Newtonian Mechanics – Why Is Kinetic Energy the Same for All Circular Objects Rolling Down an Inclined Plane?

Question

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?

Answer 1

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)