Question

Figure shows a rigid structure consisting of a circular hoop of radius $\mathbf{R}$and mass$\mathbf{m}$, and a square made of four thin bars, each of length$\mathbf{R}$and mass$\mathbf{m}$. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of$\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{}\mathbf{s}$. Assuming$\mathbf{R}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{m}$androle="math" localid="1660971946053" $\mathbf{m}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$,

(a) Calculate the structure’s rotational inertia about the axis of rotation?

(b) Calculate its angular momentum about that axis?

Short answer

1. The total rotational inertia of the rigid structure about the axis of rotation is$1.6\text{kg}.{\text{m}}^2.$
2. The angular momentum the rigid structure is$4.0\frac{\text{kg}.{\text{m}}^2}{\text{s}}.$

Step-by-step solution

Given

1. The mass of the hoop,$\mathrm{m}=2\mathrm{kg}$ .
2. The radius of the hoop,$R=0.5m$.
3. The mass of the bar, $m=2kg$.
4. The length of the bar,$\mathrm{R}=0.5\mathrm{m}$.
5. The period of the rotation of the rigid structure, $\mathrm{T}=2.5\mathrm{s}$.

To understand the concept

Find the rotational inertia of the rigid structure by adding the rotational inertia of the circular hoop and the square about the axis of rotation. Then by using the given period of rotation of the structure, we can find its angular speed. By using the relation between $\mathbf{\text{L,Iandω}}$we can find the angular momentum of the rigid structure.

Formula:

i. The M.I of the circular hoop along the axis passing through its center is,$\mathbf{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mR}}^{\mathbf{2}}$

ii. The M.I of the bar (rod) along the axis passing through its center is,$\mathbf{I}\mathbf{=}\frac{{\mathbf{ml}}^{\mathbf{2}}}{\mathbf{12}}$

iii. Parallel axis theorem is,$\mathbf{I}\mathbf{=}{\mathbf{I}}_{\mathbf{com}}\mathbf{+}{\mathbf{mh}}^{\mathbf{2}}$

iv. The angular momentum of system is$\mathbf{L}\mathbf{=}\mathbf{I\omega }$

Step 3: Calculate the structure’s rotational inertia about the axis of rotation

(a)

The rotational inertia of the circular hoop is

$\begin{array}{c}{\text{I}}_1={\text{I}}_{\text{com}}+{\text{mh}}^2\\ \Rightarrow {\text{I}}_1=\frac{1}{2}{\text{mR}}^2+{\text{mR}}^2\\ \Rightarrow {\text{I}}_1=\frac{3}{2}{\text{mR}}^2\end{array}$

The square is made up of four bars, out of which one is along the axis of rotation. So its moment of inertia is zero.

${\text{I}}_2=0$

Therotational inertiaof the other side which is parallel to the axis of rotation is

$\begin{array}{c}\text{I}={\text{I}}_{\text{com}}+{\text{mh}}^2\\ \Rightarrow {\text{I}}_3=0+{\text{mR}}^2\\ \Rightarrow {\text{I}}_3={\text{mR}}^2\end{array}$

Therotational inertiaof the bars which are perpendicular to the axis of rotation are same and are

$\begin{array}{c}{\text{I}}_4={\text{I}}_5=\frac{1}{12}{\text{mR}}^2+\text{m}{\left(\frac{\text{R}}{2}\right)}^2\\ \Rightarrow {\text{I}}_4={\text{I}}_5=\frac{1}{3}{\text{mR}}^2\end{array}$

Therefore, the total rotational inertia of the rigid structure is

$\begin{array}{c}\text{I}={\text{I}}_1+{\text{I}}_2+{\text{I}}_3+{\text{I}}_4+{\text{I}}_5\\ \Rightarrow \text{I}=\frac{3}{2}{\text{mR}}^2+0+{\text{mR}}^2+\frac{1}{3}{\text{mR}}^2+\frac{1}{3}{\text{mR}}^2\\ \Rightarrow \mathrm{I}=\frac{19}{6}=1.66~1.6{\mathrm{kgm}}^2\end{array}$

 Calculate its angular momentum about that axis

(b)

The angular speed of the rigid structure is

$\begin{array}{c}\text{ω}=\frac{\text{dθ}}{\text{dt}}\\ \Rightarrow \text{ω}=\frac{2\text{π}}{2.5}\\ \Rightarrow \text{ω}=2.5\frac{\text{rad}}{\text{s}}\end{array}$

The angular momentum of the rigid structure is,

$\begin{array}{c}\text{L}=\text{Iω}\\ \Rightarrow \text{L}=1.6\times 2.5\\ \Rightarrow \mathrm{L}=4.0\frac{{\mathrm{kgm}}^2}{\mathrm{s}}\end{array}$