91Ó°ÊÓ

The ratio of the maximum static friction force (F) to the normal (N) force is known as coefficient of static friction

$$\begin{gathered} \mathrm{Radius}\mathrm{of}\mathrm{turntable},{\mathrm{r}}_0=6\mathrm{m} \\ \mathrm{The}\mathrm{period}\mathrm{of}\mathrm{rotation}\mathrm{of}\mathrm{the}\mathrm{turntable},T_0=8\mathrm{s} \\ {\mathrm{Ó\neg }}_0=\frac{2\mathrm{Ï€}}{8} \\ \mathrm{Moment}\mathrm{of}\mathrm{inertia}\mathrm{of}\mathrm{the}\mathrm{turtable},I_t=1200\mathrm{kg}-{\mathrm{m}}^2 \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{person},\mathrm{m}=70\mathrm{kg} \\ \mathrm{The}\mathrm{point}\mathrm{at}\mathrm{which}\mathrm{the}\mathrm{feet}\mathrm{stars}\mathrm{to}\mathrm{slip},{\mathrm{r}}_1=3\mathrm{m} \end{gathered}$$

The moment of inertia of the person and the turntable combined, J, can be found by considering the person as a point mass.

Therefore,

At the rim,

$$\begin{gathered} J_0=I_1+m{r}^2 \\ =1200+\left(70\mathrm{x}{6}^2\right) \\ =3720\mathrm{kg}-{\mathrm{m}}^2 \end{gathered}$$

At 3 m from the center,

$$\begin{gathered} J_0=I_1+m{r}^2 \\ =1200+\left(70x{3}^2\right) \\ =1830\mathrm{kg}-{\mathrm{m}}^2 \end{gathered}$$

Angular momentum, $L=J-\mathrm{Ó\neg }$

The angular momentum is conserved,

localid="1667822502185" $$\begin{gathered} L_0=L_1 \\ {J}_0-{\mathrm{Ó\neg }}_0=J_1-{\mathrm{Ó\neg }}_1 \\ {\mathrm{Ó\neg }}_1=\frac{J_0-{\mathrm{Ó\neg }}_0}{J_1} \\ \mathrm{At}\mathrm{equilibrium},\mathrm{the}\mathrm{friction}\mathrm{force},F_f=\mathrm{the}\mathrm{centrifugal}\mathrm{force}{F}_c \\ {F}_f={\mathrm{μ}}_{\mathrm{s}}W={\mathrm{μ}}_{\mathrm{s}}.\mathrm{m}.\mathrm{g} \\ {F}_c=\frac{m{v}^2}{r} \\ =\frac{m{(\mathrm{Ó\neg }.r)}^2}{r} \\ =m.r.{\mathrm{Ó\neg }}^2 \end{gathered}$$

Where,

$$\begin{gathered} {\mathrm{μ}}_{\mathrm{s}}=\mathrm{coefficient}\mathrm{of}\mathrm{static}\mathrm{friction} \\ \mathrm{g}=\mathrm{acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity}\left(9.81\mathrm{m}/{\mathrm{s}}^2\right) \\ \mathrm{r}=\mathrm{radius}\mathrm{of}\mathrm{gyration} \\ \mathrm{Ó\neg }=\mathrm{angular}\mathrm{speed} \\ {F}_f=F_c \\ {F}_f={\mathrm{μ}}_{\mathrm{s}}·\mathrm{m}·\mathrm{g}={\mathrm{mrÓ\neg }}^2 \end{gathered}$$

At 3 m from the center of the turntable,

$$\begin{gathered} F_f={\mathrm{μ}}_{\mathrm{s}}mg=m{r}_1{\mathrm{Ó\neg }}_1^2 \\ {\mathrm{μ}}_{\mathrm{s}}\mathrm{g}=r_1{\mathrm{Ó\neg }}_1^2 \\ {\mathrm{μ}}_{\mathrm{s}}=\frac{{\mathrm{r}}_1{\mathrm{Ó\neg }}_1^2}{\mathrm{g}} \\ =\frac{{\mathrm{r}}_1{\left({\displaystyle \frac{j_{\mathit{0}}{\mathrm{Ó\neg }}_{\mathit{0}}}{j_1}}\right)}^2}{\mathrm{g}} \\ =\frac{3{\left({\displaystyle \frac{3720\mathrm{x}{\displaystyle \frac{2\mathrm{Ï€}}{8}}}{1830}}\right)}^2}{9.81} \\ =0.779 \end{gathered}$$

Hence, the coefficient of static friction is 0.779