Friction force will act towards right.

If no external torque is applied on the system, then angular momentum of the system remains constant. When a child sits on rotating disc, then no torque is applied (weight of child acts downward), so angular momentum will remain conserved.

$$\eqalign{ & {\text{As,}}\,\,v = \sqrt {\frac{{2gh}}{{1 + \frac{{{k^2}}}{{{r^2}}}}}} \cr & {\text{Given,}}\,\,h = \frac{{3{v^2}}}{{4g}} \cr & {\text{So,}}\,\,{v^2} = \frac{{2gh}}{{1 + \frac{{{k^2}}}{{{r^2}}}}} \cr & = \frac{{2g3{v^2}}}{{4g\left( {1 + \frac{{{k^2}}}{{{r^2}}}} \right)}} = \frac{{6g{v^2}}}{{4g\left( {1 + \frac{{{k^2}}}{{{r^2}}}} \right)}} \cr & 1 = \frac{3}{{2\left( {1 + \frac{{{k^2}}}{{{I^2}}}} \right)}} \cr & {\text{or}}\,\,1 + \frac{{{k^2}}}{{{r^2}}} = \frac{3}{2}\,\,{\text{or}}\,\,\frac{{{k^2}}}{{{r^2}}} = \frac{3}{2} - 1 = \frac{1}{2} \cr & {k^2} = \frac{1}{2}{r^2}\,\,\left( {{\text{Equation of disc}}} \right) \cr} $$
Hence, the object is disc.

If we take clockwise torque
$$\eqalign{ & {\tau _{{\text{net}}}} = {\tau _{{F_1}}} + {\tau _{{F_2}}} + {\tau _{{F_3}}} \cr & 0 = {F_1}r + {F_2}r + {F_3}r \cr & \Rightarrow {F_3} = {F_1} + {F_2} \cr} $$

$$\eqalign{ & {x_{cm}} = \frac{{m \times \ell + 2m \times 2\ell + 3m \times 3\ell + ... + nm \times n\ell }}{{m + 2m + 3m + .. + nm}} \cr & = \frac{{\left( {1 + 4 + 9 + .. + {n^2}} \right)\ell }}{{1 + 2 + .. + n}} \cr & = \frac{{\frac{{\ell n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}}{{\frac{{n\left( {n + 1} \right)}}{2}}} \cr & = \frac{{\left( {2n + 1} \right)\ell }}{3} \cr} $$

Both the sphere and the plank will slide down with same acceleration $$g\sin \theta .$$

The applied force shifts the normal reaction to one corner as shown. At this situation, the cubical block starts topping about $$O.$$  Taking torque about $$O$$

$$\eqalign{ & F \times L = mg \times \frac{L}{2} \cr & \Rightarrow F = \frac{{mg}}{2} \cr} $$

Case 1:
When they rotate in same sense $$2m\pi = 2\omega t$$
$$\eqalign{ & \frac{{3\pi }}{2} + 2n\pi = \omega t;2m\pi = 2\left( {\frac{{3\pi }}{2} + 2n\pi } \right) \cr & 2m = 3 + 4n;m = \frac{3}{2} + 2n \Rightarrow m - 2n = \frac{3}{2} \cr} $$
Not possible for $$m$$ and $$n$$ being integer.
Case 2 :
When they rotate in opposite sense $$2m\pi = 2\omega t$$
$$\eqalign{ & \frac{\pi }{2} + 2n\pi = \omega t;2m\pi = 2\left( {\frac{\pi }{2} + 2n\pi } \right) \cr & 2m\pi = \pi + 4n\pi ;2m - 4n = 1 \cr} $$
Not possible for $$m$$ and $$n$$ integer.

When angular acceleration $$\left( \alpha \right)$$ is zero then torque on the wheel becomes zero.
$$\eqalign{ & \theta \left( t \right) = 2{t^3} - 6{t^2} \Rightarrow \frac{{d\theta }}{{dt}} = 6{t^2} - 12t \cr & \Rightarrow \alpha = \frac{{{d^2}\theta }}{{d{t^2}}} = 12t - 12 = 0 \cr & \therefore t = 1\,\sec \cr} $$

$$\eqalign{ & t = \frac{1}{{\sin \theta }}\sqrt {\frac{{2h}}{g}\left[ {1 + \frac{{{K^2}}}{{{R^2}}}} \right]} \Rightarrow t \propto \left( {1 + \frac{{{K^2}}}{{{R^2}}}} \right) \cr & {\text{so,}}\,{t_1}:{t_2} = \sqrt 2 :\sqrt {1.5} \cr} $$