Why is kinetic energy of all circular objects rolling down the inclined plane of same mass is same? 
In pure rolling, solid sphere will reach the ground first with more terminal velocity, and so
$$(KE)_{\text{SolidSphere}}= \frac{1}{2}mv^2$$
since solid sphere has more final velocity it should have more $KE$
but again this contradicts $U = mgh$, all the objects have same Mass and same height, so same potential energy, and we know that 
loss in potential energy = gain in kinetic energy, does that mean $KE = \frac{1}{2}mv^2$ is useless or invalid here?
 A: When looking at such problems, note that the total kinetic energy for each object is the sum of its  translational and rotational kinetic energy. That is, $$KE=\frac{1}{2}mv^2 +\frac{1}{2}I\omega^2$$
Each object will have a different moment of inertia $I$, and so you will get different values for rotational kinetic energy.
But note that at the same time, the final total kinetic energy for each will be the same.
This must be the case if they all begin with the same potential energy, due to conservation of energy. That is, $$mgh=\frac{1}{2}mv^2 +\frac{1}{2}I\omega^2=KE_{\text{final}}$$
The point is that they all will have differing values for translational speed $v$, but they all must have the same final total kinetic energy.
A: Since these objects are rolling (except the particle), they have converted their initial potential energy to the translational kinetic energy and to the rotational kinetic energy as well. So your equation should be corrected as $$KE=\frac 12mv^2+\frac 12I\omega ^2$$ Therefore, $I$, the moment of inertia, performs a big role. It is a crucial fact that determines who will come down faster. Although the every object has the same mass, the distribution of mass about their axis of rotation matters. That is the simple definition of $I$.
Though in a parallel way, each object has the same final net kinetic energy, because there is no energy loss. That is because; although they have different $I$, they have different final $v$ as well. That makes the value you get from the above equation for each object the same. No contradiction!
Side note: The particle probably has an energy loss because the plane should be rough to provide friction (if there's no friction, other objects won't roll, they will merely slip)
