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Challenge Questions

Stop in and exercise your brain. Talk about this month's Challenge from Specs & Techs or similar puzzles.

So do you have a Challenge Question that could stump the community? Then submit the question with the "correct" answer and we'll post it. If it's really good, we may even roll it up to Specs & Techs. You'll be famous!

Answers to Challenge Questions appear by the last Tuesday of the month.

Rope Race: Newsletter Challenge (July 2016)

Posted July 01, 2016 12:00 AM
Pathfinder Tags: challenge questions

This month's Challenge Question: Specs & Techs from IHS Engineering360:

You have a rope of length L and a frictionless table with height 1.5L. Put the rope on the top of the table and pull one end to the border of the table so that a small piece hangs down. Release the rope, so that the rope starts to fall down. What is the speed of the rope at the very instant when it loses contact with the top of the table?

And the answer is:

Let’s apply the conservation of energy principle. Assume that the linear mass density of the rope is . The total mass of the rope is

For the purpose of this problem, the total mass is concentrated in the center of mass, or

The following figure shows the stages of the problem. Figure (a) shows the rope on top of the table at the moment when it starts to fall down. Figure (b) shows the rope at the instance when the rope loses contact with the top of the table. Figure (c) depicts the position of the center of mass corresponding to figure (b).

According to the conservation of energy principle, the kinetic energy of the rope in figure (b) is equal to the loss in potential energy. The loss in potential energy is the distance the center of mass moves in figure (b) multiplied by the mass and the gravitational constant (g). So, in equation form we have:

Solving for the speed, we get:

Notice that this is the same result obtained by an object of any mass that falls a distance L/2.

62 comments; last comment on 07/20/2016
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Invisible Neutrino: Newsletter Challenge (June 2016)

Posted June 01, 2016 12:00 AM
Pathfinder Tags: challenge questions

This month's Challenge Question: Specs & Techs from IHS Engineering360:

A neutrino with an energy of 14 MeV is released from a fusion reaction in the core of the sun and travels on a path directly through Earth. An ideal neutrino detector capable of detecting every neutrino of that energy passing through Earth fails to detect it. Why?

And the answer is:

The neutrino has changed to a different flavor invisible to that detector. The neutrinos released in fusion reactions in the solar core must travel through the extremely dense material of the sun before exiting its atmosphere and travelling across the vacuum of space to Earth. On their journey from the center of the sun to space, neutrinos undergo oscillations and some change from their original electron flavors to muon and tau flavors.

This oscillation effect is the cause of the apparent discrepancy between the amount of neutrinos expected to be emitted from the sun according to the standard solar model and the number of neutrinos actually detected. Early detectors were only picking up between one third and one half of the expected number of neutrinos because they were looking only for electron neutrinos, and couldn’t detect the muon and tau flavors. The Sudbury Neutrino Observatory provided definitive evidence that neutrinos undergo oscillation to different flavors as they travel through the sun.

183 comments; last comment on 07/09/2016
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Permanent Magnet: Newsletter Challenge (May 2016)

Posted May 01, 2016 12:00 AM
Pathfinder Tags: challenge questions Magnets

This month's Challenge Question: Specs & Techs from IHS Engineering360:

You have two steel bars. One is a permanent magnet and the other is not magnetized. Without using any tool, meter or instrument of any type, how can you find out which one is the permanent magnet?

And the answer is:

See the figure below. If you arrange the two bars in a T shape by putting the permanent magnet as the top of the T (second figure) and the non-magnetized bar exactly at the middle there is no way the magnet will magnetize the bar, because there is no direct unique pole in contact with the bar. So, there is no attraction between the bars.

On the other hand, if the non-magnetized bar is the top of the T structure, then a clear pole (in the first figure and in this case, it is the south pole) will be in contact with the bar and it will magnetize the bar by creating a North pole in the bottom of the bar and a South pole in the top. Therefore, in this condition the two bars will attract.

46 comments; last comment on 05/24/2016
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Gliese 581g: Newsletter Challenge (April 2016)

Posted April 01, 2016 12:00 AM
Pathfinder Tags: challenge questions

This month's Challenge Question: Specs & Techs from IHS Engineering360:

Earth has sent a team of scientists to explore Gliese 581g, a super-Earth exoplanet orbiting a Red Dwarf and located 20 light-years from Earth. Upon approach they marvel that Gliese 581g is almost a perfect sphere considering earlier data. They decide to land in the equatorial region to save fuel. Why?

And the answer is:

The scientists were surprised to learn that Gliese 581g was spherical because of previous data which had told them that it was spinning incredibly rapidly. So rapidly in fact that the centrifugal force at the equator is strong enough to measurably weaken the effect of gravity at its equator, reducing the fuel needed for landing and lift off.

A similar yet much smaller effect occurs on Earth, where the rotation at the equator generates a centrifugal force that reduces effective gravity by one third of one percent.

17 comments; last comment on 04/19/2016
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Liquid in Motion: Newsletter Challenge (March 2016)

Posted February 29, 2016 4:59 PM
Pathfinder Tags: challenge questions

This month's Challenge Question: Specs & Techs from IHS Engineering360:

Entering a laboratory, you notice a motionless clear cylinder resting on a table that is filled with a liquid circulating around inside. After hours of observation, you are puzzled to find that the liquid's motion has not changed. Knowing that no outside force or energy has entered the cylinder since you began observing, how is this possible?

And the answer is:

The liquid in the cylinder is superfluid helium-4.

Helium becomes a liquid when cooled to a temperature below its boiling point of 4.22 K (-268.93┬░ C). Taken down a couple more degrees past its "lambda point" of 2.17 K (-270.98┬░ C), liquid helium starts to exhibit unusual properties. It is at this point that a fraction of the helium has transitioned into a "superfluid." In this curious state of matter, the helium behaves like a fluid with zero viscosity. With no friction to slow its motion, the superfluid helium flows past the surface of the cylinder, continuing to circulate endlessly (or until it warms up and transitions back into its more regular liquid or gaseous states).

Unlike most liquids, helium doesn't turn into a solid when cooled down. Its atoms are light and weakly attracted to each other, so even when cooled to temperatures at which regular heat vibrations are absent, helium doesn't settle into a solid. At low temperatures its atoms wiggle with zero-point motion, a slight momentum bestowed by the quantum uncertainty principle. Instead of settling in a solid state, liquid helium undergoes a transformation known as Bose-Einstein condensation. Its atoms start acting in harmony, behaving like one big particle, no longer colliding together. It is these quantum effects that grant superfluids their remarkable properties.

For more on superfluid helium, including a detailed description of how superfluid helium is actually a mixture, with normal and superfluid components, see The physics of superfluid helium[pdf].

40 comments; last comment on 04/10/2016
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