We get the moment of inertia from the principle of summing up the moment of inertia of infinitesimal mass elements.

Let the infinitesimal mass element be **dm**.

The moment of inertia of this mass element is **dm•r ^{2}**

Here, **r ** is the distance from the mass element to the axis about where we're considering the moment of inertia.

Therefore, the moment of inertia of the door on its hinges is given by:

$\overline{)\begin{array}{rcl}{\mathbf{I}}& {\mathbf{=}}& \mathbf{\int}{\mathbf{r}}^{\mathbf{2}}\mathbf{d}\mathbf{m}\end{array}}$

The horizontal position with respect to the hinge can be taken to be **x**.

We can also define a linear density** λ = m/L** where m is the mass of the door and L is the width of the door.

We see that dm = λdx

A uniform, thin, solid door has height 2.20 m, width 0.870 m, and mass 23.0 kg.

(a) Find its moment of inertia for rotation on its hinges.

(b) Is any piece of data unnecessary?

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