"Fun" Math problems thread! (Yay)

This reminds me of M. Scott Peck’s assertion in his book “The Road Less Traveled” ( http://www.amazon.com/Road-Less-Traveled-Timeless-Edition/dp/0743243153 ). He describes that people build a mental map of the world as they go along - a ‘moving approximation’ of how the world works - and suggests that the root of all pain and mental illness results from how each individual handles the inevitable differences they discover between their map and reality. Stay flexible, friends!

Alright, here’s my attempt. I’m not convinced that it’s correct, but it was fun nevertheless. My wife thinks it’s wrong :slight_smile: She’s convinced that even in the non-inertial frame, the radial forces should not sum to zero.

(several simplification steps between the 2nd to last and last steps, but I ran out of page space :))

Been quite some time since I tackled a Lagrangian/Hamiltonian solution, but yeah! Send us one! Maybe she’ll beat me to the punch on the next upload…

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Very close! You’re correct that in the non-inertial frame the forces should sum to zero, if you include the perceived force F_c. The only errors you made were clerical ones. I highlighted your mistakes, and put my own worked solution below. :blush:

Ah, silly mkstake. Thanks for the great problem!

This is MY logic.


Since no one else has taken a shot yet…


Exactly what I was hoping for. Elegant and to the point.

More problems soon!

Took me a minute to figure out how you ended up assuming that both triangles have the same angle twice.

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I couldn’t see why those angles were duplicated, even with your solution posted. So I gave the original problem to my wife, and she had it within twenty minutes :slight_smile: (<— proud hubby). For posterity, here’s her solution. I asked her to add some descriptions to help demonstrate how she arrived to the answer.

(editors note: she says “after seeing @Kichae’s answer, it’s clear you don’t need to flip the triangle to see the radial association”)

Didn’t think that anybody out of uni actually constructed proofs…

The problems… They never end! This one is in 3 parts, with increasing difficulty.
Solve the last problem and earn a golden physics star!

an extremely cockish question, but where does the particle start the slide? Does it at the top of the curve at point x=0, or some arbitrary point x>0? I am asking because, even if there is no friction, and the velocity is negligible (which means it might as well be 0 too), if the particle starts at the point x=0 it shouldn’t slide down at all (I mean any starting velocity is negligible), as the surface should be flat in that point.

Haha! I will do this later when I get home today. We had a very similar exam problem except it was the surface of a sphere.

Are we to assume no slipping as well?

As long as the velocity isn’t exactly equal to zero, it will slide off the top… eventually. :wink:

Okay, further addendum: the particle starts at x = ε, with ε being arbitrarily small.

I really hope I don’t see any ε’s in peoples’ answers :sweat:

All of the motion is slipping motion, hah. The particle isn’t macroscopic, so it doesn’t rotate or anything. (You’ll also find this is a more difficult problem than the sphere version!)

I completely assumed it was a mass(ball) rolling and not a point particle. I need to read problems more carefully. That makes it a bit easier since there’s no rotational energy :smiley:

Because all lines connecting the centre of the circle to points on the circumference have length r, each sub-triangle has two sides of length r. That makes them both isosceles, and guarantees that the two angles between the r-length sides and the 3rd side are identical.

Thanks, :slight_smile: I did see if it eventually, and then promptly felt like an ignoramus :smile:

Yay! More problems! Must resist the urge to abandon all responsibilities at work and start solving it right now…

Honestly, I probably wouldn’t have seen it myself if it weren’t for tutoring high school math. Although, with that in mind, I should have a much easier time with Riemann sums than I do.

Remember: Hints are allowed, if you want them. Collaboration is allowed too! Tricky problems are best solved in groups.

I’ll try to think of more geometry problems too, those are pretty fun.

My math needs to be cleaned up before I post it, and I haven’t attempted #3 yet, but I get:

  1. Using energy considerations, v(x) = sqrt(2gy) = x•sqrt(2gk)

  2. The condition for leaving the curve is ay > g, where ay is the acceleration in y if confined to the curve.
    I find ay = g•(16k2y2 + 8ky)/(4ky + 1)2
    This makes the condition for leaving the curve (16k2y2 + 8ky)/(4ky + 1)2 > 1
    As (4ky + 1)2 = 16k2y2 + 8ky + 1, however, this is never >1 for positive values of k.
    Therefore, the particle never leaves the curve.


Still haven’t written the math up cleanly yet, but for 3 I find:
|N(x)| = mg/[2kx(4k2x2 + 1)]. Wups! Obvious math error in here somewhere!
|N(x)| = mg/(4k2x2 + 1)3/2. Fixed!
And you can’t make me find the direction of N! You’re not my mother!