A rod of mass `m` spins with an angular speed `omega=sqrt(g/l)`, Find its a. kinetic energy of rotation. b. kinetic energy of translation c. total kinetic energy.

Explanation

Kinetic enegy of rotation `K_("rotational") = 1/2 I_(C)omega^(2)`
`=1/2 ((ml^(2))/12)(sqrt(g)/l)^(2)=1/24mlgl`
b Kinetic energy of translation, `K_("rotational")=1/2 mv_(C)^(2)`
`=1/2 m[(omegal/2)^(2)]=1/2 m[sqrt(g/l)xxl/2]^(2)=1/3mgl`
c. total kinetic energy
`K_("total")=K_("translational")+K_("rotational")`
`=1/6mgl=1/2I_(P)omega^(2)=1/2[(ml^(2))/2][sqrt(g/l)]^(2)=1/6mgl`