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?
Best Answer
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.