Relationship btwn rotational and translational KE

[lemonpie]

Homework Statement

A 140 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.150 m/s. How much work must be done on the hoop to stop it?

Homework Equations

K = 1/2Iomega^2 + 1/2mv^2

The Attempt at a Solution

K translational = 1/2mv^2 = 0.5(140)(0.150)^2 = 1.575 J

Apparently I double this amount to get the total KE, meaning the rotational energy is also 1.575 J, and the total work that needs to be done is 3.15 J. Does this mean that the rotational energy always equals the translational energy? I don't think so. How did this business with doubling come about?

 

[davieddy]

You need to include the rotational kinetic energy about the center of mass.

Note that the speed of the point on the hoop in contact with the ground
is zero relative to the ground.

Relative to the c of m, al the mass in the hoop has the same speed as the
center has relative to the ground. That is why the rotational and translational KEs
are equal in this case.

For the hoop, I=MR^2 (all mass at distance R) but I is less for
e.g. a sphere or a solid cylinder.

So in general is is NOT true that rotational KE = translational KE

However, since omega=V/R, their ratio does not depend on radius:
R cancels out.