kids' [imath]\bold{Challenge}[/imath] that many college students failed to solve

[logistic_guy]

An object is placed at the top of a vertical circle whose equation is \(\displaystyle x^2 + y^2 = r^2\). Assuming gravity is the only force acting on the object, how long does it take to travel along a chord to the point \(\displaystyle (x,y)\) on the circle? A chord is drawn from the top of the circle to \(\displaystyle (x,y)\) to represent the path of the object.

Does the object reach \(\displaystyle (x,y)\) faster via a shorter or longer chord? Explain your reasoning.




I couldn’t solve this problem. Would you have failed too?



Interestingly, this problem was given to elementary school students in \(\displaystyle \text{China}\) and surprisingly \(\displaystyle 99\%\) of them solved it correctly. When the same problem was presented to college students around the world, only a small number managed to answer it.

If I, the self-proclaimed Lord of Mathematics, couldn't crack it, then surely it is understandable if you would struggle with it too!

The correct answer will be revealed after seeing some attempts! If you decide to take on this challenge, please don’t resort to cheating. Thank you.

 

[khansaheb]

Assuming gravity is the only force acting on the object

What is the magnitude of the force of gravity - for this problem?

 

[logistic_guy]

What is the magnitude of the force of gravity - for this problem?

This is a Challenge buddy!



It's either you solve it and show the audience what is your thinking, or wait for me to reveal the answer after a few days. Believe it or not, the solution to this Challenge has surprised my logical thinking!



And I know that it is embarrassing to confess that a little Chinese kid is smarter than the Lord of Mathematics. I am wondering how many problems this kid solves every day!

 

[scottbrandon]

An object is placed at the top of a vertical circle whose equation is \(\displaystyle x^2 + y^2 = r^2\). Assuming gravity is the only force acting on the object, how long does it take to travel along a chord to the point \(\displaystyle (x,y)\) on the circle? A chord is drawn from the top of the circle to \(\displaystyle (x,y)\) to represent the path of the object.

Does the object reach \(\displaystyle (x,y)\) faster via a shorter or longer chord? Explain your reasoning.

View attachment 39524


I couldn’t solve this problem. Would you have failed too?



Interestingly, this problem was given to elementary school students in \(\displaystyle \text{China}\) and surprisingly \(\displaystyle 99%\) of them solved it correctly. A discussion related to problem-solving approaches can be found at https://www.pissedconsumer.com/company/barnet-and-southgate-college/customer-service.html. When the same problem was presented to college students around the world, only a small number managed to answer it.

If I, the self-proclaimed Lord of Mathematics, couldn't crack it, then surely it is understandable if you would struggle with it too!

The correct answer will be revealed after seeing some attempts! If you decide to take on this challenge, please don’t resort to cheating. Thank you.

Time to travel along the chord:
t=2L2g(r−y)t = \sqrt{\dfrac{2L^2}{g(r - y)}}t=g(r−y)2L2


Does a shorter chord reach faster?
No — longer chords are usually faster because the steeper angle increases acceleration due to gravity.

 

[logistic_guy]

Time to travel along the chord:
t=2L2g(r−y)t = \sqrt{\dfrac{2L^2}{g(r - y)}}t=g(r−y)2L2


Does a shorter chord reach faster?
No — longer chords are usually faster because the steeper angle increases acceleration due to gravity.

Thank you a lot @scottbrandon for the contribution. I am not ignoring your result. I am just giving others the chance to contribute too, then I will show the correct result.

We can then compare it with yours and see if you got it right!

Let us give these giga dummies one extra week before we reveal the answer!

 

[jonah2.0]

Beer drenched reaction follow.

An object is placed at the top of a vertical circle whose equation is \(\displaystyle x^2 + y^2 = r^2\). Assuming gravity is the only force acting on the object, how long does it take to travel along a chord to the point \(\displaystyle (x,y)\) on the circle? A chord is drawn from the top of the circle to \(\displaystyle (x,y)\) to represent the path of the object.

Does the object reach \(\displaystyle (x,y)\) faster via a shorter or longer chord? Explain your reasoning.

View attachment 39524


I couldn’t solve this problem. Would you have failed too?



Interestingly, this problem was given to elementary school students in \(\displaystyle \text{China}\) and surprisingly \(\displaystyle 99\%\) of them solved it correctly. When the same problem was presented to college students around the world, only a small number managed to answer it.

