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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. {table:align=center}{tr}{td} !ballBarInelastic.png! {td}{tr}{table}.
*{+}Solution{+}{*}*:*
Consider the system to be the ball and the bar. After the collision the ball-bar system will rotate about its center of mass as well as translate. The center of mass of the system is not at the center of the bar but it is at a point between the ball and the bar's center of mass.

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