A 55-kg runner runs around the edge of a horizontal turntable mounted on a vertical, frictionless axis through its center. The runner’s velocity relative to the earth has magnitude 2.8 m/s. The turntable is rotating in the opposite direction with an angular velocity of magnitude 0.20 rad/s relative to the earth. The radius of the turntable is 3.0 m, and its moment of inertia about the axis of rotation is80kgm². Find the final angular velocity of the system if the runner comes to rest relative to the turntable. (You can model the runner as a particle.)

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 $\omega =0.20\mathrm{rad}/\mathrm{s}$, 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 $L_i=mr{v}_r-I\omega$.

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

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

Since,$L_i=L_f$then,

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

Now, substitute all the known values and simplify.

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

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