Rotational Kinetic Energy and Moment of Inertia

The total rotational energy is the sum of the rotational energies of each mass or molecule that makes up the rigid body:

$${\rm KE}_{{\rm rot}} \sum ({1\over 2} m_i v_i^2)$$ =

$$\sum({1\over2}m_ir_i^2)\left({v_i\over r_i}\right)^2$$ =

$${1\over 2} \left[\sum m_i r_i^2\right] \omega^2$$ = (1)

where we have use the fact that $\omega$ is the same for every point in the rigid body to remove it from the sum. For rigid bodies, we cannot really sum over each molecule, so we split the body into infinitesmal pieces, each of mass dm at a distance r from the axis of rotation, and the sum becomes an integral:

$${\rm KE}_{{\rm rot}}= {1\over 2} I \omega^2$$

where I is the moment of inertia of the body about the axis of rotation:

$$I=\int_{{\rm body}} dm r^2$$

 

Table 2: Table of Moments of Inertia of Various Rigid Bodies

Shape	Moment of Inertia
Uniform hoop/cylinder about center	I = M R2
Uniform Disc/Solid Cylinder about center	$I= {1\over 2} M R^2$
Solid Sphere about any axis through center	$I={2\over 5} M R^2$
Hollow Sphere about any axis through center	$I={2\over 3} MR^2$
Uniform Rod about axis through center	$I={1\over 12} M L^2$
Uniform Rod about axis through end	$I= {1\over 3} ML^2$

Note

• The moment of inertia depends not only on the mass and size of the object, but also how the mass is distributed about the axis of rotation. The more mass there is further away from the rotation axis, the greater the moment of inertia.
• When applying the work-energy theorem (energy conservation) to rigid bodies, both translational kinetic energy and rotational kinetic energy must be taken into account.