The rotational inertia of a system of point masses is given by the sum $$I = \sum m_i r_i^2$$, where $$r_i$$ is the perpendicular distance of each mass from the axis of rotation. Here, the axis is the y-axis, so the distances are the absolute values of the x-coordinates. $$I = m_1 r_1^2 + m_2 r_2^2 = (2.0 , \text{kg})(-1.0 , \text{m})^2 + (3.0 , \text{kg})(2.0 , \text{m})^2 = 2.0 , \text{kg} \cdot \text{m}^2 + 12.0 , \text{kg} \cdot \text{m}^2 = 14 , \text{kg} \cdot \text{m}^2$$.

Rotational inertias are additive. The total rotational inertia is the sum of the inertia of the hoop and the inertias of the two beads. The hoop's inertia is $$I_{hoop} = MR^2$$. Each bead is a point mass at a distance R from the axis, so each has an inertia of $$I_{bead} = mR^2$$. The total inertia is $$I_{total} = I_{hoop} + I_{bead1} + I_{bead2} = MR^2 + mR^2 + mR^2 = MR^2 + 2mR^2$$.

Let the axis pass through mass 1. Its distance from the axis is $$r_1 = 0$$. The other two masses, mass 2 and mass 3, are at a distance $$s$$ from mass 1. The total rotational inertia is the sum of the individual inertias: $$I = \sum m_i r_i^2 = m(0)^2 + m(s)^2 + m(s)^2 = 0 + ms^2 + ms^2 = 2ms^2$$.

The radius of gyration $$k$$ is a conceptual distance. It is defined as the radial distance from the axis of rotation at which the entire mass ($$M$$) of the body could be concentrated to produce the same rotational inertia ($$I$$) as the actual body. This concept simplifies comparisons of rotational inertia for different objects.

Rotational inertia is given by the sum or integral of $$mr^2$$ or $$r^2 dm$$. Since the inertia depends on the square of the distance $$r$$ from the axis of rotation, placing the mass as far as possible from the axis will maximize the rotational inertia for a given total mass. This is why flywheels are often designed like spoked wheels with a heavy rim.

Rotational inertia depends on how mass is distributed relative to the axis of rotation. The farther the mass is from the axis, the greater the rotational inertia. The formulas are: Hoop ($$MR^2$$), Hollow Sphere ($$\frac{2}{3}MR^2$$), Solid Disk ($$\frac{1}{2}MR^2$$), and Solid Sphere ($$\frac{2}{5}MR^2$$). Since $$1 > 2/3 > 1/2 > 2/5$$, the thin hoop has the greatest rotational inertia because all of its mass is concentrated at the maximum possible radius $$R$$.

Using the parallel axis theorem, $$I' = I_{cm} + Md^2$$. The axis tangent to the surface is parallel to the axis through the center, and the distance $$d$$ between these axes is the radius $$R$$. Therefore, $$I' = \frac{2}{5}MR^2 + M(R)^2 = (\frac{2}{5} + 1)MR^2 = \frac{7}{5}MR^2$$.

Rotational inertia ($$I$$) is the rotational analog of mass (inertia). Just as mass measures an object's resistance to a change in its linear velocity (i.e., its resistance to linear acceleration), rotational inertia measures an object's resistance to a change in its angular velocity (i.e., its resistance to angular acceleration).

The parallel axis theorem states that $$I = I_{cm} + Md^2$$, where $$d$$ is the distance between the center of mass axis and the parallel axis of rotation. For a rod rotating about its end, the distance $$d$$ from the center to the end is $$L/2$$. Thus, $$I = \frac{1}{12}ML^2 + M(\frac{L}{2})^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = (\frac{1}{12} + \frac{3}{12})ML^2 = \frac{4}{12}ML^2 = \frac{1}{3}ML^2$$.

The formula for rotational inertia of a continuous body is $$I = \int r^2 ,dm$$. For this rod rotating about the origin, the distance from the axis is $$r=x$$. The differential mass element is $$dm = \lambda(x) ,dx = (\beta x^2) ,dx$$. Substituting these into the integral gives $$I = \int_0^L x^2 (\beta x^2 ,dx) = \int_0^L \beta x^4 ,dx$$.