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If a ball rolls downhill, will the linear KE equal the rotational KE at the bottom of the hill?

I think that it will since Ei=Ef. Could someone explain just to make sure?

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  • $\begingroup$ About which point are you considering the rotation?? It is the same problem that the flanges of a train wheel go backward (in the frame of reference of the rail, and the point of contact). If the reference moves with the point of contact then it's all rotational energy! (the point of contact never moves from its origin;-) $\endgroup$ Jul 2, 2017 at 11:23

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I assume that by $E_i=E_f$ you mean energy conservation.

If you neglect friction (with air/ground) in the example you describe, energy is conserved, which means that the potential energy which the ball had being on the hill initially is transformed into kinetic energy as it rolls down the hill. At any moment in time, the total energy, i.e. the sum of kinetic (linear and rotational) and potential energy will be the same and equal to the potential energy at the start (when there was no motion, i.e. no kinetic energy).

As an equation:

$$E_\mathrm{tot}=mgh+\frac{1}{2}mv^2+\frac{1}{2}I\omega^2=\mathrm{const.},$$

where $h$ is the vertical height (dependent on time), $m$ the mass of the ball, $v$ its linear velocity, $I$ the moment of inertia and $\omega$ the angular/rotational velocity. The first term is the potential energy, second term the linear kinetic energy and the third term the rotational energy.

If the ball does not slip, $v=\omega r$, and taking into account that the moment of inertia for a sphere of radius $r$ is: $I=\frac{2}{5}mr^2$, you get that:

$$\frac{E_\mathrm{kin, lin}}{E_\mathrm{kin, rot}}=\frac{\frac{1}{2}m v^2}{\frac{1}{2}I\omega^2}\\ =\frac{mv^2}{\frac{2}{5}mv^2}\\ =\frac{5}{2}$$

So the linear kinetic energy is at any time 2.5 times larger than the rotational kinetic energy.

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This is not true.

The linear energy $K_L$ is given by \begin{equation} K_L=\frac{1}{2}mv^2 \end{equation} while the rotational energy $K_R$ is given by \begin{equation} K_R=\frac{1}{2}I\omega^2 \end{equation} where $m$ is the mass of the ball, $I$ the moment of inertia, $v$ the linear velocity and $\omega$ the angular velocity; in particular, $v$ and $\omega$ are related by \begin{equation} v=\omega r \end{equation} so we have \begin{equation} K_R=\frac{1}{2}\frac{I}{r^2}v^2 \end{equation} $K_L$ is equal to $K_R$ if \begin{equation} \frac{1}{2}\frac{I}{r^2}v^2=\frac{1}{2}mv^2 \end{equation} The previous equation is true if \begin{equation} v=0 \end{equation} or \begin{equation} I=mr^2 \end{equation} which is the momento of inertia of the cylindrical shell, but you were considering a sphere, so the answer to your question is negative.

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