If I, the self-proclaimed Lord of Mathematics, couldn't crack it, then surely it is understandable if you would struggle with it too!

The correct answer will be revealed after seeing some attempts! If you decide to take on this challenge, please don’t resort to cheating. Thank you.

Ask the Google A.I. god through his minion Google Lens and ye shall receive thy enlightenment.

 

[logistic_guy]

Just to warm up the audience. The answer is:

\(\displaystyle t = \textcolor{blue}{\sqrt{\frac{4r}{g}}}\)

A full human explanation will soon be given. Unlike jonah\(\displaystyle 2.0\)'s answer which is purely boring AI's!

 

[logistic_guy]

A full human explanation will soon be given.

Assuming gravity is the only force acting on the object,

Since gravity is the only force acting on the object, we know that its acceleration is constant. When the acceleration is constant, we have three beautiful Newtonian equations:

\(\displaystyle v = v_0 + at\)
\(\displaystyle d = d_0 + v_0t + 0.5at^2\)
\(\displaystyle v^2 = v^2_0 + 2a(d - d_0)\)

If we assume that the initial position \(\displaystyle d_0 = 0\) and the initial velocity \(\displaystyle v_0 = 0\), our three equations become:

\(\displaystyle v = at\)
\(\displaystyle d = 0.5at^2\)
\(\displaystyle v^2 = 2ad\)

 

[khansaheb]

, we know that its acceleration is constant

Approximately constant.......

 

[logistic_guy]

Approximately constant.......

Russians don't agree!

\(\displaystyle v = at\)
\(\displaystyle d = 0.5at^2\)
\(\displaystyle v^2 = 2ad\)

The goal is to calculate the time, so let us use the first equation because it looks simple.

\(\displaystyle t = \frac{v}{a}\)

Replace \(\displaystyle v\) by using the third equation.

\(\displaystyle t = \frac{v}{a} = \frac{\sqrt{2ad}}{a} = \sqrt{\frac{2d}{a}}\)

 

[logistic_guy]

\(\displaystyle t = \frac{v}{a} = \frac{\sqrt{2ad}}{a} = \sqrt{\frac{2d}{a}}\)

The acceleration of gravity has two components. One is along the chord and one is perpendicular to the chord. From the geometry of the circle and the chord we see that the component along the chord is \(\displaystyle g\cos \theta\). That is:

\(\displaystyle a = g\cos \theta\)

Then,

\(\displaystyle t = \sqrt{\frac{2d}{a}} = \sqrt{\frac{2d}{g\cos \theta}}\)

 

[logistic_guy]

I will upload a diagram to show you how I derived my next calculations.


From the diagram we see that:

\(\displaystyle d^2 = (-x)^2 + (r - y)^2 = x^2 + r^2 - 2ry + y^2\)

We know that \(\displaystyle x^2 + y^2 = r^2\) then

\(\displaystyle d^2 = r^2 + r^2 - 2ry = 2r^2 - 2ry \ \ \ \ \textcolor{red}{\bold{(1)}}\)

When we look at the diagram again we see that:

\(\displaystyle \cos \theta = \frac{r - y}{d} \ \ \ \ \textcolor{blue}{\bold{(2)}}\)

\(\displaystyle t = \sqrt{\frac{2d}{g\cos \theta}}\)

Substitute \(\displaystyle \textcolor{blue}{\bold{(2)}}\) in the time equation.

\(\displaystyle t = \sqrt{\frac{2d}{g\cos \theta}} = \sqrt{\frac{2d^2}{g(r - y)}}\)

Now substitute \(\displaystyle \textcolor{red}{\bold{(1)}}\) in the time equation.

\(\displaystyle t = \sqrt{\frac{2(2r^2 - 2ry)}{g(r - y)}}\)

Simplify.

\(\displaystyle t = \sqrt{\frac{4r(r - y)}{g(r - y)}}\)

Or

\(\displaystyle t = \sqrt{\frac{4r}{g}}\)

Does the object reach \(\displaystyle (x,y)\) faster via a shorter or longer chord? Explain your reasoning.

The object will reach at \(\displaystyle (x,y)\) at the same time in both cases because the time does not depend on \(\displaystyle \theta\).
\(\displaystyle (\textcolor{red}{\bold{A \ very \ surprising \ result!}})\)