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Conservation of angular momentum (view source)

Revision as of 03:19, November 18, 2007

11 bytes removed ,  03:19, November 18, 2007
L -> l
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:<math>\mathbf{U_f}= \mathbf{U_{rod}}+\mathbf{U_{block}}</math>  
 
:<math>\mathbf{U_f}= \mathbf{U_{rod}}+\mathbf{U_{block}}</math>  
   −
:<math>\mathbf{U_{rod}}= \frac{L}{2}\times(1-cos\theta)\times{M}\times{g}</math>
+
:<math>\mathbf{U_{rod}}= \frac{l}{2}\times(1-cos\theta)\times{M}\times{g}</math>
   −
:<math>\mathbf{U_{block}}= L\times(1-cos\theta)\times{m}\times{g}</math>  
+
:<math>\mathbf{U_{block}}= l\times(1-cos\theta)\times{m}\times{g}</math>  
    
:<math>\mathbf{KE_i}= \frac{I}{2}\times\omega^2</math>  
 
:<math>\mathbf{KE_i}= \frac{I}{2}\times\omega^2</math>  
Line 29: Line 29:  
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>\mathbf{cos\theta} = 1 - \frac{I\times{\omega^2}}{g\times{l}\times{(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^2}}{I\times{g}\times{L}\times{(2m+M)}}</math>
+
:<math>\mathbf{cos\theta} = 1 - \frac{{m^2}{v^2}{l}}{I\times{g}\times{(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 ''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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