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...


1 Answer 1


Yes, rotational speed (not acceleration) should be included in the K.E. calculation. The general (2D) form of K.E. when tracking the center of mass is

$$ K = \tfrac{1}{2} m v_{\rm COM}^2 + \tfrac{1}{2} I_{\rm COM} \omega^2 \tag{1} $$

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.

  • $\begingroup$ So if I had a non point mass block on a hill at rest top and wanted to determine its speed at the bottom (assuming no friction and the hill is curved), I would have to account for rotational kinetic energy gained by the mass falling, correct? $\endgroup$
    – user256872
    Sep 13, 2020 at 15:49
  • $\begingroup$ 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! $\endgroup$
    – user256872
    Sep 13, 2020 at 15:53
  • $\begingroup$ 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. $\endgroup$ Sep 13, 2020 at 20:45

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