A ball hits a bar and sticks to it.

A uniform bar of mass M and length d is initially at rest on a horizontal and frictionless table. A ball of mass m and very small radius is moving towards the bar with speed v_o, hits and sticks to the bar at a distance h measured from on end of the bar. The bar is not attached to the table and its moment of inertia with respect to its center of mass is known to be I_cm. The goal of the problem is to find the angular velocity and the velocity of the center of mass of the ball - bar system after the collision.

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Solution:
System: The ball is treated as a point mass and the bar as a rigid body with its mass uniformly distributed. The external forces on the system are due to the interaction between the elements and the Earth (gravitational force on ball and on bar) and the elemnts and the table's surface (as there is no fricition it is only the normal force exerted by the surface on the ball and on the bar).

Model: 1-D Angular Momentum and Angular Impulse and Linear Momentum and Impulse.

Approach:

After the collision the ball-bar system will translate and rotate about its center of mass which is at a point between the position of the ball and the center of the bar. The motion of the two elemnts of the system is in the plane of the surface defined as the (x,y) plane. To solve the problem we will identify the external forces and the torques due to the external forces about a given point. Then we will apply

The forces acting on the elements of the system are the gravitational force on the ball and on the bar and the contact force exerted by the table's surface on the ball and on the bar. Because there is no friction thesi contact force has only one component p[erpendicular to the surface.
Modeling the objects:
Ball:
Because the ball has a small radius we will neglect any possible rotation of the ball an we will treat it as a point mass translating with initiall speed v_o.
Bar:
We will treat the bar as a rigid body with its mass uniformly distributed therfore, its center of mass is in the middle of the bar.
Modeling the interactions:
Ball:
The ball interacts with the Earth, the surface and the bar. As a resullt of these three interactions there are three forces acting on it:
#The gravitational force exerted by the Earth on the ball, F_EonBall. This is an external force and is applied at the ball's center of mass.
#The contact force exerted by the ground on the ball. Because there is no friction this force has only one component perpendicular to the surface, F_{N on Ball}. This force is an external force and is applied at the ball's center of mass.
#The force exerted by the bar on the ball, F_{Bar on Ball}. This force is an internal force and is applied at th eball's center of mass. It is the action reaction pair of the force exerted by the ball on the bar.
Bar:
The bar interacts with the Earth, the surface and the ball. As a resullt of these three interactions there are three forces acting on it:
#The gravitational force exerted by the Earth on the bar, F_{E on Bar}. This force is an external force and is applied at the bar's center of mass.
#The contact force exerted by the ground on the bar. Because there is no friction this force has only one component perpendicular to the surface, F_{N on Bar}. This force is an external force and is applied at the bar's center of mass.
#The force exerted by the ball on the bar, F_{Ball on Bar}. This force is an internal force and is applied at the point where the ball hits the bar, which is the point at a distance h from the lower end of the bar.

Because there is no motion of the ball and the rod in the direction perpendicular to the table's surface then Newton's second law applied on the ball and the stick implies that F_{E on Ball}+ F_{N on Ball} = 0 and F_{E on Bar}+ F_{N on Bar} = 0