# Conservation of angular momentum

The conservation of angular momentum holds that the angular momentum just prior to a collision is the same as the angular momentum just after a collision.

## Example

Imagine a rod of length l and mass M suspensed from one end vertically, and that a small block mass having velocity v and mass m collides with the other end and sticks to it. The maximum angle of displacement of the rod from the vertical axis can be calculated using the conservation of angular momentum:

$\mathbf{L_i}=\mathbf{L_f}$
$\mathbf{m}\times{v}\times{l}= \mathbf{I}\times{\omega}$

Thus

$\mathbf{\omega}= \frac{\mathbf{m}\times{v}\times{l}}{\mathbf{I}}$

After the collision, there is conservation of energy such that the final potential energy of the rod (with the sticking block mass) equals the initial kinetic energy just after the collision:

$\mathbf{U_f}= \mathbf{KE_i}$

Now solve the the two sides of the above equation separately:

$\mathbf{U_f}= \mathbf{U_{rod}}+\mathbf{U_{block}}$
$\mathbf{U_{rod}}= \frac{l}{2}\times(1-cos\theta)\times{M}\times{g}$
$\mathbf{U_{block}}= l\times(1-cos\theta)\times{m}\times{g}$
$\mathbf{KE_i}= \frac{I}{2}\times\omega^2$

Solving the above four equations yields:

$\mathbf{cos\theta} = 1 - \frac{I\times{\omega^2}}{g\times{l}\times{(2m+M)}}$

Plugging in for angular velocity from the initial equations above yields:

$\mathbf{cos\theta} = 1 - \frac{{m^2}{v^2}{l}}{I\times{g}\times{(2m+M)}}$

Calculating the moment of inertia I now becomes necessary for a rod of length l and mass M, with a small block of mass m at its end.

An ordinary rod of length l has the following moment of inertia relative to an axis of rotation at one end:

$\mathbf{I}= \frac{M}{3}\times{l^2}$

A moment of inertia is additive, defined as follows:

$I \ \stackrel{\mathrm{def}}{=}\ \sum_{i=1}^{N} {m_{i} r_{i}^2}\,\!$

where m is the mass at each (perpendicular) distance r from the axis of rotation.

Thus the the moment of inertia I for a rod of length l and mass M, with a small block of mass m at its end is simply this:

$\mathbf{I}= \frac{M}{3}\times{l^2} + m\times{l^2}$

which is:

$\mathbf{I}= \frac{1}{3}\times(3m+M)\times{l^2}$

Plugging this back into the unsolved equation above yields:

$\mathbf{cos\theta} = 1 - \frac{{3m^2}{v^2}}{l\times{g}\times{(2m+M)(3m+M)}}$

If we complicate the problem further by assuming the small block began with velocity zero from an incline of height h, then applying conservation of energy to the moment in time just prior to its collision with the rod yields the following velocity of impact:

$\mathbf{\frac{m\times{v^2}}{2}}=m\times{g}\times{h}$

and hence

$\mathbf{v^2}=2gh$

and thus the solution is:

$\mathbf{cos\theta} = 1 - \frac{{6m^2}{h}}{l\times{(2m+M)(3m+M)}}$

or

$\mathbf{\theta} = arccos(1 - \frac{{6m^2}{h}}{l\times{(2m+M)(3m+M)}})$