# Calculating Moment Of Inertia

The problem I am working on is:

A uniform, thin, solid door has height 2.10 m, width 0.835 m, and mass 24.0 kg.

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

(b) Is any piece of data unnecessary?

I am not really certain of what I am doing at all. So, here is my attempt, following the outline given in the book.

I know from the book that the height isn't necessary to know for this problem, but I am not really certain why. As for part (a), I don't really know how to approach the problem. Do I need calculus techniques? Is it one of the integrals that I am suppose to use?

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You're right about not needing the height. You then calculate the moment of inertia by in principle summing up the moment of inertia of infinitesimal mass elements. The moment of inertia of an infinitesimal mass element $\mathrm dm$ is $\mathrm dm\cdot r^2$, in which $r$ represents the distance from the mass element to the axis about which we are considering the moment of inertia. The moment of inertia of the door is then $\int r^2 \,\mathrm dm$.

So in this case, we can call $x$ the horizontal position with respect to the axis (the hinge) and we remark that we can define a linear density $\lambda=\frac{m}{L}$, in which $m$ is the mass and $L$ is the width of the door. Then we see $\mathrm dm=\lambda\, \mathrm dx$ as the mass $\mathrm dm$ of an infinitesimally thin vertical sliver of width $\mathrm dx$ of the door is given by the product of the linear density and width.

We get $$I_{\text{about hinge}}=\int r^2 \,\mathrm dm=\int\limits_0 ^L x^2\lambda \,\mathrm dx=\left[\frac{\lambda}{3} x^3\right]_0^L=\frac{\lambda L^3}{3}$$

But we know that $\lambda=\frac{m}{L}$ so $I_{\text{about hinge}}=\frac{mL^2}{3}$. You can then fill in the numbers you have to get the answer. Note that I didn't do this to do your homework for you, but I wanted to show the general thoughts behind calculating moment of inertia.

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Why is the height not needed, though? – Mack Nov 25 '12 at 22:14
In fact the moment of inertia is calculated as the integral of the square of the distance wrt to mass. But for every infinitesimal mass element that we consider, the distance to the axis is simply given by $x$ (by definition of distance between a line and a point as the length of a line segment perpendicular from the point to the line) so we don't need the height. I suppose you could make it a double integral involving both the width and height and an area density instead of linear density, but this should make no difference to your final answer and only leaves you with more work to do. – user16228 Nov 25 '12 at 22:26
So, there isn't really any intuitive way of seeing that the height isn't required? – Mack Nov 26 '12 at 0:19
That depends on what you call intuitive. But one way to see it is that you could, for each "sliver" I describe, compress it vertically into a single point. For each sliver all the points are still at the same distance to the axis and they still have the same mass so the moment of inertia about the given axis is the same. – user16228 Nov 26 '12 at 0:27
Well the mass is already uniformly concentrated over the width by assumption. I suppose you could look at the example I give in the comment in that way. What I was trying do with it was show that what I did in the answer was equivalent to a calculation which more obviously doesn't involve the height of the door. – user16228 Nov 26 '12 at 0:40