Circular motion 
A small block slides down from the top of a hemisphere of radius $r$. It is assumed that there is no friction between the block and the hemisphere. At what height $h$ will be the block lose contact with the surface of sphere.

 A: Assuming that the block was given a few small tap to let it start falling, $v_i$=0, the initial energy is:
$$
E_i=K_i+U_i
$$
where $K_i$ and $U_i$ are the initial kinetic and potential energy respectively. With the above assumption $K_i=0$. The gravitational potential is given by 
$$
U_i = mgh_i
$$
where $m$ is the mass, $g$ is the acceleration due to gravity, and $h_i$ is the initial height from the floor. This means the box is at a height $d_i=2r$. Thus the potential energy is $U_i=2mgr$. Thus the initial total energy, 
$$E_i=2mgr\tag1$$
Later, when the block looses contact with the surface, the total energy is:
$$
E_f = K_f + U_f
$$
where $K_f$ and $U_f$ are the final kinetic and potential energy, respectively. The final kinetic energy is
$$
K_f = \dfrac{1}{2}mv_f^2
$$ 
where $v_f$ is the final velocity. 

The final height the the box is $$h_f=r\cos\theta+r\tag2$$ as shown in the picture above. Thus the total final energy is:
$$
E_f = \dfrac{1}{2}mv_f^2+mg(r\cos\theta+r)
\tag3
$$
Because of the conservation of energy $E_i=E_f$ and using Eq.(1) and Eq.(3)
$$
2mgr=\dfrac{1}{2}mv_f^2+mg(r\cos\theta+r)
$$
With some algrebra:
$$
gr(1-\cos\theta)=\dfrac{1}{2}v_f^2
\tag4
$$
Now looking at a free body diagram of the final position:

where the Y-axis is along the normal force. Normally there would be a normal force on the block, but since the block is just about to fall off, the normal force is 0, i.e. $N=0$. Thus the sum of the forces in the $y$ direction shown in the picture is:
$$
ma_y = mg\cos\theta
\tag5
$$
And since the block was rotating while resting on the hemisphere:
$$
a_y = \dfrac{v_f^2}{r}
$$
Putting this into Eq.(5):
$$
m\dfrac{v_f^2}{r}=mg\cos\theta
$$
Then some algebra:
$$
v_f^2=gr\cos\theta
\tag6
$$
Pluggin Eq.(6) into Eq.(4):
$$
gr(1-\cos\theta)=\dfrac{1}{2}gr\cos\theta
$$
Once again, after some algebra:
$$
\cos\theta=\dfrac{2}{3}
$$
which means 
$$
\theta=\cos^{-1}\left(\dfrac{2}{3}\right)= 48.18^o
$$
Finally putting this angle into Eq.(2):
$$
h = r\cos\theta+r=r\dfrac{2}{3}r+r=\dfrac{5}{3}r
$$
