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Why does a rotating wheel have kinetic energy $K=\frac{1}{2}mv^2$ associated with its movement?

I will clarify.

Let's say I have a rotating wheel in an empty void which has an angular velocity $\omega$, then its total energy will be $E=\frac{1}{2} I \omega^2$, as the wheel is not moving. Now suppose I put this wheel on a surface. The wheel starts moving and, from what I understand, the total energy associated to the wheel is $E=\frac{1}{2} m v^2+ \frac{1}{2} I\omega^2$. How is that possible since energy cannot be created nor destroyed? The wheel is the same, so how can it have two different amounts of energy?

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You are correct that before contact with the surface the wheel has $KE=\frac{1}{2}I\omega^2$ and after contact with the surface the wheel has $KE=\frac{1}{2}I\omega^2+\frac{1}{2}mv^2$. Since energy is conserved and since $m$ and $I$ are unchanged that implies that $\omega$ must decrease in order to compensate for the linear KE.

As the spinning wheel contacts the stationary surface there will be a frictional force between the surface and the wheel which will act to oppose the slipping. In this case the direction opposing slipping is forward. The forward force will provide both a linear force to increase linear momentum and a rotational torque to decrease the rotational motion. This leads to a decrease in $\omega$ and an increase in $v$

Ideally the decrease in rotational KE will be exactly equal to the increase in linear KE. However, in practice there will be some skidding at the beginning and therefore some dissipation to heat.

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  • $\begingroup$ Thanks. Could you clarify how the transition from rotational kinetic energy to linear kinetic energy happens? $\endgroup$ – Xfgh Jul 10 at 22:36
  • $\begingroup$ Sure. I will modify the answer with details. $\endgroup$ – Dale Jul 10 at 22:42
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Find the rotational and translational kinetic energy of a ball with rotational energy $\frac{1}{2} I\omega_0^2$ initially if put on a slant.

\begin{align*} \frac{1}{2} m v^2+ \frac{1}{2} I\omega^2&=\frac{1}{2} I\omega_0^2\\ \frac{1}{2} m v^2+ \frac{1}{2} \alpha mR^2\omega^2&=\frac{1}{2} \alpha mR^2\omega_0^2&(\text{dimensions of }I\rightarrow [M][L]^2\Rightarrow \alpha\text{ is dimensionless})\\ \frac{1}{2} m v^2+ \frac{1}{2} \alpha mv^2&=\frac{1}{2} \alpha mR^2\omega_0^2&(\because v=wr\text{ if the ball is rolling without slipping})\\ \end{align*} For $\alpha=1$, exactly half the rotational kinetic energy $\frac{1}{2} I\omega_0^2$ goes into translational kinetic energy in no rolling case.

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