Waste less time on Facebook — follow Brilliant.
×

Interesting Mechanics Problem - Spinning Disk

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.

Note by Lokesh Sharma
4 years, 1 month ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

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 - 4 years, 1 month ago

Log in to reply

Very clear solution. Thank you so much Josh!

Lokesh Sharma - 4 years, 1 month ago

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?

Jess Smith - 4 years, 1 month ago

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.

Josh Silverman Staff - 4 years, 1 month ago

Log in to reply

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

Jess Smith - 4 years, 1 month ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...