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Conservation of Angular Momentum

27 bytes removed, 20:47, 13 December 2016
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Solving the above four equations yields:
:$\cos{\theta } = 1 - \frac{I \omega^2}{gl(2m+M)}$
Plugging in for angular velocity from the initial equations above yields:
:$\cos{\theta } = 1 - \frac{m^2 v^2 l}{Ig(2m+M)}$
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 The [[moment of inertia]] is additive, and 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 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:
:$\mathbf{cos{\theta} = 1 - \frac{{3m^2}{v^2}}{l\times{g}\times{lg (2m+M)(3m+M)}}$
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:
:$\mathbf{\frac{m\times{v^2}}{2}}=m\times{g}\times{h}mgh$
and hence
:$\mathbf{v^2}=2gh$
and thus the solution is:
:$\cos{\theta } = 1 - \frac{{6m^2}{h}}{l(2m+M)(3m+M)}$
or
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