To solve this problem it is necessary to apply the concepts related to energy conservation as well as centripetal acceleration.
By conserving energy we know that
[tex]\Delta KE = \Delta PE[/tex]
Where,
KE = Kinetic Energy
PE = Potential Energy
[tex]\Delta KE = \Delta PE[/tex]
[tex]\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 = mgh_i-mgh_f[/tex]
Initial Kinetic Energy according the statement is zero, same as final potential energí, therefore
[tex]\frac{1}{2}mv_f^2 = mgh_i[/tex]
Re-arrange for v,
[tex]v_f = \sqrt{2gh_i}[/tex]
Where h here represent the radius of hemispherical bowl.
We have also the definition of centripetal acceleration, which is
[tex]a_c = \frac{v^2}{R}[/tex]
But we have that the radius is equal to the height, then
[tex]a_c = \frac{v^2}{h_i}[/tex]
Replacing the previous value of velocity found,
[tex]a_c = \frac{(\sqrt{2gh_i})^2}{h_i}[/tex]
[tex]a_c = \frac{\sqrt{2gh_i}}{h_i}[/tex]
[tex]a_c = 2g[/tex]
Substituting the value for gravitational acceleration
[tex]a_c = 2*9.8[/tex]
[tex]a_c = 19.6m/s^2[/tex]
Therefore the radial acceleration of ice cube at bottom is [tex]19.6m/s^2[/tex]
