Translational vs. Rotational Kinetic Energy

[kash25]

Hi,
Suppose I am trying to find the work done in bringing a resting cylinder to an angular speed of 8 rad/s.
Why is it INCORRECT to find the corresponding tangential velocity at a point on the outer surface of the cylinder (using angular speed * radius = tangential speed) and use the translational (0.5mv^2) work-kinetic energy theorem?
Why MUST we use the rotational version with I and angular speed?
Thank you.

[Vanadium 50]

Because there is rotational kinetic energy as well.

[Doc Al]

Why is it INCORRECT to find the corresponding tangential velocity at a point on the outer surface of the cylinder (using angular speed * radius = tangential speed) and use the translational (0.5mv^2) work-kinetic energy theorem?
Realize that the tangential velocity depends on the distance from the axis--the cylinder does not have a uniform tangential velocity. But if you're willing to add up the translational KE of each piece (dm) of the cylinder, that's just fine. (You'll get the same answer.)

KE = Σ½dm v² = Σ½dm r²ω² = ½(Σdm r²)ω² = ½Iω²
Why MUST we use the rotational version with I and angular speed?
It's just much easier. :wink: