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