The kinetic energy of a uniform rod of mass m rotating with constant angular velocit `omega` about one of its end which is fixed as shown in figure is `(momega^(2)l^(2))/(N)` find N

Answer

6

Explanation

`dm=(dx)/(l)`m
`dE=(1)/(2)dmv^(2)=(1)/(2)(dx)/(l)m(wx)^(2)`
`dE=(mw^(2))/(2l)x^(2).dx`
`E=intdE=(mw^(2))/(2l)int_(0)^(l)x^(2).dx`
`E=(mw^(2)l^(2))/(6)`