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A rod of length L rotates about an axis passing through its centre and normal to its length with an angular velocity omega. If A is the cross section and D is the density of material of rod, its rotational K.E. is : |
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Answer» `(1)/(2)AL^(3)DOMEGA^(2)` `=(1)/(2)Iomega^(2)=(1)/(2).(ML^(2))/(12).omega^(2)=(1)/(24)XX(ALD)L^(2)omega^(2)` `=(1)/(24)AL^(3)Domega^(2)` |
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