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TopNewestHey 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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– Lokesh Sharma · 3 years, 1 month ago

Very clear solution. Thank you so much Josh!Log in to reply

– Jess Smith · 3 years, 1 month ago

Why does it stop slipping when \(v = w/R\)? Is this just something that is shown experimentally or is it derived mathematically?Log in to reply

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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– Jess Smith · 3 years, 1 month ago

Yep, \( v = \omega / R \) confused me. Thanks!Log in to reply