I want to view the solution to this problem but nobody has ever posted any solution to this problem. Please do if you know the solution.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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 \(\displaystyle I = \frac{MR^2}{2}\)

Log in to reply

Very clear solution. Thank you so much Josh!

Log in to reply

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

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.

Log in to reply

Log in to reply