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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 bar's center of mass.{table:align=center}{tr}{td} !ballBarInelastic.png! {td}{tr}{table}
*{+}Solution{+}{*}*:*
System: The ball is treated as a [point mass|point particle] and the bar as a rigid body with its mass uniformly distributed. The external forces on the system are the result of the interactions between the elements of the system and the Earth (gravitational force on ball and on bar) and with the table's surface (as there is no fricition it is 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].
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Approach:
The motion of the objects before and after the collision is on the table's surface, _(x,y)_ plane. After the collision, the ball-bar system will translate to the right and rotate counter-clockwise about its center of mass which is not at the center of the bar but at some point between the ball and the center of the bar. 
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To solve this problem we apply linear momentum and impulse and angular momentum and angular impulse principles to the ball-bar system. For that purpose we need to calculate the sum of the external forces exerted on the system and the sum of the torques due to the external forces about a given point.
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+The sum of the external forces+:
The external forces on the system are the gravitational and the normal force applied at the center of mass of ball and the bar. Because there is no motion in the direction perpendicular to the table's surface (z-axis) the normal forces on both objects have the same magnitude but opposite direction than the repective gravitational forces. Therefore, the sum of the external forces on the system is zero and the linear momentum of the system is constant:
{latex}$\vec P^}\begin{large}\[ \vec{P}^{before} = \vec P^{after}${P}^{after} \]\end{large}{latex}{

table:align=center}{tr}{td} !FreBodyBall.png|width=32,height=32! {td}{tr}{table}

+The sum of the external torques about a given point Q+:
The net torque due to the external forces about a given point Q is given by:
{latex}\begin{large} $\sum \tau_{Q}^{ext} = \tau_{Q}^{mg} +\tau_{Q}^{N_{ball}} + \tau_{Q}^{Mg} +\tau_{Q}^{N_{bar}}$ \end{large}{latex}
where the superscripts _mg_ and _Mg_ indicate the torques due to the gravitational force on the ball and the bar, respectively, and the superscripts _N~ball\~_ and _N~bar\~_ the torques due to the normal force on the ball and bar, respectively.

{latex} 
We We chose an arbitratry point *{-}Q{-}* on the table and we measured the position of the ball's center of mass with the vector $\vec r$of magnitude *{_}r{_}*, the red arrow in the figure. The black arrows are the normal force and the gravitational force exerted on the ball,$\vec N_{ball}$ an the gravitational force on the ball,$\vec{F}_{ of magnitudes *{_}N{~}ball{~}{_}* and *{_}F{~}E on ball}${~}{_}*, respectively. Because these two forces are applied at the same point, the ball's center of mass, the corresponding [moment arms|moment arm of these two forces] are the same. In addition, these forces have the same magnitude but opposite direction, therefore the torques due to each of them also have the same magnitude and opposite direction and they add to zero. The torques are indicated with the purple vectors in the figure.{latex}$ \tau_{Q}^{mg} +\tau_{Q}^{N_{ball}}=\vec{r} x \vec{F}_{E on ball}+ \vec{r} x \vec{N}_{ball}= 0${latex}{table:align=right}{tr}{td} !FreBodyBall.png! {td}{tr}{table}


In the same way, we can calculate the torques due to the normal and the gravitational force exerted on the bar. These two forces are also applied at the same point, the centre of mass of the bar, therefore the moment arm about point Q is the same and because these two forces have the same magnitude and opposite direction the resulting torques about point Q will add to zero. We conclude that the sum of the torques due to the external forces on the system about point Q is zero, therefore that angular momentum of the ball-bar system about point Q is constant.
{latex}
$ \vec L\_
{Q}^{before} =\vec L_{Q}
^
{after}$ {latex}


{note:title=Note: You can show that in this particular problem the net torque is zero about any point Q in space (not necessarily a point in the plane of the table). As a result, the angulat momentum of the bal - bar system is constant about any point in space.
}

{note}
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Apply linear momentum and Impulse:

Before the collision only the ball is moving then the linear momentum of the system is the linear momentum of a point particle of mass m moving with speed v in the \+x \-direction:
{latex}$\vec P^{before} = m v\hat{x}${latex}
After the collision the ball is attached to the bar and they are both translation with the same velocity, the velocity of the center of mass of the system:
{latex}$\vec P^{after}
=(m+M)\vec v\_
{cm}${latex}
Because there are no external forces exerted on the system the linear momentum is constant,we obtain that the velocity of the center of mass is:
{latex}$
\vec v_{cm}
=\frac
{mv}
{m+M}\hat{x}
${latex}
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For the same reason, the acceleration of the center of mass is zero (no external forces), therefore the velocity of the center of mass is constant and the center of mass of the system moves along a stratight line and increases linearly with time as shown in the figure below, where the center of mass of the ball-bar system is indicated with tha black X.{table:align=center}{tr}{td} !BallBarInelastic2.png|width=600,height=300! {td}{tr}{table}
Apply angular momentum and Angular Impulse:

Because the net torque is zero about any point in space we have the freedom to choose it anywhere we want. Let's consider the point Q to be somewehre along the straight line through which the ball is moving before hitting the bar. The point Q is shown in red in the figure below.{table:align=center}{tr}{td} !BallBarInelasticQ1.png|width=600,height=300! {td}{tr}{table}
Before the collision only the ball is moving, therefore the angular momentum of the system is the [angular momentum of a point particle moving along a straight line|Angular Momentum of a point particle moving along a straight line]. By choosing the point Q along the line of motion of the ball the angular momentum about point Q is zero(the angle between the ball's position vector and ball's velocity vector is either 180^o^ or 0^o^ depending on the location of point Q being to the right or to the left of the ball, respectively):
{latex}$\vec L_{Q}^{before}=\vec r$ x $m\vec v = 0${latex}
After the collision the angular momentum of the ball-bar system about point Q is the angular momentum of a rigid body rotating clockwise with angular speed w and translating to the right with speed *_v{~}cm{~}_*:
{latex}$\vec L_{Q}^{after} = \vec r_{cm} x (m+M)\vec v_{cm} + I_{cm}^{system}\vec \omega$ (eq. 1){latex}
where *_r{~}cm{~}_* is the position vector of the center of mass of the system measured from point Q and _I{~}cm{~}{^}sys^_ is the moment of inertia of the ball-bar system measured from the center of mass of the system and is given by:
{latex}&I_{cm}^{system}=I_{cm}^{bar} + M(d_cm-D/2)^2 + m(h-d_cm)^2$(eq.2){latex}
In the above expression, _d{~}cm~_ is the distance between the lower end of the bar and the position of the center of mass of the ball-bar system
and is given by:
{latex} $d_{cm}=\frac{MD/2+mh}{m+M}
$ (eq. 3)
{latex}.

The angular momentum about point Q after the collision in (eq.1) becomes:
{latex}
$ \vec L\_
{Q}
^
{after}
= (m+M)(h-d\_
{cm})v_{cm}
\hat
{z} - I_{cm}^{system}\omega\hat{z}
$ (eq.4)
{latex}
The angular momentum about point Q before and after the collision are equal, therefore setting (eq.4) to zero we can obtain the angular velocity:


{latex}
$\vec \omega = - \frac{(m+M)(h-d\_
{cm}
)v_{cm}}{I_{cm}}\hat
{z}