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Conservation of angular momentum (view source)

Revision as of 03:46, November 18, 2007

452 bytes added ,  03:46, November 18, 2007
done!
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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>\mathbf{cos\theta} = 1 - \frac{{3m^2}{v^2}}{l\times{g}\times{(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:
 +
 
 +
:<math>\mathbf{\frac{m\times{v^2}}{2}}=m\times{g}\times{h}</math>
 +
 
 +
and hence
 +
 
 +
:<math>\mathbf{v^2}=2gh</math>
 +
 
 +
:<math>\mathbf{cos\theta} = 1 - \frac{{6m^2}{h}}{l\times{(2m+M)(3m+M)}}</math>
    
[[category:physics]]
 
[[category:physics]]
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