An insect lying on the bottom of the hemi-spherical bowl tries to come out from it. The coefficient of static friction between insect and bowl is $0.5$. How high up does the insect go without slipping? Now if the bowl starts rotating about axis as shown in figure. At what angular speed $w$ will the insect just be able to come out of the bowl?(Radius of the bowl $5 cm$) 

First when the bowl is not rotating

$$\begin{gathered} f=mg \sin \theta \\ N=mg \cos \theta \\ f=\mu N \\ \mu mg \cos \theta =mg \sin \theta \\ \tan \theta =\mu \\ \theta ={\tan }^{-1}\mu \end{gathered}$$

The height up to which insect will reach 

$$\begin{gathered} =R\left(1-\cos \theta \right) \\ =5\left(1-\frac{1}{\sqrt{1+{\mu }^2}}\right) \\ =5\left(1-\frac{1}{\sqrt{1.25}}\right) cm=0.53 cm \end{gathered}$$

Now, the bowl start rotating with angular velocity $w.$

$$\begin{gathered} N-mg \cos \theta =\frac{m{v}^2}{R^{\text{'}}}\sin \theta \\ f+\frac{m{v}^2}{R^{\text{'}}} \cos \theta =mg \sin \theta \\ N=mg \cos \theta +\frac{m{v}^2}{R^{\text{'}}}\sin \theta \\ \mu N+\frac{m{v}^2}{R^{\text{'}}}\cos \theta =mg \sin \theta \\ \mu \left(mg \cos \theta +\frac{m{v}^2}{R^{\text{'}}}\sin \theta \right)+\frac{m{v}^2}{R^{\text{'}}}\cos \theta =mg \sin \theta \end{gathered}$$

Now, when the angular displacement of the insect will be $90°$ then it will just come out from the bowl put $\theta =90°$ and $R^{\text{'}}=R$ in above equation 

$$\begin{gathered} \mu \frac{m{v}^2}{R}=mg \\ {v}^2=\frac{Rg}{\mu } \\ {R}^2{w}^2=\frac{Rg}{\mu } \\ {w}^2=\frac{g}{\mu R} \\ w=\sqrt{\frac{g}{\mu R}} \\ =\sqrt{\frac{10}{0.5\times 5\times {10}^{-2}}} \\ =\frac{{10}^2}{5}=20 rad/s \end{gathered}$$