tex].

Since the mass is M= [tex]\lambda\frac{4}{3}R^3[/tex], the moment of inertia is [tex]\frac{2}{5}\pi R^2[/tex].
 
  • #4
thanks that helps a lot. now i know i was missing something important..
 

1. What is moment of inertia for a rectangular plate?

Moment of inertia for a rectangular plate is a physical quantity that describes the rotational inertia of an object around an axis. It is a measure of an object's resistance to changes in its rotation.

2. How is moment of inertia for a rectangular plate calculated?

The moment of inertia for a rectangular plate can be calculated by multiplying the mass of the object by the square of its distance from the axis of rotation. For a rectangular plate, this distance is typically measured from the center of the plate to the axis.

3. What factors affect the moment of inertia for a rectangular plate?

The moment of inertia for a rectangular plate is affected by the mass of the object, the shape of the plate, and the distance of the plate from the axis of rotation. A larger mass, a more elongated shape, and a greater distance from the axis will result in a higher moment of inertia.

4. Why is moment of inertia important in physics?

Moment of inertia is important in physics because it helps us understand and predict the behavior of rotating objects. It is a key factor in determining an object's angular momentum and its response to external forces.

5. How is moment of inertia used in real-world applications?

Moment of inertia is used in various real-world applications, such as designing vehicles, buildings, and machines. It is also used in sports, such as figure skating and diving, to optimize the performance of athletes. In engineering, moment of inertia is important for designing structures that can withstand rotational forces, such as bridges and cranes.

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