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Interesting Mechanics Problem - Spinning Disk

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Note by Lokesh Sharma
3 years, 1 month ago

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Hey Lokesh, please excuse my brevity:

The spin of the wheel slows down due to friction with the table, so

\(\displaystyle I\dot{\omega } =\displaystyle \tau=\displaystyle-RMg\mu_k\rightarrow \omega(t) = \displaystyle\omega_0 - \frac{RMg\mu_k}{I}t\)

Meanwhile, the disc's linear velocity accelerates due to the same frictional force

\(\displaystyle M\dot{v} =\displaystyle Mg\mu_k \rightarrow v(t) =\displaystyle g\mu_kt\)

The wheel stops slipping when \(v=\omega R\) so we have

\(\displaystyle\frac{g\mu_k}{R}t =\displaystyle\omega_0 - \frac{RMg\mu_k}{I}t \rightarrow t =\displaystyle \frac{R \omega}{3 g \mu}\)

evaluating \(\omega\) at this point we find \(\omega_f = \omega_i/3\)

In all of the above, we take \(\displaystyle I = \frac{MR^2}{2}\) Josh Silverman Staff · 3 years, 1 month ago

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@Josh Silverman Very clear solution. Thank you so much Josh! Lokesh Sharma · 3 years, 1 month ago

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@Josh Silverman Why does it stop slipping when \(v = w/R\)? Is this just something that is shown experimentally or is it derived mathematically? Jess Smith · 3 years, 1 month ago

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@Jess Smith When that happens, there is no longer any motion between the bottom of the disk and the surface and so there is no kinetic friction to slow the thing down anymore.

In reality there will be rolling friction, which is more complex, but for the purposes of this problem, friction fully disappears when \(v=\omega R\) and the spinning as well as the forward motion is sustainable by the conservation of linear and angular momentum.

**Edit: I see that in the original post I mistakenly wrote \(v=\omega/R\), it is actually \(v=\omega R\), if that was the only source of confusion then you probably, already understand the condition, hopefully this clarification can help someone else. Josh Silverman Staff · 3 years, 1 month ago

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@Josh Silverman Yep, \( v = \omega / R \) confused me. Thanks! Jess Smith · 3 years, 1 month ago

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