Changes

Solving the above four equations yields:

Solving the above four equations yields:

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

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

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.

Calculating the moment of inertia <math>I</math> now becomes necessary for a rod of length <math>l</math> and mass <math>M</math>, with a small block of mass <math>m</math> 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 <math>l</math> has the following moment of inertia relative to an axis of rotation at one end:

:<math>\mathbf{I}= \frac{M}{3}\times{l^2}</math>  

:<math>I = \frac{M}{3} l^2</math>  

A [[moment of inertia]] is additive, defined as follows:

The [[moment of inertia]] is additive and defined as:

::<math>I \ \stackrel{\mathrm{def}}{=}\  \sum_{i=1}^{N} {m_{i} r_{i}^2}\,\!</math>

::<math>I \ \stackrel{\mathrm{def}}{=}\  \sum_{i=1}^{N} {m_{i} r_{i}^2}\,\!</math>

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

where <math>m</math> is the mass at each (perpendicular) distance <math>r</math> 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:

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

:<math>\mathbf{I}= \frac{M}{3}\times{l^2} + m\times{l^2}</math>  

:<math>I = \frac{M}{3} l^2 + m l^2</math>  

which is:

which is:

:<math>\mathbf{I}= \frac{1}{3}\times(3m+M)\times{l^2}</math>  

:<math>I = \frac{1}{3} (3m+M) l^2</math>  

Plugging this back into the unsolved equation above yields:

Plugging this back into the unsolved equation above yields:

:<math>\mathbf{cos\theta} = 1 - \frac{{3m^2}{v^2}}{l\times{g}\times{(2m+M)(3m+M)}}</math>

:<math>\cos{\theta} = 1 - \frac{3m^2 v^2}{lg (2m+M)(3m+M)}</math>

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:

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:

:<math>\mathbf{\frac{m\times{v^2}}{2}}=m\times{g}\times{h}</math>

:<math>\frac{m v^2}{2} = mgh</math>

and hence

and hence

:<math>\mathbf{v^2}=2gh</math>

:<math>v^2 = 2gh</math>

and thus the solution is:

and thus the solution is:

:<math>cos\theta = 1 - \frac{{6m^2}{h}}{l(2m+M)(3m+M)}</math>

:<math>\cos{\theta} = 1 - \frac{{6m^2}{h}}{l(2m+M)(3m+M)}</math>

or

or