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.

 

[nlightner]

I would like to clarify that the relationship between rotational and translational kinetic energy depends on the specific scenario being studied. In the case of a hoop rolling along a horizontal floor, the rotational energy and translational energy are indeed equal. This is because the hoop is rolling without slipping, meaning that the point of contact between the hoop and the floor is always at rest, and therefore no work is being done on the hoop due to friction. In this scenario, the work done to stop the hoop would simply be the sum of the translational and rotational kinetic energies, which is why the total work is double the value of the translational kinetic energy.

However, in other scenarios where there may be external forces or friction acting on the object, the relationship between rotational and translational kinetic energy may not be equal. In these cases, the work done to stop the object would be the sum of the translational and rotational kinetic energies, but the values of these energies may not be equal.

In conclusion, the relationship between rotational and translational kinetic energy depends on the specific situation being studied and cannot be generalized to always be equal. It is important to consider all factors and forces acting on the object to accurately determine the work needed to stop it.