Where to hit the pool ball such that it satisfies the given condition? The question Where to hit the ball in such a way that after covering a certain distance it rolls back?
You are given a pool ball and you have to find where to hit it such that it rolls back after covering a certain distance.
My Attempt
Given x is the distance from centre of mass where you have to target
I = Linear momentum
w = omega
The condition for such a motion should be that Vcm=0 and w>0.
Vcm = 0 because the ball has to stop translation motion and roll back
omega>0 as it has to roll back
We also know that change in momentum = Pfinal - Pinitial = 0 - mVcm
0 as it stops.
Now by angular momentum
I * x = 2/5mR^2 * omega
this give x = 2/5* R^2 * omega
as Vcm = Rw = 0
x = 0
I get x = 0. Is this correct?
I have added a picture so that it is easier to understand my question.
Take a pen and place your fingers like I have in the picture and both the fingers at same time slip your fingers on the pen's body in such a way that it moves forward. The aim is to send it foward while rolling back. Similarly the pool ball if hit at a particular distance will cover some distance and then roll back. 
 A: Consider the following diagram:

Left is the pool ball being hit by the cue and right an instant after the cue hit.
If we hit the pool ball below the centre point of the pool ball, two things will happen:
1) An impulse force causes temporary acceleration to the right, resulting in the pool ball acquiring a rightward translation velocity $v$. From the impulse force needs to be subtracted the friction force $F_f$.
2) The Normal component of the impulse force causes a momentary friction force and assorted torque $T$ to act on the cue ball:
$$T= \text{Normal force} \times R \times \mu_0,$$
with $R$ the radius of the ball and $\mu_0$ the friction coefficient between cue and pool ball.
This 'impulse torque' causes momentary angular acceleration in the anti-clockwise direction:
$$T=I\dot{\omega},$$
With $I$ the inertial moment of the ball and $\dot{\omega}$ the angular acceleration. After a short time interval $\delta t$ the ball has angular velocity, anti-clockwise as:
$$\omega=\dot{\omega} \delta t.$$

Edit:
In reality a second torque has to be taken into account: $-\mu mgR$, stemming from friction between the ball and the plane and which diminishes $T$.

So an instant after the cue's hit the ball has now some rightward translational speed $v$ and some anti-clockwise angular speed $\omega$.
Assuming there is now friction between the ball and the surface with a friction coefficient $\mu$, the Normal force $F_N=mg$ produces a friction force $F_f$:
$$F_f=\mu mg$$
This causes deceleration:
$$a=-\frac{\mu mg}{m}=-\mu g$$
Translational speed will now decrease according:
$$v(t)=v-\mu gt$$
$v(t)$ will become zero for:
$$t=\frac{v}{\mu g}$$
Similarly $F_f$ causes a torque $F_fR=\mu mgR$, which causes angular deceleration according to:
$$\dot{\omega}=-\frac{\mu mgR}{I}$$
Angular speed will now decrease according:
$$\omega(t)=\omega-\frac{\mu mgR}{I}t$$
For a solid sphere we have:
$$I=\frac{2mR^2}{5}$$
So:
$$\omega(t)=\omega-\frac{5\mu g}{2R}t$$
Inserting $t=\frac{v}{\mu g}$ into this we get:
$$\omega(t)>\frac{5v}{2R}$$
What does this mean? It means that if this condition is fulfilled then when the ball stops translating  ($v=0$) it will still be rotating anti-clockwise and will effectively start rolling back.
Unfortunately establishing the post-cue translational speed $v$ and angular speed $\omega$ is no sinecure. Later on I will make an attempt at that calculation.
Now let's have a look at Case II in the OP's photo, where an inclined finger presses down on the round object:

We'll make the following assumptions:
1) There's enough friction between finger and object to prevent slippage between them.
2) The applied force $F$ acts perpendicularly to the surface of the object.
3) Friction between object and floor allows slippage.
4) Point H acts as a frictionless hinge.
Let's try and determine $F_{net}$ as it will determine horizontal acceleration. The balance of $x$-forces is:
$$F\sin\alpha-F_{f2}-F_{f1}\cos\alpha=F_{net}$$
$$F\sin\alpha-\mu_2 F_N-\mu_1F\cos\alpha=F_{net}$$
The balance of $y$-forces is:
$$F_N=mg+F\cos\alpha+F_{f1}\sin\alpha$$
$$F_N=mg+F\cos\alpha+\mu_1 F\sin\alpha$$
So we get:
$$F_{net}=F\sin\alpha-\mu_2 (mg+F\cos\alpha+\mu_1 F\sin\alpha)-\mu_1F\cos\alpha$$
And from there the rightward acceleration $a$ can be calculated:
$$a=\frac{F_{net}}{m}$$
Regards the torques acting on the object, this given by:
$$T_{net}=(F_{f1}-F_{f2})R$$
$$T_{net}=[\mu_1F-\mu_2(mg+F\cos\alpha+\mu_1 F\sin\alpha)]R$$
The angular acceleration $\dot{\omega}$ is given by:
$$\dot{\omega}=\frac{T_{net}}{I}$$
The problem is that as the object starts moving forward due to acceleration, $\alpha$ changes and thus also $F_{net}$ and $T_{net}$, so that again there's no easy way to calculate $v$ and $\omega$.
