91Ó°ÊÓ

It is given that the mass of the person as $m=55\mathrm{kg}$, runners speed relative to the earth as $v_f=2.8\mathrm{m}/\mathrm{s}$, angular velocity of the turntable as $\mathrm{Ó\neg }=0.20\mathrm{rad}/\mathrm{s}$, radius of the turntable as $r=3.0\mathrm{m}$ and the moment of inertia as$I=80{\mathrm{kgm}}^2$ .

By the conservation of angular momentum, $L_i=L_f$. The initial angular momentum due to the person and the turntable is $L_i=mr{v}_r-I\mathrm{Ó\neg }$.

The moment of inertia of the system is $I\text{'}=\left(m{r}^2+I\right)$ then the final angular momentum is given by,

$$\begin{gathered} L_f=I\text{'}\mathrm{Ó\neg }\text{'} \\ =\left(m{r}^2+I\right)\mathrm{Ó\neg }\text{'} \end{gathered}$$

Since,$L_i=L_f$then,

role="math" localid="1667991534420" $$\begin{gathered} mr{v}_r-I\mathrm{Ó\neg }=\left(m{r}^2+I\right)\mathrm{Ó\neg }\text{'} \\ ⇒\mathrm{Ó\neg }\text{'}=\frac{mr{v}_r-I\mathrm{Ó\neg }}{m{r}^2+I} \end{gathered}$$

Now, substitute all the known values and simplify.

$$\begin{gathered} {\mathrm{Ó\neg }}^{\mathrm{′}}=\frac{mr{v}_r−I\mathrm{Ó\neg }}{m{r}^2+I} \\ =\frac{\left(55\mathrm{kg}\right)\left(3.0\mathrm{m}\right)(2.8\mathrm{m}/\mathrm{s})−\left(80{\mathrm{kgm}}^2\right)(0.20\mathrm{rad}/\mathrm{s})}{\left(55\mathrm{kg}\right)(3.0\mathrm{m}{)}^2+80{\mathrm{kgm}}^2} \\ =\frac{462{\mathrm{kgm}}^2/\mathrm{s}−\left(80{\mathrm{kgm}}^2\right)(0.20\mathrm{rad}/\mathrm{s})}{165{\mathrm{kgm}}^2+80{\mathrm{kgm}}^2} \\ =0.776\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Therefore, the angular velocity of the given system is $\mathrm{Ó\neg }\text{'}=0.766\mathrm{rad}/{\mathrm{s}}^2$.