A sheet of plywood 1.37 cm thick is used to make a cabinet door 42.1 cm wide by 77.3 cm tall, with hinges mounted on the vertical edge. A small 207-g handle is mounted 45 cm from the lower hinge at the same height as that hinge. If the density of the plywood is 550 kg/m3, what is the moment of inertia of the door about the hinges? Neglect the contribution of hinge components to the moment of inertia.

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anonymous · Answer

I will proceed to do everything in generic variables for clarity.  Plug things in at the end if you want a numeric answer.  Let the width of the door be $w$, height (tallness) be $h$, and the thickness be $t$.  Let the distance from the hinge of the handle be $R$ and the mass of that handle be $M$.  Let the density of the plywood be $\delta$.

The moment of Inertia for the whole system can be express as the sum of the moments of inertia for each relevant component of the door.$$I=I_{\text{plywood}}+I_{\text{handle}}$$We will treat the handle as a point mass, so thus $I_{\text{handle}}=MR^2$.  To find $I_{plywood}$, we will need to perform an integral, since the mass elements $dm$ are different distances $r$ from the axis of rotation.  We'll simplify the situation by saying that the door is infinitely thin (won't affect the answer too much since it's relatively much wider and taller than it is thick).$$I_{\text{plywood}}=\int r^2dm$$Now, $dm$ is related to an infinitesimal volume element $dV$ by the density.  Thus $dm=\delta\ dV$.  Additionally, we can express $dV$ in terms of the variables we have already for the three dimensions for the door by $dm=\delta\ dw\ dh\ dt$.  Let us substitute this into our integral.$$I_{\text{plywood}}=\iiint r^2 \delta\ dw\ dh\ dt$$The convenient thing, however, is that the distance to the axis of rotation doesn't vary with position along the height or thickness axes (not along the thickness axis because we're approximating the door to be infinitely thin).  Thus, we can drop two integrals right off the bat.$$I_{\text{plywood}}=\delta ht\int r^2dw$$Now, integrating along the width axis is the tricky part.  Let us integrate from the hinges of the door to the outer edge.  Since this is actually the same as the way we measure $r$ we can actually do the simple substitution $dw=dr$ and set our bounds to be from $r=0$ (hinge-side of the door) to $r=w$ (outside of the door).  At this point, evaluating the integral is easy.$$I_{\text{plywood}}=\delta ht \int\limits_0^w r^2 dr=\delta ht\left.\frac{r^3}{3}\right|_{r=0}^w=\frac{\delta htw^3}{3}$$We can thus express the entire moment of inretia as follows.$$\boxed{\displaystyle I=\frac{\delta htw^3}{3}+MR^2}$$PS - you can click on any equation to enlarge it if it renders too small.