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
@Harsh Shrivastava
–
Irodov(that's too good a book as I felt for rotational mech),and hc verma is good as well.For more tough prob you can refer to 300 creative physics problems by Laszlo Holics.
@A Former Brilliant Member
–
Then try this "A guide to physics problems"... May be the second part be a bit tough as its Mostly Quantum Electrodynamics,Quantum Mechanics and all those stuffs but part 1 is good...... You will surely like it..And also try the integral I have posted
Use the fact that since we assume strings to be inextensible, velocity of M at bottommost point will be only in horizontal direction, and use constraint relation between the velocity of M and m by equating their velocities along the thread connecting them.Finally use energy conservation.
Which part...I guess straight lines and circles you must be well acquainted with......Coordinate needs some heavy practice....Formulas and all those stuffs should be known all the time..?As such I don't follow any special book only fiitjee package.
@Harsh Shrivastava
–
OK ok.That will lead to answer of I=1/ab.If a and b are of opposite sign then the integrals value is Negative (ab<0).But the integrand be a square the area can never be negative.How do you resolve this...Here you integrating through Singularities.Pretty much the same thing as dx/x2 integral from (-1 to 1) isn't -2.The function blows up when x-->0...That's why its a Feynman's Puzzle...Try using limits in integrand.
Hey in your q you talked abt Max area....I think it should be min.I did it as the req point of intersection is apq,a(p+q) where the points P and Q are (ap2,2ap)and (aq2,2aq).Then the cond yields apq=−1.Vertex being (0,0)The Area of PCQ is a2∣pq(p−q)∣.Puting pq=−1/a And later applying AM-GM yields the ans as Min=1.a=1/4 from y2=4ax
@Harsh Shrivastava
–
Ya I see the problem has been changed.Thanks....I was surprised because of the title "its right"....I accepted my ans to be wrong....
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
@Spandan Senapati
Log in to reply
Ya I know this q.Follow Harsh s solution.And you will get it.And if any problem then tag me and I will post it.Ok
Log in to reply
Hey Spandan ,AE can you guyz tell me good source for practicing mechanics questions, especially rotational mechanics?
Log in to reply
Log in to reply
Log in to reply
Log in to reply
Log in to reply
Log in to reply
Log in to reply
Log in to reply
Log in to reply
Log in to reply
Log in to reply
Log in to reply
I got it. Will post a solution by night.
Use the fact that since we assume strings to be inextensible, velocity of M at bottommost point will be only in horizontal direction, and use constraint relation between the velocity of M and m by equating their velocities along the thread connecting them.Finally use energy conservation.
Log in to reply
Thanks Harsh & Spandan.
Any suggestion on coordinate geometry, I am very weak in it?
Log in to reply
TMH and archive will suffice.
Which part...I guess straight lines and circles you must be well acquainted with......Coordinate needs some heavy practice....Formulas and all those stuffs should be known all the time..?As such I don't follow any special book only fiitjee package.
Hey you like integrals.I have a good one1)int1/(a+b(1−x))2dxwith x varying from 0−1.This is a Feynman's Puzzle.Looks easy but needs some more....
Log in to reply
use the transformation x-> 1-x and then differentiate the function 1/(a+bx) wrt a and then done.
Log in to reply
I=1/ab.If a and b are of opposite sign then the integrals value is Negative (ab<0).But the integrand be a square the area can never be negative.How do you resolve this...Here you integrating through Singularities.Pretty much the same thing as dx/x2 integral from (-1 to 1) isn't -2.The function blows up when x-->0...That's why its a Feynman's Puzzle...Try using limits in integrand.
OK ok.That will lead to answer ofLog in to reply
Hey in your q you talked abt Max area....I think it should be min.I did it as the req point of intersection is apq,a(p+q) where the points P and Q are (ap2,2ap)and (aq2,2aq).Then the cond yields apq=−1.Vertex being (0,0)The Area of PCQ is a2∣pq(p−q)∣.Puting pq=−1/a And later applying AM-GM yields the ans as Min=1.a=1/4 from y2=4ax
Log in to reply
Yeah you are right, my mistake, edited!
Log in to reply