The beam has $7$ holes as shown and the diameter of each hole is $\frac{5}{8^{\text{'}\text{'}}}.$ With the beam in the upright position

$($holes on top$),$ each hole serves as a pivot.

Weight of beam: $,29\mathrm{.}4oz$

Length of beam: $,{36}^{\text{'}\text{'}}($tip to tip$)$

Thickness of beam: $\frac{1}{5^{\text{'}\text{'}}}$

Width of beam: $1^{?}$

REPORT DUE:

Submit next class.

Missing the deadline means your report will not be graded, therefore, it is a

zero. So$,$ do not miss the deadline!

Moment of inertia of a body at its mass center, $G,$ is ${\displaystyle \underset{\text{b}\text{o}\text{d}\text{y}}{\overset{\phantom{}}{}}}{r}^{2}dm.$

For a uniform rectangular parallelepiped of length, $l,$

width, $w,$ and mass, $m,$ the moment of inertia at the center

of mass is $\frac{1}{12}m(l^2+w^2).$

For a uniform disc of radius, $R,$ and mass, $m,$ the moment of inertia at the mass

center for the axis perpendicular to the disc is $\frac{m{R}^2}{2}.$

For a uniform semi$-$circular disc of radius, $r,$ and mass, $m,$ the mass center

is located at $\frac{?}{\text{b}\text{a}\text{r}}(y)=\frac{4r}{3}.$ The moment of inertia at the mass center, $I_0,$

for the axis perpendicular to the disc is

$m{r}^2(\frac{1}{2}-\frac{16}{9{}^2})=\frac{m{r}^2}{18{}^2}(9{}^2-32).$