The proportion of kinetic energy of a rotating rigid body?

[Jason Ko]

The kinetic energy of a rotating rigid body is given by K=1/2mv^2 + 1/2Iw^2 but how to determine the proportion of translational energy and rotational energy? I know that if the mass distribution is more concentrated at its center mass, then more energy goes to the translational part. But is there actually a theorem account for it and how do we actually calculate the proportion?

 

[gleem]

The kinetic energy of a rotating rigid body is given by K=1/2mv^2 + 1/2Iw^2 but how to determine the proportion of translational energy and rotational energy?

The proportion of translational and rotational kinetic energy depends on the object and how the energy is imparted to it.

I know that if the mass distribution is more concentrated at its center mass, then more energy goes to the translational part.

No not in general. Take the extreme example of a hoop that has no mass at its center.

 

[A.T.]

But is there actually a theorem account for it and how do we actually calculate the proportion?

The proportion depends on the motion of the body, which you have not specified, so it's completely arbitrary.

 

[jbriggs444]

The kinetic energy of a rotating rigid body is given by K=1/2mv^2 + 1/2Iw^2 but how to determine the proportion of translational energy and rotational energy? I know that if the mass distribution is more concentrated at its center mass, then more energy goes to the translational part. But is there actually a theorem account for it and how do we actually calculate the proportion?

As @A.T. says, you have to know how the body is moving -- how is $\omega$ related to $v$?

Is it rolling without slipping on its [round] perimeter? So that $v = \omega r$?

Is it rolling without slipping on some extended framework? So that $v = \omega R$ with $R > r$?

Is it rolling without slipping on a slender axle while the bulk of the body extends further out? So that $v = \omega R$ with $R < r$?

Is it a flywheel mounted on a vehicle with a gear drive so that $v = k \omega$ for some $k$ that depends on the gear ratio?

Is it even round?

Is it flying through space on a ballistic trajectory so that $v$ and $\omega$ have no relationsip at all?

 

[Jason Ko]

The proportion of translational and rotational kinetic energy depends on the object and how the energy is imparted to it.

No not in general. Take the extreme example of a hoop that has no mass at its center.

A wheel rolling down an inclined plane without slipping. The GPE converts to KE.

 

[Nugatory]

A wheel rolling down an inclined plane without slipping. The GPE converts to KE.

You will still need the moment of inertia which depends on how the mass of the wheel is distributed. But once you have that, if the wheel is rolling without slipping then the translational and rotational speeds are related as @jbriggs444 said above: $v=\omega r$ and a bit of algebra will find the relationship between them.

 

[robphy]

It's useful to write the moment of inertia as $I=f\cdot MR^2$,
where $f$ is determined by the distribution of mass.

 

[Lnewqban]

The kinetic energy of a rotating rigid body is given by K=1/2mv^2 + 1/2Iw^2 but how to determine the proportion of translational energy and rotational energy?

We don't determine that proportion, nature does.

A wheel rolling down an inclined plane without slipping. The GPE converts to KE.

In the case of a wheel rolling down an inclined plane without slipping, the rate of the translational movement is restricted or limited by the rate of the rotational one.
It will take more energy to initiate the rotation of any wheel that has a huge moment of inertia; therefore, the translational movement will be necessarily slow.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/sphinc.html#sph

http://hyperphysics.phy-astr.gsu.edu/hbase/hoocyl.html#hc2