91Ó°ÊÓ

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,

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

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

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

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

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

$\mathrm{Δ}{y}_{com}=4R$

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,

$$\begin{gathered} 0+2\text{mg}\left(\mathrm{Δ}{y}_{com}\right)=\frac{1}{2}\text{I}{\mathrm{Ó\neg }}^2+0 \\ ⇒2\text{mg}\left(\mathrm{Δ}{y}_{com}\right)=\frac{1}{2}10.833{\text{mR}}^2{\mathrm{Ó\neg }}^2 \\ ∴{\mathrm{Ó\neg }}^2=\frac{4\text{g}\left(\mathrm{Δ}{\text{y}}_{com}\right)}{10.833{\text{R}}^2} \\ ⇒\mathrm{Ó\neg }=\sqrt{\frac{4(\left(9.8\right)\left(4\right)\left(0.150\right)}{\left(10.833\right){\left(0.150\right)}^2}} \\ ⇒\mathrm{Ó\neg }=9.823~9.82\text{rad}/\text{s} \end{gathered}$$