A semicircular rigid cylindrical tube is placed on a horizontal floor. Two identical balls that loosely fit inside the tube are simultaneously released from the top of the best described the total force of normal reactions between the tube and the floor?

 

initially, the normal force will be 

$\mathrm{N}=\left(\mathrm{M}+2\mathrm{m}\right)\mathrm{g}$

After a while 

The balls stick to the outer wall due to centripetal force and the normal force will become

$N=Mg-2\left(\frac{m{v}^2}{r}\cos \theta \right)$

where $\theta$ is the angle made by the position of the balls with the vertical.  

Finally when the balls reach the ground normal becomes

$N=Mg$

Therefore normal force first decreases, acquires a minimum value then increases.