1. ## Physics meets Geometry

I could not find a solution to the following problem:

UNTAMED (I am walking 8 foxes at the top of my head and trying to keep their tails apart ! (*))

I am the one who posted the problem. I vaugely remembering seeing something similar somewhere years ago. But just couldn't solve it.
Has anyone seen anything like that?

2. Originally Posted by oswaldo
I could not find a solution to the following problem:

UNTAMED (I am walking 8 foxes at the top of my head and trying to keep their tails apart ! (*)) [Thumbnail below.]

I am the one who posted the problem. I vaguely remembering seeing something similar somewhere years ago. But just couldn't solve it.
Has anyone seen anything like that?
Potential energy lost = kinetic energy gained: $\displaystyle mgr(1-\sin\theta) = \tfrac12mv^2$.

Radial component of weight just sufficient to provide centripetal acceleration: $\displaystyle mg\sin\theta = mv^2/r$.

Put those two equations together to get $\displaystyle \sin\theta = 2(1-\sin\theta)$, from which $\displaystyle \sin\theta = 2/3$ (independent of m, r and g !).

3. Good work! Thanks Opalg.

I had the conservation of Energy part before. For the second equation I was thinking of equation of the tangent line at point A, but yours makes more sense.

BUT I still have a question:
"Radial component of weight just sufficient to provide centripetal acceleration
$\displaystyle mg\cos\theta = mv^2/r$ "

Radial component is towards the center, so the equation should be:

$\displaystyle mg\sin\theta = mv^2/r$

which should give a different answer!!! But 3/5 makes more sense!

Thanks.

-O

4. Originally Posted by oswaldo
"Radial component of weight just sufficient to provide centripetal acceleration
$\displaystyle mg\cos\theta = mv^2/r$ "

Radial component is towards the center, so the equation should be:

$\displaystyle mg\sin\theta = mv^2/r$

which should give a different answer!!!
You're absolutely right! It should be $\displaystyle \sin\theta$, not $\displaystyle \cos\theta$. That makes the equation $\displaystyle \sin\theta = 2(1-\sin\theta)$, with solution $\displaystyle \sin\theta=2/3$. I have edited my previous comment to correct it. (I think I was misled by the fact that θ is decreasing from π/2 towards 0, rather than increasing.)

5. Thank you for the correction Opalg!
$\displaystyle \sin\theta = 2/3$ gives:
$\displaystyle \theta$ is about 42 degrees.

To me, this is intriguing. I would expect to object remain on course, say until 30 degrees. But apparently my intuition is wrong.

One more details: $\displaystyle mg\sin\theta = mv^2/r$
holds only at point A, when the object is airborne. Not before nor after. Interesting!

Having the answer not depending on m, g, r makes it more beautiful.

Thanks for a great solution!

-O