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 :

Options

A. `(1)/(2)AL^(3)Domega^(2)`
B. `(1)/(6)AL^(3)Domega^(2)`
C. `(1)/(24)AL^(3)omega^(2)D`
D. `(1)/(12)AL^(3)Domega^(2)`

Answer

C

Explanation

K.E. of rotation
`=(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)`