Changes

== Example ==

== 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''':

Imagine a rod of length <math>l</math> and mass <math>M</math> suspended from one end vertically, and that a small block mass having velocity <math>v</math> and mass <math>m</math> 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''':

:<math>\mathbf{L_i}=\mathbf{L_f}</math>  

:<math>L_i = L_f </math>  

:<math>mvl= I \omega</math>

:<math>mvl= I \omega</math>

Thus

Thus:

:<math>\mathbf{\omega}= \frac{\mathbf{m}\times{v}\times{l}}{\mathbf{I}}</math>

:<math>\omega = \frac{mvl}{I}</math>

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:

[[Conservation of Energy|Conservation of energy]] can be applied after the collision, such that the final potential energy of the rod, <math>U_f</math> (with the sticking block mass) equals the initial kinetic energy, <math>T_i</math> just after the collision:

:<math>\mathbf{U_f}= \mathbf{KE_i}</math>

:<math>U_f = T_i</math>

Now solve the two sides of the above equation separately:

Now solve the two sides of the above equation separately:

:<math>\mathbf{U_f}= \mathbf{U_{rod}}+\mathbf{U_{block}}</math>  

:<math>U_f = U_{rod} + U_{block}</math>  

:<math>\mathbf{U_{rod}}= \frac{l}{2}\times(1-cos\theta)\times{M}\times{g}</math>

:<math>U_{rod} = \frac{mgl}{2} (1-cos\theta)</math>

:<math>\mathbf{U_{block}}= l\times(1-cos\theta)\times{m}\times{g}</math>  

:<math>U_{block}= mgl (1-cos\theta)</math>  

:<math>\mathbf{KE_i}= \frac{I}{2}\times\omega^2</math>  

:<math>T_i= \frac{1}{2} I \omega^2</math>  

Solving the above four equations yields:

Solving the above four equations yields:

:<math>\mathbf{cos\theta} = 1 - \frac{I\times{\omega^2}}{g\times{l}\times{(2m+M)}}</math>

:<math>cos\theta = 1 - \frac{I \omega^2}{gl(2m+M)}</math>

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

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

:<math>\mathbf{cos\theta} = 1 - \frac{{m^2}{v^2}{l}}{I\times{g}\times{(2m+M)}}</math>

:<math>cos\theta = 1 - \frac{m^2 v^2 l}{Ig(2m+M)}</math>

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.

Calculating the moment of inertia <math>I</math> 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:

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