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Here's a challenge: Prove that when a solid disc is set sliding at some speed on a horizontal surface with friction and also is given an arbitrary rotation (set spinning) , both motions will cease simultaneously. In other words, prove that you will never find that the disc is still translating while the rotation already stopped or that the disc stops translating but after that is still rotating for a while.

I thought it would be fun to post this question. I know the answer but just to keep you guys busy on a question I post instead of me being busy on answering your questions... :-)

Imagine a funnel that has a surface of revolution obtained from revolving y = 1/x , between x=1 and x=infinity - about the x-axes.

The volume of that funnel of course is

V = Integral_1^infinity pi y^2 dx

= pi * Integral_1^infinity 1/x^2 dx

= pi .

The surface area is

A = integral_1^infinity 2 pi y sqrt(1+ y'^2) dx

= 2 pi * integral_1^infinity 1/x * sqrt(1+ 1/x^4) dx

But this integral diverges.

Hence the volume is finite but the surface area is infinite.

This begs the question: does this mean that you can fill this funnel with a finite amount of paint but that finite amount of paint does not suffice to paint the inside wall...?

Since today, if I use the addition sign in an answer, it doesn't show after submitting the answer. Any clues on what's going on?

Suppose a submarine, floating at some depth in the ocean, starts traveling horizontally at relativistic speeds.

From the point of view of an outside observer, the sub will be shorter (length contraction), displace less water and it will therefore (less buoyancy) move down....

From the point of view of the sailors on the sub however, the water is moving towards them, each volume element of water will be length contracted and therefore the density of the water will be higher than when they where at relative rest. This increased density of the water will increase the buoyancy and will cause the submarine to move up... .

What is is, up or down?

Here's a problem I solved this weekend but for which I would like to see possible other solution methods.

You have a river of width W, where the water flow has parabolic velocity profile:

v(x) = 4 v_middle (x/W) ( 1 - x/W)

[[The river flows in the +y direction]]

Your boat has a constant speed C relative to the water, where C > v_middle .

Determine the path y(x), where (0, y(0) ) is the departure point, that minimizes the time for crossing the river

(i) If you have to arrive at a point directly opposite to your point of departure

(ii) If you do not need to end directly oposite, but just reach the other bank.

I solved it using Variational Calculus, but am curious about other possible solution methods.

So, top contributors, consider this to be your challenge of the day...

Here's a problem I solved this weekend but for which I would like to see possible other solution methods.

You have a river of width W, where the water flow has parabolic velocity profile:

v(x) = 4 v_middle (x/W) ( 1 - x/W)

[[The river flows in the +y direction]]

Your boat has a constant speed C relative to the water, where C > v_middle .

Determine the path y(x), where (0, y(0) ) is the departure point, that minimizes the time for crossing the river

(i) If you have to arrive at a point directly opposite to your point of departure

(ii) If you do not need to end directly oposite, but just reach the other bank.

I solved it using Variational Calculus, but am curious about other possible solution methods.

So, top contributors, consider this to be your challenge of the day...

Mount Mathematica is a cone with top angle alpha. The surface is smooth and therefore frictionless. If we throw a lasso (with a fixed non-sliding knot) over the top of the cone and pull ourselves up parallel to the mount's surface, the lasso stays over the top if the angle is not too blunt. So, what is the maximum top angle that allows us to climb up?

Why does hot air rise (because it is lighter is an answer known to me, I am looking for a "microscopic" explanation? And since it does, how come it is colder higher up in the atmosphere...?