Figure$\mathbf{10}\mathbf{-}\mathbf{48}\mathbf{}$ shows a rigid assembly of a thin hoop (of mass m and radius$\mathbf{\text{R}}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{150}\mathbf{}\mathbf{}\mathbf{\text{m}}$) and a thin radial rod (of mass m and length$\mathbf{\text{L}}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{R}}$). The assembly is upright, but if we give it a slight nudge, it will rotate around a horizontal axis in the plane of the rod and hoop, through the lower end of the rod. Assuming that the energy given to the assembly in such a nudge is negligible, what would be the assembly’s angular speed about the rotation axis when it passes through the upside-down (inverted) orientation?

1. Radius of a thin hoop is$R=0.150\text{m}$
2. Mass of a thin hoop and a thin radial rod is m.
3. Length of the rod is,$\text{L}=2\text{R}=2\left(0.150\right)=0.3\text{m}.$

We can find the rotational inertia of the hoop and the rod about thehorizontal axis using the parallel axis theorem. Then using the formula for position of the COM and motion of an assembly we can find change in the position of the center of mass of the assembly. Then using this and the given values in the equation for law of conservation of energy we can find angular speed of the assembly about the rotation axis when it passes through the upside-down orientation

Formulae:

$\begin{array}{rcl}\text{I}& =& {\text{I}}_{com}+{\text{mR}}^2\\ {\text{E}}_i& =& {\text{E}}_f\\ {\text{y}}_{com}& =& \frac{{\text{m}}_1{\text{y}}_1+{\text{m}}_2{\text{y}}_2}{{\text{m}}_1+{\text{m}}_2}\\ & & \end{array}$

According to the parallel axis theorem,

$\text{I}={\text{I}}_{com}+{\text{mR}}^2$

In this case, The M.I of the hoop about the horizontal axis is,

${\text{I}}_{hoop}=\frac{1}{2}{\text{mR}}^2+\text{m}{\left(\text{R}+\text{L}\right)}^2$

The M.I of the rod about the horizontal axis is,

${\text{I}}_{rod}=\frac{1}{12}{\text{mL}}^2+\text{m}{\left(\frac{\text{L}}{2}\right)}^2$

Total M.I of the assembly is,

$\begin{array}{rcl}\text{I}& =& {\text{I}}_{hoop}+{\text{I}}_{rod}\\ & \Rightarrow & \text{I}=\frac{1}{2}{\text{mR}}^2+\text{m}{\left(\text{R}+2\text{R}\right)}^2+\frac{1}{12}\text{m}{\left(2\text{R}\right)}^2+\text{m}{\left(\frac{\text{2R}}{\text{2}}\right)}^{\text{2}}\\ & \Rightarrow & \text{I}=\frac{1}{2}{\text{mR}}^2+9{\text{mR}}^2+\frac{1}{3}{\text{mR}}^2+{\text{mR}}^2\\ & \Rightarrow & \text{I}=\left(10.833{\text{mR}}^2\right){\text{kgm}}^2\\ & & \end{array}$

The position of the center of mass of the assembly is,

${\text{y}}_{com}=\frac{{\text{m}}_{hoop}{\text{y}}_{hoop}+{\text{m}}_{rod}{\text{y}}_{rod}}{{\text{m}}_{hoop}+{\text{m}}_{rod}}$

$$\begin{gathered} \Rightarrow {\text{y}}_{com}=\frac{\text{m}\left(\text{R}+\text{L}\right)+\frac{\text{mL}}{2}}{2\text{m}} \\ \Rightarrow {\text{y}}_{com}=\frac{\text{m}\left(\text{R}+2\text{R}\right)+\text{m}\left(\text{R}\right)}{2\text{m}} \\ \Rightarrow {\text{y}}_{com}=2\text{R} \end{gathered}$$

After rotation, the change in the position of the center of mass of the assembly is,

$\Delta y_{com}=4R$

The assembly is initially.Soits initial K.E is zero.

According to the conservation of energy,

$$\begin{gathered} {\text{E}}_i={\text{E}}_f \\ {\text{K.E}}_i+{\text{P.E}}_i={\text{K.E}}_f+{\text{P.E}}_f \end{gathered}$$

In this case,

$$\begin{gathered} 0+2\text{mg}\left(\Delta y_{com}\right)=\frac{1}{2}\text{I}{\omega }^2+0 \\ \Rightarrow 2\text{mg}\left(\Delta y_{com}\right)=\frac{1}{2}10.833{\text{mR}}^2{\omega }^2 \\ \therefore {\omega }^2=\frac{4\text{g}\left(\Delta {\text{y}}_{com}\right)}{10.833{\text{R}}^2} \\ \Rightarrow \omega =\sqrt{\frac{4(\left(9.8\right)\left(4\right)\left(0.150\right)}{\left(10.833\right){\left(0.150\right)}^2}} \\ \Rightarrow \omega =9.823~9.82\text{rad}/\text{s} \end{gathered}$$