# Change in direction imply angular acceleration

Does direction change imply angular acceleration. When a non-point mass object changes direction (like a block sliding down a hill of changing slope), why do we not account for rotational $$K.E$$ when using conservation of energy? It is technically rotating around its center of mass...

$$K = \tfrac{1}{2} m v_{\rm COM}^2 + \tfrac{1}{2} I_{\rm COM} \omega^2 \tag{1}$$
with $$m$$ the mass, and $$I_{\rm COM}$$ the mass moment of inertia about the center of mass. The above is invariant to the location, meaning that when measured at a different point A for example, the same value is returned by $$K=\tfrac{1}{2} m v_{\rm A}^2 + \tfrac{1}{2} I_{\rm A} \omega^2$$.
You can only ignore the rotational part if $$\omega = 0$$ or $$I_{\rm COM} = 0$$ as is the case with a point mass.
• Also, assuming I had the shape of the hill, how would I apply conservation of energy? Since there is slipping, the $v_{\text{cm}} = r \omega$ relation does not hold... How would I solve such a problem? Thanks again! Sep 13, 2020 at 15:53
• If the hill is curved at the bottom then yes, include its rotation. But how? The path has a certain radius of curvature $r_{\rm curve}$, which would give you the rotation $\omega = v \ r_{\rm curve}$. To find this curvature look up the subject of differential geometry. Sep 13, 2020 at 20:45