A uniform thin rod of length l is suspended from one of its ends and is rotated at f rotations per second. The rotational kinetic energy of the rod will be

Options

A. `2mL^(2)pi^(2)n^(2)`
B. `1/2mL^(2)pi^(2)n^(2)`
C. `2/3mL^(2)pi^(2)n^(2)`
D. `1/6mL^(2)pi^(2)n^(2)`

Answer

C

Explanation

`KE_("rot")= 1/2 Iomega^(2)`
` =1/2 ((ml^(2))/3) (4pi^(2)n^(2))`