Interesting Mechanics Problem - Spinning Disk

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Note by Lokesh Sharma
7 years, 5 months ago

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Comments

Hey Lokesh, please excuse my brevity:

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

Iω˙=τ=RMgμkω(t)=ω0RMgμkIt\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

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

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

gμkRt=ω0RMgμkItt=Rω3gμ\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 ω\omega at this point we find ωf=ωi/3\omega_f = \omega_i/3

In all of the above, we take I=MR22\displaystyle I = \frac{MR^2}{2}

Josh Silverman Staff - 7 years, 5 months ago

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Why does it stop slipping when v=w/Rv = w/R? Is this just something that is shown experimentally or is it derived mathematically?

Jess Smith - 7 years, 5 months ago

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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=ωRv=\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=ω/Rv=\omega/R, it is actually v=ωRv=\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 - 7 years, 5 months ago

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@Josh Silverman Yep, v=ω/R v = \omega / R confused me. Thanks!

Jess Smith - 7 years, 5 months ago

Very clear solution. Thank you so much Josh!

Lokesh Sharma - 7 years, 5 months ago
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