91影视

diameter of cylinder $d=5m$

mass $m=30kg$

static coefficient of friction${渭}_{s,min}=0.60;{渭}_{s,max}=1.0$

kinetic coefficient of friction${渭}_{k,min}=0.40;{渭}_{k,max}=0.70$

The force of static friction must be no lower than the weight of the person.

The normal force that cylinder wall applies to person is equal to the centripetal force and it is given by, $N=\frac{m{V}^2}{r}$

The friction force that keeps the person from sliding down is given by,$f_f={渭}_{s}N$

The weight of the person is given by $F_G=mg$

When the net force is zero, we got,

role="math" localid="1649596502522" $$\begin{gathered} f_f=F_G \\ 鈬�{渭}_{s}N=mg \\ 鈬�{渭}_s\frac{m{V}^2}{r}=mg \\ 鈬�V=\sqrt{\frac{rg}{{渭}_s}}\left(I\right) \end{gathered}$$

Use the lowest coefficient of static friction in the above equation $\left(I\right)$, to get the minimum speed.

$V_{min}=\sqrt{\frac{rg}{{渭}_{s,min}}}=\sqrt{\frac{2.5脳9.81}{0.60}}=6.39m/s$

The minimum angular speed is given by,${蝇}_{min}=\frac{V_{min}}{r}=\frac{6.39}{2.5}=2.6rad/s=2.6脳9.5493\left(rpm\right)=24rpm$