Question

A 25 kg solid door is 220 cm tall, 91 cm wide. What is the door's moment of inertia for (a) rotation on its hinges and (b) rotation about a vertical axis inside the door, 15 cm from one edge?

Short answer

The moment of inertia about the edge of slab is $6.9\text{kg}{\text{m}}^2$

The moment of inertia about 15 cm from the edge of slab is$4.1\text{kg}{\text{m}}^2$

Step-by-step solution

Step 1. Given information

A mass of solid door $m=25\text{kg}$, height of door $a=220\text{cm}$, , and $91\text{cm}$ wide.

Part (a)

The door is a slab of uniform density.

The hinges are at the edge of the door, So moment of inertia about the edge of slab is

$\begin{array}{l}I=\frac{1}{3}M{a}^2\\ I=\frac{1}{3}\left(25\text{kg}\right){\left(0.91\text{m}\right)}^2\\ =6.9\text{kg}{\text{m}}^2\end{array}$

The moment of inertia about the edge of slab is$6.9\text{kg}{\text{m}}^2$

Part (b)

The distance from the axis through the center of mass along the height of the door is

$d=\left(\frac{0.91\text{m}}{2}-0.15\text{m}\right)=0.305\text{m}$

. Using the parallel-axis theorem,

The moment of inertia about 15 cm from the edge of slab is

$\begin{array}{l}I=I_{\text{cm}}+M{d}^2\\ =\frac{1}{12}\left(25\text{kg}\right){\left(0.91\text{m}\right)}^2+\left(25\text{kg}\right){\left(0.305\text{cm}\right)}^2\\ =4.1\text{kg}{\text{m}}^2\end{array}$