Show that kinetic energy of rotation is 1/2 w.l?

Question

this question falls under the topic of infinitesimal rotations were w- omega and l-length are vectors

Answer 1

No, rotational kinetic energy is

E=(1/2).I.w^2

where

I - moment of inertia (thats I as in 'eye')

w - angular velocity

the moment of inertia is dependant on how the mass is distributed about the centre of rotation.

Hope this helps : )

Answer 2

If a piece of mass m is at a distance r from axis of rotation it is moving in a circle  wirh speed v=rw(omega)

Thus KE(rotational) =1/2mv^2= 1/2mr^2w^2

now Moment of inertia=I=mr^2(for hoop)

hence KE(rotational)=1/2Iw^2