This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
When posting on Brilliant:
Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
Markdown
Appears as
*italics* or _italics_
italics
**bold** or __bold__
bold
- bulleted - list
bulleted
list
1. numbered 2. list
numbered
list
Note: you must add a full line of space before and after lists for them to show up correctly
# 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"
Math
Appears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
Let m be the mass of the ring so 2m is the mass of each loop.Now by symmetry its clear that the ring moves downwards so let its acceleration be a then each of the loops moves with acceleration w .But consider a loop+ring.Both must have a common acceleration along the common normal (here radius).So w/√2=a/√2 so w=a.The normal reaction between them beN and friction be f then we have (N/√2−f)/2m=Rα=f/2m.so f=N/2√2,a=w=N/4√2m.For the ring mg−√2N=ma.Yields N=4√2mg/9.So normal given by ground if F=2mg+Ncosπ/4=22mg/9.And accn of the ringa=g/9.Friction is f=2mg/9=2/9.Note that we don't take into account the radial acceleration since t=0 there is nt any given to the system.
I have included that.If the friction you are taking about is due to ground then I have included it.And if it's friction between both ring and hoop then the problem says to neglect that.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Let m be the mass of the ring so 2m is the mass of each loop.Now by symmetry its clear that the ring moves downwards so let its acceleration be a then each of the loops moves with acceleration w .But consider a loop+ring.Both must have a common acceleration along the common normal (here radius).So w/√2=a/√2 so w=a.The normal reaction between them beN and friction be f then we have (N/√2−f)/2m=Rα=f/2m.so f=N/2√2,a=w=N/4√2m.For the ring mg−√2N=ma.Yields N=4√2mg/9.So normal given by ground if F=2mg+Ncosπ/4=22mg/9.And accn of the ringa=g/9.Friction is f=2mg/9=2/9.Note that we don't take into account the radial acceleration since t=0 there is nt any given to the system.
Log in to reply
There is going to be torque by friction about the centre of the ring right?
Log in to reply
I have included that.If the friction you are taking about is due to ground then I have included it.And if it's friction between both ring and hoop then the problem says to neglect that.
Log in to reply
Log in to reply
Log in to reply
A,D are correct.simple mechanics q.