Can The Cat Get His Dinner?

Clearly, angular velocity of cat is same as that of mouse (i.e. $\omega =\frac{u}{R}$

At a radial distance x, cat's velocity has two components, $\omega x$ tangential and $v_{r} = \sqrt{u^2 - \omega^2 x^2} = \sqrt{u^2 \big(1-\frac{x^2}{R^2}\big)}$ radial.

Hence, $v_{r} = \frac{dx}{dt} = \frac{u}{R} \sqrt{R^2 - x^2}$

$\Rightarrow \displaystyle R\int_{0}^{R} \frac{dx}{\sqrt{R^2-x^2}} = u \int_{0}^{t} dt$

$\Rightarrow t = \frac{R}{u} \frac{\pi}{2} \approx 11 s$

Okay here is what I did. It seems a bit abstract but gets the job done. Simply try to make the velocity or ie direction vector of the cat and you come to realize that it is tending to a half circle shape with a radius of R/2. Now I assumed that the mouse started from point A at the top of the circle. And now after certain time has completed a quarter circle in clockwise revolution to end at point B. Now as the cat traced a semi cricle with end points centre of original circle and B. To test if this is true and if the cat really caught the mouse we just need to see if they covered the same path length as their velocities are anyways equal. And true enough they are equal. Try it yourself. Thus ultimately time taken is equal to the time taken by mouse to make a quarter circle which is nothing but 11 secs. So thats the answer put short. Just see in the comments below, how I came to the conclusion its a semi circular arc.

$\omega_{cat}=\omega_{mouse}=\frac{1}{7} \\ v_{cat}=\frac{r}{7}\\ \text{let } u_{cat} \text{be the component of velocity perpendicular to linear velocity} \\ \text{(don't know if it has a special name)} \\ 4^2=v_{cat}^2+u_{cat}^2 \\ u_{cat} = \sqrt{16-\frac{r^2}{49}}=\frac{\sqrt{28^2-r^2}}{7} \\ \text{but } u_{cat}=\frac{dr}{dt} \\ \frac{dr}{d\theta} = \frac{dt}{d\theta} \times \frac{dr}{dt} \\ \frac{dr}{d\theta} = 7 \times \frac{\sqrt{28^2-r^2}}{7} \\ \frac{dr}{d\theta} = \sqrt{28^2-r^2} \\ \theta = arcsin{\frac{r}{28}} \\ \text{when r = 28,} \\ \theta = \frac{\pi}{2} = \frac {11}{7} \\ t= \frac{\theta}{\omega} = 11$

I am not posting the actual solution; but try to use this hint. Angular velocities of both the cat and mouse are equal. In the frame rotating at this angular velocity, the cat covers distance equal to the radius of the circle with its radial component of velocity to catch the mouse. By solving we get, t = (R/u)arc sin(u/v) = 11s