Question

Figure $11-45$ shows a rigid structure consisting of a circular hoop of radius $R$ and mass $m$ and a square made of four thin bars, each of length $R$ and mass $m$. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of $2.5s$. Assuming $R=0.50m$ and $m=2.0kg$, calculate the structure rotational inertia about the axis of rotation.

Answer 1

For the hoop, we use the parallel axis theorem to obtain
$I_1=I_{com}+m{h}^2=\frac{1}{2}m{R}^2+m{R}^2=\frac{3}{2}m{R}^2$
Of the thin bars (in the form of a square) the member along the rotation axis has (approximately) no rotational inertia about that axis (since it is thin), and the member farthest from it is very much like it (by being parallel to it) except that it is displaced by a distance $h$; It has rotational inertia given by the parallel axis theorem:
$I_2=I_{com}+m{h}^2=0+m{R}^2=m{R}^2$
Now the two members of the square perpendicular to the axis have the same rotational inertia (that is $I_3=I_4$). We find $I_3$ using Table $10-2\left(e\right)$ and the parallel axis theorem:
$I_3=I_{com}+m{h}^2=\frac{1}{12}m{R}^2+m{\left(\frac{R}{2}\right)}^2=\frac{1}{3}m{R}^2$
Therefore, the total rational inertia is
$I_1+I_2+I_3+I_4=\frac{19}{6}m{R}^2=1.6kg.m^2$