Why do we add individual kinetic energy for particles of a rotating object but we don't do the same with objects in linear motion? I'm studying Rotational Kinetic Energy of an Object and the section reads:
Each particle of mass mi in the object has kinetic energy, Ki:
$$ K_{i} = \frac{1}{2}m_{i}v_{i}^2 $$
We can sum the kinetic energy of each particle together to get the total rotational kinetic energy, $ K_{rot} $, of the object:
\begin{aligned} K_{rot} = \sum_i \frac{1}{2}m_iv_i^2\end{aligned}
It seems that with linear motion we treat the object as a single particle and then the kinetic energy of the object is the kinetic energy of the particle. Not the sum of the particles of the object.
Why, with a rotating body, we add the particles of the body. But with, for example, a cube moving across a straight line we don't do the same?
 A: We do the same, if by object we understand a rigid body.
For linear motion of a rigid body all $v_i$ are equal and can be taken out of the sum: $$\sum \frac{1}{2}m_iv_i^2 = v_i^2\sum \frac{1}{2}m_i = \frac{1}{2}mv^2$$
A: 
Why, with a rotating body, we add the particles of the body. But with,
for example, a cube moving across a straight line we don't do the
same?

We do add the kinetic energies of the particles that comprise the cube. We do it by calculating the kinetic energy of the center of mass of the cube. We can do this in the case of a cube in linear motion because all the particles move in the same direction with the same velocity in the case of linear motion.
A rotating body is different because the we deal with angular velocity instead of linear velocity and rotational moment of inertia instead of translational moment of inertia (mass). The rotational kinetic energy is given by
$$KE_{rot}=\frac{1}{2}Iω^2$$
Where
$I$ = rotational moment of inertia. It is analogous to translational inertia (mass) for linear kinetic energy. You can look up the rotational moment of inertia for various types of shapes (solid spheres, solid cylinders, rods, etc.)
$ω$ = the angular velocity in radians per second. It is analogous to the linear velocity for translational kinetic energy.
Hope this helps.
A: For a system of particles, the kinetic energy is the kinetic energy of the center of mass plus the kinetic energy of all the particles relative to the center of mass; this is true for any system, such as a gas, not just a rigid body.  (See Goldstein, Classical Mechanics for the derivation.)
For a rigid body the relative positions of the particles in the body are fixed.  For pure translational motion of a rigid body, there is no relative motion of the particles in the body with respect to the center of mass, and the total kinetic energy is just that of the center of mass.  For rotational motion, the kinetic energy of the particles in the body is the rotational kinetic energy with respect to the center of mass.  For translation plus rotational motion of a rigid body the total kinetic energy is the kinetic energy of the center of mass plus the rotational kinetic energy of the particles with respect to the center of mass.
A: There's essentially no difference.
If we divide into $n$ particles with mass $m_1,\dots,m_n$ with distance $r_1,\dots,r_n$ with respect to the rotational axis, with speed $v_1,\dots,v_n$ then
$$KE = \sum_i \frac{1}{2}m_iv_i^2 = \sum_i \frac{1}{2}m_i \omega^2 r_i^2= \frac{1}{2}\omega^2 \sum_i m_i r_i^2 = \frac{1}{2}I\omega^2$$
where $\sum_i m_i r_i^2$ is equal to $I$, by the definition of inertia.
